Answer:
you use your sink and spray the little hose thingy and boom. or you can make it rain with mooney
Explanation:
no u can not make it rain in the house but if u want in the jar, here are the steps
Question: How can you make it rain indoors? 1) First, fill a glass jar with very hot tap water. 2) Then, flip the lid over and fill it with ice. ... In about 15 minutes you will see "rain" fall inside the jar .
hope this helped you
Increasing temperature can
Answer:
increases reaction rates
Explanation:
Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.
Arrange these compounds by their expected boiling point.Highest boiling pointLowest boiling pointCH3OH, CH3Cl, CH4
Answer:
CH3OH > CH3Cl > CH4
Explanation:
The boiling point of a compound depends on the nature and magnitude of intermolecular forces acting between its molecules.
Methanol contains the -OH group, hence hydrogen bonds are dominant intermolecular forces in the compound. Presence of hydrogen bonds increases the boiling point of a substance, hence methanol has the highest boiling point followed by chloromethane which has dipole-dipole interaction as dominant intermolecular forces and methane having only weak dispersion forces as dominant intermolecular forces.
A 75.0 mL sample of water is heated to its boiling point. How much heat (in kJ ) is required to vaporize it?
Given :
A 75.0 mL sample of water is heated to its boiling point.
To Find :
The heat (in kJ ) is required to vaporize it .
Solution :
Heat of vaporization of water , [tex]\Delta H_{vap}=40.65\ kJ/mol[/tex].
Molar mass of [tex]H_2O[/tex] , [tex]M=18\ g/mol[/tex] .
Density of water , [tex]\rho=1\ g/ml[/tex] .
Mass of 75 ml of water sample , [tex]m=75\ g[/tex] .
Moles of 75 g water = [tex]\dfrac{75}{18}=4.17\ mol[/tex] .
Now , heat required to vaporize 4.17 mol of water :
[tex]H=4.17\times 40.65\ kJ\\\\H=169.51\ kJ[/tex]
Therefore , heat required to boil 75 ml sample of water is 169.51 kJ.
Hence , this is the required solution .
2.643 grams of potassium butanoate (KCH3(CH2)2CO2 ) is fully dissolved in 50.00 mL of water, which is carefully transferred to a conical flask. Then 100.00 mL of 0.120 M HCℓ is added dropwise to this solution from a burette. Given: Ka(butanoic acid) = 1.5 × 1O−5 . 2.1 Showing all your calculations and reasoning, determine the pH of the solution that results after the addition of all the acid mentioned above.
Answer:
The pH of the solution is 4.69
Explanation:
Given that,
Mass of potassium = 2.643 grams
Weight of water = 50.00 mL
Weight of HCl=100.00 ml
Mole = 0.120 M
We know that,
[tex]KCH_{3}(CH_{2})_{2}CO_{2}[/tex] is a basic salt.
Let's write it as KY.
The acid [tex]HCH_{3}(CH_{2}CO_{2})[/tex] would become HY.
We need to calculate the moles of KY
Using formula of moles
[tex]moles\ of\ KY=\dfrac{m}{M}\times1000[/tex]
[tex]moles\ of\ KY=\dfrac{2.643}{126}\times1000[/tex]
[tex]moles\ of\ KY=20.97\ m\ mole[/tex]
The reaction is
[tex]KY+HCl\Rightarrow HY+ KCl[/tex]
The number of moles of KY is 20.98 m
initial moles = 20.98
Final moles [tex]m=20.98-0.120\times100= 8.98[/tex]
We need to calculate the value of pKa(HY)
Using formula for pKa(HY)
[tex]pKa_{HY}=-log Ka[/tex]
[tex]pKa_{HY}=-log(1.5\times10^{-5})[/tex]
[tex]pKa_{HY}=4.82[/tex]
We need to calculate the pH of the solution
Using formula of pH
[tex]pH=pKa+log(\dfrac{[KY]}{[KH]})[/tex]
Put the value into the formula
[tex]pH=4.82+log(\dfrac{8.98}{12})[/tex]
[tex]pH=4,69[/tex]
Hence, The pH of the solution is 4.69
a. When 25.4 and 40.433 are added, the answer should be based on how significant figures or decimal
Answer:
65.8 and it is shown considering the significant figures of the shortest number.
Explanation:
Hello,
In this case, when we add 25.4 and 40.433 the result is 65.833, nevertheless, the result is shown based on the initial number having the fewest amount of significant figures that in this case is 25.4 that has three, therefore, the result must be 65.8 and it is shown considering the significant figures of the shortest.
Regards.
The weak acid HA is 4 % ionized (dissociated) in a 0.30 M solution. A. What is Ka for this acid?B. What is the pH of this solution?
Answer:
[tex]Ka=5x10^{-4}[/tex]
[tex]pH=1.92[/tex]
Explanation:
Hello,
In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:
[tex]\% ionization=\frac{[H^+]}{[HA]}*100\%[/tex]
[tex][H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M[/tex]
Therefore the Ka is computed by using the equilibrium expression:
[tex]Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}[/tex]
And the pH:
[tex]pH=-log([H^+])=-log(0.012)\\\\pH=1.92[/tex]
Regards.
describethe differences in the ways the sand castle is changed by an ocean wave and by someone stompie on it
What is the name of the ionic compound Li3PO4?
Answer:
lithium phosphate
Explanation:
The name of the given ionic compound is lithium phosphate. The correct option is D.
What is an ionic compound?Ionic compounds are neutral compounds composed of positively charged cations as well as negatively charged anions.
For binary ionic compounds, the identity of the cation is written first, accompanied by the title of the anion.
Ionic compounds are made up of ions that are held together through the force of attraction between oppositely charged ions.
One of the most well-known ionic compounds is common salt. Molecular compounds are made up of discrete molecules that are held together by sharing electrons, a process known as covalent bonding.
Lithium phosphate (Li3PO4) is an ionic compound that is primarily used as a conductive thin-film electrolyte in the fabrication of high performance energy storage devices.
Thus, the correct option is D.
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The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in millimeters? A) 1.15x 10-8 mm B) 1.15 1013 mm C) 1.15 10-7 mm D) 1.15 x 1017 mm
Answer:
C) 1.15 × 10⁻⁷ mm
Explanation:
Step 1: Given data
Average distance between nitrogen and oxygen atoms: 115 pm
Step 2: Convert the distance to meters (SI base unit)
We will use the conversion factor 1 m = 10¹² pm.
115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m
Step 3: Convert the distance to millimeters
We will use the conversion factor 1 m = 10³ mm.
1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm
A silver ring has a mass of 138.45 g. How many calories of heat are required to increase the
temperature from 11.8 °C to 162.5 °C?
Answer:
150.7
Explanation:
162.5 - 11.8 = 150.7
What should you do if you spill a chemical on your hand?
Answer:
You should rinse your head thoroughly ASAP, don't try to blow on the chemical or anything. Its going burn, you have to let your head cool off.
Explanation:
why rinse off your hand? you're getting the fluid away fast, so I can't infect your skin that much.
What mass of aluminum has a total nuclear charge of 1.6 CC? Aluminum has atomic number 13. Suppose the aluminum is all of the isotope with 14 neutrons. slader
Answer:
aluminium is all of isotope with 14 neutrons .
What would the force of the Earth on the moon be when the moon is 2.0D from the Earth?
Answer:
Force, [tex]F=7.3\times 10^{36}\ N[/tex]
Explanation:
We need to find the force of the Earth on the moon be when the moon is 2.0D from the Earth.
The force that act between Earth and the Moon is gravitational force. It is given by :
[tex]F=\dfrac{Gm_1m_2}{r^2}[/tex]
We have,
[tex]m_1=\text{mass of Earth}=5.972 \times 10^{24}\ kg[/tex]
[tex]m_2=\text{mass of Moon}=7.34\times 10^{22}\ kg[/tex]
So,
[tex]F=\dfrac{6.67\times 10^{-11}\times 5.972 \times 10^{24}\times 7.34\times 10^{22}}{(2)^2}\\\\F=7.3\times 10^{36}\ N[/tex]
So, the force of the Earth on the Moon is [tex]7.3\times 10^{36}\ N[/tex].
Liquid octane (C8H18), a component of gasoline, reacts with gaseous oxygen to form gaseous carbon dioxide and water vapor. Write the balanced chemical reaction for this process.
Answer:
[tex]2C_{8}H_{18 (l)} + 25O_{2 (g)}[/tex] → [tex]16CO_{2(g)} +18H_{2}O_{(l)}[/tex]
Explanation:
Describe how the Rube Goldberg device design that you created follows the law of conservation of energy. Be sure to include both parts of the law of conservation of energy in your description.
Answer:
Following are the explanation of the Rube Goldberg device:
Explanation:
According to the Rube Goldberg devices, which conform with "the energy conservation law," choose a chain of events to carry out such a basic task differently, for this unit, a range of instant theatrical resources are converted into possible energy. It is also responsive to an energy conservation law.
It is the unit that produces theatrical power when quartz falls off from the roof. Instead, its marble hits or falls into a plastic bottle, its velocity converted into potential energy which lifts its soda bottle. Its next marble, when the stone came in contact with it, can pass angular momentum to some other marble. Its gravitational force of quartz drops. It has been built into kinetic energy, distributed only to Nando. All energy of every domino becomes passed to another in a domino effect before the last one takes its contraption off because as the mousetrap gets off, all of the stored potential energy is released and the cutters were also removed. Its sequence which binds its puck is cut, which also frees the ballon as well as stores its electricity.What is the mass grams of 0.0135 mol of the element sodium, Na?
Answer:
Tvfbggbtb
Explanation:
If 33.6 g of NO and 26.9 g of O₂ react together, what is the mass in grams of NO₂ that can be formed via the reaction below? 2 NO (g) + O₂ (g) → 2 NO₂ (g)
Answer:
If 33.6 g of NO and 26.9 g of O₂ react together, 51.52 grams of NO₂ can be formed.
Explanation:
You know: 2 NO (g) + O₂ (g) → 2 NO₂ (g)
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and product participate in the reaction:
NO: 2 moles O₂: 1 moleNO₂: 2 molesBeing:
N: 14 g/moleO: 16 g/molethe molar mass of the compounds participating in the reaction is:
NO: 14 g/mole + 16 g/mole= 30 g/moleO₂: 2*16 g/mole= 32g/moleNO₂: 14 g/mole + 2*16 g/mole= 46 g/moleThen, by stoichiometry of the reaction, the following amounts of reactant and product participate in the reaction:
NO: 2 moles* 30 g/mole= 60 gO₂: 1 mole* 32g/mole= 32 gNO₂: 2 moles* 46 g/mole= 92 gThe limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
You can apply the following rule of three to determine the limiting reagent: if by stoichiometry 60 g of NO react with 32 g of O₂, 33.6 g of NO with how much mass of O₂ will it react?
[tex]mass of O_{2}=\frac{33.6 grams of NO*32 grams of O_{2} }{60 grams of NO}[/tex]
mass of O₂= 17.92 grams
But 17.92 grams of O₂ are not available, there are 26.9 grams available. Since it has more mass than it needs to react with 33.6 grams of NO, NO will be the limiting reagent.
Then you apply the following rule of three: if by reaction stoichiometry 60 grams of NO form 92 grams of NO₂, 33.6 grams of NO how much mass of NO₂ does it form?
[tex]mass of NO_{2}=\frac{33.6 grams of NO*92 grams of NO_{2} }{60 grams of NO}[/tex]
mass of NO₂= 51.52 grams
If 33.6 g of NO and 26.9 g of O₂ react together, 51.52 grams of NO₂ can be formed.
The most common use of the ammonia manufactured is in A. fertilizers for crops. B. plastics manufacturing. C. household cleaners. D. dyes.
One dose of a liquid medication is 2.5 tablespoons. If a patient takes a dose of this medication four times a day for a week, how many milliliters of this medication is consumed in one week? (1 tbsp = 15 mL)
Answer:
1050 mL
Explanation:
Step 1: Given data
Dosage of the liquid medication: 2.5 tablespoons (tbsp)
Times that the patient takes the medication: 4 times a day for a week
Step 2: Calculate the number of times the patient takes the medication
We will use the relationship 1 week = 7 days.
[tex]1 week \times \frac{7days}{1week} \times \frac{4dose}{1day} = 28 dose[/tex]
Step 3: Calculate the number of tablespoons
We will use the relationship 2.5 tbsp/1 dose.
[tex]28dose \times \frac{2.5tbsp}{1dose} = 70tbsp[/tex]
Step 3. Convert 70 tbsp to mL
We will use the relationship 1 tbsp = 15 mL.
[tex]70tbsp \times \frac{15mL}{1tbsp} = 1050 mL[/tex]
A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 298 K. The final volume of the air in the tire is Vf= 1.50 L and the final pressure is Pf = 5.00 bar. Calculate the final temperature of the air in the tire.
Complete Question
A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 298 K. The final volume of the air in the tire is Vf= 1.50 L and the final pressure is Pf = 5.00 bar. Calculate the final temperature of the air in the tire.
Assume that Cv,m = 5R/2
Answer:
The value is [tex]T_f = 471.978 \ K[/tex]
Explanation:
From the question we are told that
The initial pressure is [tex]P_i = 1.00 \ bar[/tex]
The initial temperature is [tex]T_i = 298 K[/tex]
The final volume is [tex]V_f = 1.50 \ L[/tex]
The final pressure is [tex]P_f = 5.0 \ bar[/tex]
Generally the equation for reversible adiabatic compression is
[tex]P_i V_i^{\gamma} = P_f V_f^{\gamma} [/tex]
Generally from the ideal gas equation we have that
[tex]PV = n RT[/tex]
=> [tex] V = \frac{n RT}{P}[/tex]
So
[tex]P_i *[ \frac{n RT_i}{P_i}]^{\gamma} = P_f [ \frac{n RT_f}{P_f}]^{\gamma}[/tex]
=> [tex]P_i ^{1 - \gamma} * T_i ^{\gamma} = P_f ^{1-\gamma} * T_f^{\gamma}[/tex]
=> [tex]T_f ^{\gamma} = \frac{ P^{1 - \gamma * T_i^{\gamma}}}{ P_f ^{1 - \gamma}}[/tex]
=> [tex]T_f =[ \frac{ P_i ^{1 - \gamma} * T_i^{\gamma}}{P_f^{1 - \gamma}}]^{\frac{1}{\gamma} }[/tex]
=> [tex]T_f = T_i * [\frac{P_i}{P_f} ]^{\frac{1- \gamma}{\gamma}[/tex]
Here [tex]\gamma[/tex] is a constant mathematically represented as
[tex]\gamma = \frac{C_{P,m}}{C_{V, m}}[/tex]
Here
[tex]C_{P,m}} [/tex] is the molar heat capacity at constant pressure
and [tex]C_{V,m}} [/tex] is the molar heat capacity at constant volume given as
[tex]C_{V,m}} = \frac{5R}{2} [/tex]
Generally [tex]C_{P,m}} [/tex] for an ideal gas is mathematically represented as
[tex]C_{P,m}} = R +C_{V,m}} [/tex]
So
[tex]\gamma = \frac{\frac{5R}{2} + R}{ \frac{5R}{2} }[/tex]
=> [tex]\gamma = \frac{7}{5} [/tex]
So
=> [tex]T_f = T_i * [\frac{P_i}{P_f} ]^{\frac{1- \frac{7}{5}}{ \frac{7}{5}}[/tex]
=> [tex]T_f = 298 * [\frac{1}{5} ]^{\frac{1- \frac{7}{5}}{ \frac{7}{5}}[/tex]
=> [tex]T_f = 471.978 \ K[/tex]
The Final temperature of the air in the tire = 471.978k
Given data :
Initial pressure ( [tex]P_{i}[/tex] ) = 1.00 bar
Initial temperature ( [tex]T_{i}[/tex] ) = 298 K
Final volume of air ( [tex]V_{f}[/tex] ) = 1.50 L
Final pressure ( [tex]P_{f}[/tex] ) = 5.00 bar
Assume Cvm = 5r / 2 ( missing data )
Determine the final temperature of the air in the tire
Applying the equation for a reversible Adiabatic compression and general ideal gas equation
[tex]Pi * [ \frac{nRTi}{Pi} ] \alpha = Pf * [ \frac{nRTf}{Pf}] \alpha[/tex] ---- ( 1 )
Resolving equation ( 1 )
[tex]final temperature = Ti * [ \frac{Pi}{Pf}]^{\frac{1-\alpha }{\alpha } }[/tex] ------- ( 2 )
where [tex]\alpha[/tex] ( mathematical constant ) = ( Cpm / Cvm ) --- ( 3 )
Cpm = molar heat capacity at constant pressure = R + Cvm
Cvm = molar heat capacity at constant volume = 5r / 2
∴ Equation ( 3 ) becomes
[tex]\alpha[/tex] = [tex]\frac{7}{5}[/tex]
Final step : Determine the final temperature of the air in the tire
Input values back into equation ( 2 )
[tex]final temperature = 298 + [\frac{1}{5}] ^{\frac{1-\frac{7}{5} }{\frac{7}{5} } }[/tex]
= 471.978 K
Hence we can conclude that the final temperature of the air in the tire is 471.978 K
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123.0 x 12.35 / (0.05 x 6.049) significant figures
Answer:
123.0*12.35/(0.05*6.049)
first do in small bracket 123.0*12.35/0.30245
and then divide 123.0*40.83319557
and multiply 5022.483055
Fill in the blank to complete each statement.
5. A(n)________ change releases energy.
6. New substances are produced by a(n)______
change.
7._________ is related to the energy of motion of the particles of matter.
8. The law of________ of mass states that in any physical or chemical
change, matter is neither created nor destroyed.
9.______energy naturally flows from warmer matter to cooler matter.
10. The form or appearance of matter is altered during a(n)______ change.
Answer:
5.physical change
6.chemical change
7.physical change
8.conservation of mass
9.thermal energy
10.physical change
I honeslty dont know if this is right
explanation:
The efficiency of the wedge is low because of _____.
Answer:friction
Explanation:
Consider 55 mL of water (H2O) in a beaker and 55 mL of acetone |(CHs)2CO in an identical beaker under identical conditions Complete the sentences to explain the relationship between the vapor pressure of water and acetone. Match the words to the appropriate blanks in the sentences.
larger than
more quickly than
the same as
weaker
smaller than
stronger
more slowly than
at the same rate as
1. Water will evaporate______ acetone
2. The vapor pressure of acetone is__________ than that of water at the same temperature.
3. This is because the intermolecular forces between acetone molecules are ___________than those between water molecules.
Answer:
1. more slowly than
2. larger than
3. weaker
Explanation:
Acetone molecules are bonded by very weak intermolecular forces when compared to that of the hydrogen bond between water molecules. This makes it very easy for acetone molecules to vaporize easily into its gaseous state, much more faster than water molecules (since the acetone molecules need a lesser amount of energy to break these bonds). Also, the boiling point of liquid acetone is much lower than that of water, meaning that it has a higher vapor pressure than that of water.
an engineering team is conducting a trial launch of a few rocket . which part is the engineering process is the team in ?
Answer:
Evaluate the result (Apex)
Explanation:
At a resting pulse rate of 81 beats per minute, the human heart typically pumps about 65 mL of blood per beat. Blood has a density of 1060 kg/m3. Circulating all of the blood in the body through the heart takes about 1 min for a person at rest.Required:a. How much blood is in the body? b. On average, what mass of blood does the heart pump with each heart beat?
Answer:
5.265L/body ; 68.9×10³ kg / beat
Explanation:
Given the following :
Resting pulse rate = 81 beats / minute
Amount of blood pumped per beat = 65mL = (65 / 1000)L = 0.65L
Density of blood = 1060 kg/m^3
Circulation time = 1 minute for a person at rest
A.) How much blood is in the body?
Resting pulse rate × amount of blood pumped per beat * circulating time
(81 beat / min) × 0.65L/beat * 1minute / body = 5.265L / body
B.) b. On average, what mass of blood does the heart pump with each heart beat?
0.065L/beat * 1m³/1000L= 0.000065 m³ / beat
From density = mass / volume ;
Average mass = density * volume
= 1060kg/m³ * 0.000065 m³ / beat
= 0.0689kg / beat
= 68.9 × 10³ kg / beat
How many liters of plasma are present in 8.5 pints?
Answer:
there are 4.0205 Litres of plasma in 8.5 pints
Explanation:
Recall that:
1 quart qt = 2 pints pt and 1 quart = 946 mL
∴ two pints will contain 946 mL
and one pint will be = 946 mL/2
= 473 mL
Similarly, 8.5 pints will contain 473 mL × 8.5 = 4020.5 mL
We know that :
1000 mL = 1 Litres
Hence 4020.5 mL = (4020.5 /1000)L
= 4.0205 Litres
Therefore, there are 4.0205 Litres of plasma in 8.5 pints
There are 4.0205 Litres of plasma in 8.5 pints
The calculation is as follows:1 quart qt = 2 pints pt
and
1 quart = 946 mL
So, two pints will contain 946 mL
and one pint will be = [tex]946 mL\div 2[/tex]= 473 mL
Now
8.5 pints will contain
[tex]= 473 mL \times 8.5[/tex]
= 4020.5 mL
We know that :
1000 mL = 1 Litres
Hence 4020.5 mL = (4020.5 /1000)L
= 4.0205 Litres
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Which weak acid would be best to use when preparing a buffer solution with a pH of 9.80?A. An acid with Ka = 7.9 times 10-12.B. An acid with Ka = 2.0 times 10-5.C. An acid with Ka = 2.5 times 10-7.D. An acid with Ka = 1.6 times 10-6.E. An acid with Ka = 2.0 times10-8.F. An acid with Kg = 1.3 times 10-10.
Answer:
An acid with Kg = 1.3 times 10-10.
Explanation:
The formula for pH of a buffer solution is as follows
pH = pKa + log [ A⁻ ] / [ HA ]
HA is acid concentration and A⁻ is concentration of conjugate base .
Generally the the second term is very small so neglecting this figure
pH = pKa
9.8 = pKa
If Ka = 10⁻¹⁰ x 1.3
pKa = - log ( 10⁻¹⁰ x 1.3 )
= 10 - log 1.3
= 9.8 approx .
Hence option F is correct .
Soil is a heterogeneous mixture. Which of the following
statements best explains why?
a
It is uniformly mixed, so it has an even composition
throughout
b
The components are chemically bound together in
a disordered shape, so it does not have the same
composition throughout the mixture.
c С
The particles are large enough to see, so it looks
the same throughout.
d
It is unevenly mixed, so it does have the exact
same composition throughout the mixture,
S
Answer: D
Explanation:
i just took the ck-12 quiz and that was the right answer
The measurement 0.020 L is the same as?
Answer:
20 mL
2 cL
Explanation:
We can use the base 10 decimal system to get equivalent measurements:
1 L = 1000 mL
1 L = 100 cL