A gas syringe contains 56.05 mL of a gas at 21.50C. Determine the volume that the gas will occupy if the temperature is increased to 44.30C at the same pressure.

Answers

Answer 1

Answer:

The volume the gas will occupy is 60.39mL

Explanation:

The gas law to be followed here is the Charles' law which states that the volume of a given mass of gas is inversely proportional to its temperature (in Kevin) provided that pressure remains constant. That is V∝T.

Equation thereof = V₁/T₁ = V₂/T₂

where V₁ is the initial volume  (56.05mL)

V₂ is the final volume (unknown)

T₁ is the initial temperature (21.5°C + 273 = 294.5K)

T₂ is the final temperature (44.3°C + 273 = 317.3K)

Thus,

56.05/294.5 = V₂/317.3

V₂ = 56.05 x 317.3/294.5

V₂ = 60.39mL


Related Questions

An unknown organic compound composed of carbon, hydrogen and oxygen was analyzed and found to be 50.84% C, 8.53% H and the rest being oxygen. Which of the following represents the correct empirical formula for the compound?
(a) CH2O
(b) C3H602
(c) C4H803
(d) C2H40
(e) C5H1003

Answers

Answer:

e) C5H10O3

Explanation:

According to the information given in this question;

50.84% represents C

8.53% represents H

[100% - (50.84 + 8.53)]

= 100 - 59.37

= 40.63% of Oxygen (O)

This percentage means that;

Carbon = 50.84g

Hydrogen = 8.53g

Oxygen = 40.63g

By dividing by their respective atomic masses, we convert to moles:

C = 50.84g ÷ 12 = 4.24mol

H = 8.53g ÷ 1 = 8.53mol

O = 40.63g ÷ 16 = 2.54mol

Next we divide by the smallest no of the values (2.54mol)

C = 4.24 ÷ 2.54 = 1.669

H = 8.53 ÷ 2.54 = 3.358

O = 2.54 ÷ 2.54 = 1

To get a simple whole number ratio, we multiply the results by 3

C = 1.669 × 3 = 5.007

H = 3.358 × 3 = 10.074

O = 1 × 3 = 3

Simple whole number ratio of carbon, hydrogen and oxygen is 5:10:3. Hence, the empirical formula is C5H10O3.

I NEED THIS NOW NO LINKS OR ILL REPORT PLZZZ
What is an example of a chemical property?
Odensity

O reactivity

Omalleability

O solubility

Answers

#2 reactivity is the answer

anuvia, the trade name for sitagliptin, was introduced in 2006 for the treatment of type 2 diabetes. In what type of orbital does the lone pair on each N atom reside.

Answers

Answer: hello your question is poorly written below is the complete question

answer:

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital

Explanation:

Determining the type of orbital in which the lone pair on each N atom will reside.

From the configuration attached below we can determine the type of orbital and they are ;

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital

Which is an example of poor safety practices when working outdoors

Answers

Answer:

C

Explanation:

u can't touch a chemical with bare skin

Answer:

touching a chemical with his or her bare skin.

The value of ΔH° for the reaction below is -6535 kJ. ________ kJ of heat are released in the combustion of 16.0 g of C6H6 (l)?

2C6H6 (l) + 15O2 (g) → 12CO2 (g) + 6H2O (l)

Answers

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

For the given reaction, 2 moles of C₆H₆ the heat energy released is - 6535 KJ. Then, for 16 g of the compound or 0.205 moles needs 669.83 KJ of heat released in combustion.

What is combustion ?

Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent, typically oxygen, resulting in the release of heat, light, and various combustion products, such as carbon dioxide and water vapor.

The process of combustion involves a rapid and exothermic (heat-releasing) oxidation reaction that produces a flame, which is visible in many cases.

Here, 2 moles of the hydrocarbon releases - 6535 KJ of energy.

molar mass of C₆H₆ = 78 g/mol

then no.of moles in 16 g = 16 /78 = 0.205 moles.

then energy released by 0.205 moles  = 0.205 moles × 6535 KJ /2 moles = 669.83 kJ

Therefore, the heat energy released by 16 g of the compound in combustion is 669.83 kJ.

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g Select the statement that best answers the following question What effect does the cation of an ionic compound have on the appearance of the solution? The cation affects the intensity of the color more than the color of the solution. The cation affects the color of the solution more than the intensity of the color. The cation does not affect the color or color intensity of the solution. The cation only affects the intensity of the color in a solution.

Answers

Answer:

The cation affects the intensity of the color more than the color of the solution.

Explanation:

According to Beer Lambert law, the intensity of the colour of the solution depends on the concentration of the specie responsible for the colour in the solution.

Let us recall that transition metal compounds are coloured in solution due to electronic transitions.

Therefore, the cation affects the intensity of the color more than the color of the solution.

a. Giải thích vì sao tính bazơ tăng từ LiOH đến CsOH?

Answers

i cant understand what you’re saying

Chrysanthenone is an unsaturated ketone. If Chrysanthenone has M+ = 150 and contains 2 double bond(s) and 2 ring(s); what is its molecular formula? Enter the formula in the form CH first, then all other atoms in alphabetical order; do not use subscripts. The formula is case-sensitive.

Answers

Answer:

the Molecular formula will be; C10H14O

Explanation:

Given the data in the question;

Chrysanthenone is an unsaturated ketone,

it has M+ = 150 and contains 2 double bond(s) and 2 ring(s).

molecular formula = ?

we know that ketone contain 1 oxygen and mass of oxygen is 16

so mass of the C and H remaining will be;

⇒ 150 - 16 = 134

Now we determine the number of C atoms;

⇒ 134 / 13 = 10

hydrocarbon with 10 hydrogen atom have CnH2n+2 means

⇒ ( 10 × 2 ) +2 = 22 hydrogens

But then we have 3 unsaturation meaning 6 hydrogens less and also we have ring meaning 2 more hydrogens

⇒ 22 - 6 - 2 = 14

Hence the Molecular formula will be; C10H14O

What causes an ice cube to melt when removed from a freezer?

Answers

Answer:

the melting process begins right away because the air temperature around the ice cubes is warmer than the temperature in the freezer

The temperature in the room is higher than the temperature in the freezer, thus causing the melting state of the ice cube when it is removed from the freezer.

I WILL GIVE BRAINLIEST IF YOU ANSWER CORRECTLY


How many grams in 3.75 x 1024 atoms of F?​

Answers

Answer:

96 grams

Explanation:

The information below describes a redox reaction.
Ag+ (aq) + Al(s) — Ag(s) + Al3+ (aq)
Ag+ (aq) + -> Ag(s)
Al(s)->A3+ (aq) + 3e-
What is the coefficient of silver in the final, balanced equation for this reaction?

Answers

Answer:

Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

Explanation:

Oxidation:                            Al°(s) =>   Al⁺³(aq) + 3e⁻

Reduction:           3Ag⁺(aq) + 3e⁻ => 3Ag°(s)

_________________________________________

Net Rxn:           Al°(s)  + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)

One mole of neutral aluminum atoms (Al°(s)) undergo oxidation delivering 3 moles  of electrons to 3 moles silver ions (3Ag⁺³(aq)) that are reduced to 3 moles of neutral silver atoms (3Ag°(s)) in basic standard state 25°C; 1atm.

A balanced equation obeys the law of conservation of mass. According to the law of conservation of mass, mass can neither be created nor be destroyed. The coefficient of silver is 3.

What is a balanced equation?

A balanced chemical equation can be defined as the chemical equation in which the number of reactants and products on both sides of the equation are equal. The amount of reactants and products on both sides of the equation will be equal in a balanced chemical equation.

The numbers which are used to balance the chemical equation are called the coefficients. The coefficients are the numbers which are added in front of the formula.

The balanced chemical equation for the given redox reaction is given as:

Al (s) + 3 Ag⁺ (aq)  → Al³⁺ (aq) + 3Ag (s)

Thus the coefficient of silver is 3.

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Guanidine is a neutral compound but is an extremely powerful base. In fact, it is almost as strong a base as a hydroxide ion. Identify which nitrogen atom in guanidine is so basic, and explain why guanidine is a much stronger base than most other amines.

Answers

Answer:

See explanation

Explanation:

The molecule called guanidine is shown in the image attached to this answer. It contains three nitrogen atoms. Two among them are sp3 hybridized while one of them is sp2 hybridized.

Guanidine is more basic than other amines because its protonanation leads to the formation of three equivalent resonance structures thereby making its protonated form quite stable. This effect is not observed in other amines.

Also, the sp2 hybridized nitrogen atom is more basic and more easily protonated because when it is protonated, three equivalent resonance structures are obtained.

How many grams of KCl solid do I need to make 500 ml of 8% KCl solution

Answers

Answer:

40 grams of KCI are required to make 500 ml of 8% solution.

A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.
B) The rate of effusion of Xe gas through a porous barrier is observed to be 7.03×10-4 mol / h. Under the same conditions, the rate of effusion of SO2 gas would be ______ mol / h

Answers

Answer:

For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.

For B: The rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

Explanation:

For A:

The average molecular speed of the gas is calculated by using the formula:

[tex]V_{gas}=\sqrt{\frac{8RT}{\pi M}}[/tex]

     OR

[tex]V_{gas}\propto \sqrt{\frac{1}{M}}[/tex]

where, M is the molar mass of gas

Forming an equation for the two gases:

[tex]\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}[/tex]          .....(1)

Given values:

[tex]V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol[/tex]

Plugging values in equation 1:

[tex]\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s[/tex]

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.

For B:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

[tex]Rate\propto \frac{1}{\sqrt{M}}[/tex]

Where, M is the molar mass of the gas

Forming an equation for the two gases:

[tex]\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}[/tex]          .....(2)

Given values:

[tex]Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol[/tex]

Plugging values in equation 2:

[tex]\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr[/tex]

Hence, the rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]

If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give your reason.​

Answers

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, since the reaction taking place between sodium hydroxide and chlorine has is:

[tex]NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Which must be balanced according to the law of conservation of mass:

[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]

Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:

[tex]n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH[/tex]

Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.

Regards!

A sample of titrations of 0.200 M sodium hydroxide (NaOH) being added to 30.0 mL of hydrochloric acid (HCl) of unknown concentration. Measure the volume of sodium hydroxide needed to neutralize the HCl, and calculate the concentration of the HCl.

Answers

Answer:

Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.

Note: The volume of NaOH used isnot given and issp to be determined by titration.

Explanation:

The question is a lab activity. Therefore, the volume of 0.2 M NaOH required for the neutralization of 30.0 mL of hydrochloric acid of unknown concentration is to be determined from titration. However, assuming a certain volume of the base is used for the complete neutralization of the acid, the concentration of the acid can be calculated.

Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid can be calculated thus:

Step 1: Balanced chemical equation of reaction:

HCl + NaOH ---> NaCl + H₂O

From the equation of the reaction, 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and water.

Step 2: Determine the number of moles of NaOH in 20.0 mL of 0.200 M solution

Molarity = number of moles/volume in Litres

Number of moles = molarity × volume in litres

Volume of NaOH in litres = 20 ml × 1 litre/1000ml = 0.02 litres

Number of moles NaOH = 0.02 L × 0.2 M = 0.004 moles

Therefore, 0.00 M of NaOH reacted with 0.004 moles of HCl

Step 3: Determine concentration of HCl

Molarity or concentration = number of moles / volume in litres

Number of moles of HCl = 0.004 moles

Volume of HCl in litres = 30.0 mL × 1 L/1000 mL = 0.03 L

Concentration of HCl = 0.004 moles / 0.03 L = 0.133 M

Therefore, assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.

All bases dissociate
True or false

Answers

Answer:

verdadero

Explanation:

porque esoo [tex]\lim_{n \to \infty} a_n x_{123} \frac{x}{y} \sqrt[n]{x} x^{2} \sqrt{x} \pi \neq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right.[/tex] 

Three acid samples are prepared for titration by 0.01 M NaOH:

1. Sample 1 is prepared by dissolving 0.01 mol of HCl in 50 mL of water.
2. Sample 2 is prepared by dissolving 0.01 mol of HCl in 60 mL of water.
3. Sample 3 is prepared by dissolving 0.01 mol of HCl in 70 mL of water.

a. Without performing a formal calculation, compare the concentrations of the three acid samples (rank them from highest to lowest).
b. When the titration is performed, which sample, if any, will require the largest volume of the 0.01 M NaOH for neutralization?

Answers

B is the correct answer

during the process of photosynthesis, green plants produce...

Answers

Answer: photosynthesis

Explanation:

they use carbon dioxide and energy from the sun to make sugar molecules and oxygen

TIME REMAINING
49:56
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Scientists use different types of microscopes to observe objects that are not visible to the naked eye. A scientist is
reviewing various samples of pond water to determine what species of microorganisms live in the pond. The scientist
wishes to make drawings of the structure of each microorganism and study each one's method of movement. Which of
the following microscopes would be best for the scientist to use?
transmission electron microscope
b. scanning electron microscope
c. compound light microscope
d. dissecting microscope
a.
Please select the best answer from the choices provided
ОА
ОВ
D
Nox
Submit
Save and Exit
Mark this and retum
Sono

Answers

Answer:

compound light microscope

QUESTIONS :
1.
Many of the flavours and smells of fruits are esters. A learner prepared an ester with a sme
Ilke banana in the school laboratory using pentanol and ethanoic acid. She set up the
apparatus as shown in the diagram below.
PAPER TOWEL DIPPED
-WATER BATH
IN COLD WATER
PEITANOL ETHANOIC
ACID+ 4 DROPS OF
SULPHURIC ACID
1.1 Which property of sulphuric acid makes it suitable to use as a catalyst for the
preparation of esters?
1.2 Why do we heat the test tube in a water bath and not directly over a flame?
1.3 With reference to the characteristic smells of esters, name TWO examples where
esters are used in different industries.
1.4 State ONE function of the wet paper towel in the opening of the test tube.
1.5 Write down the IUPAC name of an ester fomed.​

Answers

Answer:

See explanation

Explanation:

Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.

Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.

The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.

Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.

The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.

The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.

Guys! I need a long inforation about...

"Colloidal solutions
Types of colloids
Formation of colloids"

Please! And thank you ​

Answers

A colloid is a mixture in which one substance of microscopically dispersed insoluble particles are suspended throughout another substance. However, some definitions specify that the particles must be dispersed in a liquid, and others extend the definition to include substances like aerosols and gels.

The types of colloids includes sol, emulsion, foam, and aerosol.

Sol is a colloidal suspension with solid particles in a liquid.

Emulsion is between two liquids.

Foam is formed when many gas particles are trapped in a liquid or solid.

Aerosol contains small particles of liquid or solid dispersed in a gas.

Explanation:

everything can be found in the picture

Gold is a precious metal that is chemically unreactive. It is used mainly in jewelry, dentistry, and electronic devices. A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3. Calculate the density of gold.

Answers

Answer:

Density = 19.3 g/cm³

Explanation:

In order to answer this question we need to keep in mind the following definition of density:

Density = Mass / Volume

As both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:

Density = 301 g / 15.6 cm³Density = 19.3 g/cm³

an alkane group has a formula of CxH6, determine the value of x​

Answers

Answer:

[tex]x = 2[/tex]

Explanation:

General formula for alkanes

[tex]C _{n}H _{2n + 2}[/tex]

since H = 6

[tex]2n + 2 = 6 \\ 2n = 4 \\ n = 2 \\ since \: x = n\\ \therefore \: x = 2[/tex]

Answer:

x=2

Explanation:

Because alkanes have a general formula of CnH(2n+2).

We known that 2n+2=6, so we solve for n.

2n+2=6

2n=4

n=2

Thus, there are 2 carbon atoms.

1.2.1. Name and explain the purpose of one law /legislation that protects citizens against cyberbullying ?​

Answers

Answer:

Federal law is a law that may protect people from digital attacks such as cyberbullying.

Explanation:

Although it is mostly based on the situation, federal law does protect those who suffer from cyberbullying. The attackers may be charged with defamation, since cyberbullying consists of spreading false information about someone on the internet, that's when the law comes in.

A sample of oxygen gas occupies a volume of 2.,0cm3 at pressure of 700K pa. what will be pressure of the same sample occupies a volume of 150cm, assume temperature remains constant​

Answers

Answer:

The pressure will be 933.33 Kpa

Explanation:

Given that:

Volume V₁ = 200 cm³  (note, there is a mistake in the volume. It is supposed to be 200 cm³)

Pressure P₁ = 700 Kpa

Pressure P₂ = ??? (unknown)

Volume V₂ = 150 cm³

Temperature = constant

Using Boyle's law:

PV = constant

i.e.

P₁V₁ = P₂V₂

700 Kpa × 200 cm³ = P₂ × 150 cm³

P₂ = (700 Kpa × 200 cm³)/150 cm³

P₂ = 933.33 Kpa

Calculate the molality of each of the following solutions:
(a) 14.3 g of sucrose (C12H22O11) in 685 g of water,
(b) 7.15 moles of ethylene glycol (C2H6O2) in 3505 g of water.

Answers

Answer:

(a) The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]

(b)The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]

Explanation:

The molality (m) of a solution is defined as the number of moles of solute present per kg of solvent.

The Molality of a solution is determined by the expression:

[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

Molality is expressed in units [tex]\frac{moles}{kg}[/tex]

(a) You have 14.3 g of sucrose (C₁₂H₂₂O₁₁), the solute. With the molar mass of sucrose being 342 [tex]\frac{g}{mole}[/tex], then 14.3 grams of the compound represents the following number of moles:

[tex]14.3 grams*\frac{1 mole}{342 grams} =[/tex] 0.042 moles

Having 685 g= 0.685 kg (being 1000 g= 1 kg) of water, the solvent, molality can be calculated as:

[tex]molality=\frac{0.042 moles}{0.685 kg}[/tex]

Solving:

molality= 0.0613[tex]\frac{moles}{kg}[/tex]

The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]

(b) In this case you have 7.15 moles of ethylene glycol (C₂H₆O₂), the solute, in 3505 g (equal to 3.505 kg) of water, the solvent, molality can be calculated as:

[tex]molality=\frac{7.15 moles}{3.505 kg}[/tex]

Solving:

molality= 2.04[tex]\frac{moles}{kg}[/tex]

The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]

You combine 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4 what is the pH of the solution

Answers

Answer:

pH = 3.68

Explanation:

We can solve this problem by using Henderson-Hasselbach's equation:

pH = pKa + log[tex]\frac{[Formate]}{[Formic Acid]}[/tex]Where pKa = -log(Ka)pKa = -log(1.8x10⁻⁴) = 3.74

Assuming we have 1 L of the buffer solution then the molar concentrations of formate and formic acid would be:

[Formate] = 0.75 mol / 1 L = 0.75 M[Formic Acid] = 0.85 mol / 1 L = 0.85 M

We now have all required data to calculate the pH:

pH = 3.74 + log[tex]\frac{0.75}{0.85}[/tex]pH = 3.68

what's the ph of 0.0000067 m hcl solution

Answers

Answer:

[tex]pH = - log[H {}^{ + } ] \\ = - log(0.0000067) \\ pH = 5.17[/tex]

Can someone help me with this one

Answers

Answer:

Easy my dude let me help you out

A.In

B.27

C.73

D.49

E.56

F.56

G.114

H.180

Also with protons and electrons they equal the same atomic number

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