Rank the compounds below in order of decreasing base strength.

a. HPO4^2-
b. NH3
c. NO2
d. SO4^2-

Answer:

36704 $

Explanation:

First we calculate how much the worker gets paid in one week:

Then we calculate how many weeks does the worker work in 1.7 years:

Finally we calculate how much does the worker make in 1.7 years:

Answer:

a) 2.9 mol

b) 19 atm

Explanation:

Step 1: Given data

Step 2: Calculate the number of moles (n) corresponding to 82 g of CO

The molar mass of CO is 28.01 g/mol.

82 g × 1 mol/28.01 g = 2.9 mol

Step 3: Calculate the pressure (P) inside the container

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T / V

P = 2.9 mol × (0.0821 atm.L/mol.K) × 298 K / 3.7 L = 19 atm

Answer:

67 L

Explanation:

Step 1: Write the balanced equation

Na₂S(aq) + 2 HCl(aq) → 2 NaCl(aq) + H₂S(g)

Step 2: Calculate the moles corresponding to 234 g of Na₂S

The molar mass of Na₂S is 78 g/mol.

234 g × 1 mol/78 g = 3.0 mol

Step 3: Calculate the moles of H₂S produced from 3.0 moles of Na₂S

The molar ratio of Na₂S to H₂S is 1:1. The moles of H₂S formed are 1/1 × 3.0 mol = 3.0 mol.

Step 4: Calculate the volume occupied by 3.0 moles of H₂S at STP

At STP, 1 mole of H₂S occupies 22.4 L.

3.0 L × 22.4 L/1 mol = 67 L

Answer:

a) 0.15 mol.

b) 8.95 g.

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:

[tex]PV=nRT[/tex]

And proceed as follows:

a) Here, we solve for the moles, n,  as follows:

[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]

b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:

[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]

Regards!

The correct answer is mixture

Answer:

32.0 kJ

General Formulas and Concepts:

Thermochemistry

Specific Heat Formula: q = mcΔT

Explanation:

Step 1: Define

Identify variables

[Given] m = 1.00 g

[Given] ΔT = 1.48 °C

[Given] c = 21.6 kJ/g °C

[Solve] q

Step 2: Find Heat

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

31.968 kJ ≈ 32.0 kJ

Answer:

a

Explanation:

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

[tex]rate = k[P]^{2} [Q][/tex]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]

Second row:

2. Rate value:

[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]

3.Third row:

[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]

4. Fourth row:

[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]

Answer:

it is...

Explanation: Pure copper(II) sulfate is white. It is also known as anhydrous copper(II) sulfate because it has no water in it. When water is present in a sample of copper(II) sulfate it turns blue.

Answer:

put it in a beaker then tne beaker in a boiling water bath and when it temperature has reached 100°c condense that gas ... if there is no gas evaporating by 100° then there is no water

Explanation:

water evaporates at 100°c

Answer:

d. 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Explanation:

Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coats the metal preventing it from further oxidation: aluminum oxide. Aluminum is a cation with charge 3+ (Al³⁻) and oxide is an anion with charge 2- (O²⁻). Thus, the neutral compound aluminum oxide has the chemical formula Al₂O₃. The unbalanced chemical equation is:

Al(s) + O₂(g) → Al₂O₃(s)

We can balance using the trial and error method. First, we will balance O atoms by multiplying Al₂O₃ by 2 and O₂ by 3.

Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Finally, we get the balanced equation by multiplying Al by 4.

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Answer:

The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive.

Explanation:

The given substances are:

barium chloride(BaCl2),

glucose(C6H12O6),

butanoic acid (C3H7COOH) which is a weak acid,

sodium bromide (NaBr).

The conductivity of a solution is proportional to the number of ions present in a particular solution.

1mol. of BaCl2 in water produces a total three mol. of ions.

[tex]BaCl_2 (aq) -> Ba^2^+(aq) + 2Cl^-(aq)[/tex]

Gluocse is a covalent compound and it does not dissociate into ions in water.

So, it does not conduct electricity.

Butanoic acid is a weak acid. But due to the release of H+ ions it can conduct a very less amount of electricity.

NaBr is an ionic compound and in 1mol. of NaBr in water gives two mol. of ions.

NaBr (aq)  -> Na+ (aq)  + Br- (aq)

Hence, the order of conductivity among the given substances in aqueous solution is:

BaCl2 > NaBr > butanoic acid > glucose

Answer:

A. Mass percent composition of oxygen = 58.5%

B. Mass percentage composition of oxygen = 42.1%

C. Mass percentage composition of oxygen = 72.7 %

Explanation:

Percentage mass composition of an element in a compound is given by the formula below:

Percentage mass composition = mass of element/ molar mass of compound × 100%

Percentage mass of oxygen in the given compounds are then calculated.

a. Calcium nitrate, Ca(NO₃)₂: molar mass of compound is obtained first.

Molar mass of Ca(NO₃)₂ = 40 + 14 × 2 + 16 × 2 × 3 = 164 g

Mass of oxygen = 16 × 6 = 96 g

Mass percent composition of oxygen = 96/164 × 100% = 58.5%

b. Iron (ii) sulfate, FeSO₄: molar mass of compound is obtained first.

Molar mass of FeSO₄ = 56 + 32 + 16 × 4 = 152 g

Mass of oxygen = 16 × 4 = 64 g

Mass percentage composition of oxygen = 64/152 × 100% = 42.1%

c. Carbon dioxide, CO₂: molar mass of compound is obtained first.

Molar mass of CO₂ = 12 + 16 × 2 = 44 g

Mass of oxygen = 16 × 2 = 32 g

Mass percentage composition of oxygen = 32/44 × 100% = 72.7 %

Organic compround is octane C8H18

The fluorine atom is partially negatively charged while chlorine is partially positively charged.

Polar molecules are molecules whose molecules are partially charged due to the electronegative differences between the atoms in the molecule of the compound.

Chlorine monofluoride is a polar molecule.

Fluorine is more electronegative than the chlorine atom.

Therefore, the fluorine atom is partially negatively charged while chlorine is partially positively charged.

Learn more about polar molecules at: https://brainly.com/question/1433127

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Answer:

One structure shows all the bonds whereas the other shows some of the bonds.

Explanation:

Expanded structure shows all of the bonds connecting all of the atoms in the compound while on the hand, condensed structural shows all atoms, but exclude some or all of the vertical and horizontal bonds. condensed structure makes it easier to write the formula in a line as compared to expanded structure. A bond-line representation is a kind of representation of molecular structure of compounds on a line. In this representation, covalent bonds are represented with one line for each level of bond order.

Answer:

1.104 J/g°C

Explanation:

Using Q = m × c × ∆T

Where;

m = mass of substance (g)

c = specific hear capacity (J/g°C)

∆T = change in temperature (°C)

For a colorimeter,

Q(water) = - Q(metal)

m. c. ∆T (water) = - m. c. ∆T (metal)

According to the information provided;

For water:

m = 28.0g

c = 4.184 J/g°C

∆T = (21.23 - 19.73°C)

For the metal:

m = 2.05g

c = ?

∆T = (21.23 - 98.88°C)

m. c. ∆T (water) = - m. c. ∆T (metal)

[28 × 4.184 × (21.23 - 19.73°C)] = -[2.05 × c × (21.23 - 98.88°C)]

[117.152 × 1.5] = -[2.05 × c × (-77.65)]

175.728 = -[-159.1825c]

175.728 = 159.1825c

c = 175.728 ÷ 159.1825

c = 1.104

c = 1.104 J/g°C

Answer:

[tex]m_{CO_2}=75.6gCO_2[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out mandatory for us to calculate the reacting moles of both C and O2 because we are given grams and pressure, temperature and volume, respectively:

[tex]n_C=36gC*\frac{1molC}{12gC}=3.0molC \\\\n_{O_2}=\frac{3.0atm*14L}{0.08206\frac{atm*L}{mol*K}*298K}=1.72molO_2[/tex]

Thus, since C and O2 react in a 1:1 mole ratio, we infer C is in excess, and the grams of CO2 can be calculated with the moles of O2:

[tex]m_{CO_2}=1.72molO_2*\frac{1molCO_2}{1molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\ m_{CO_2}=75.6gCO_2[/tex]

Best regards!

Answer:

Al2o3 is 101

(nh4)2O IS 52

S8 is 256.56

Ba(oh)2 is 171.35

Cacl2 is 110.98

H2O is 18.01

Explanation:

Answer:

potentiol or kenetic

Explanation:

Answer:

The answer is C... I am almost positive.

Answer:

Substance 1 will diffuse at a faster rate.

Explanation:

We can solve this problem by keeping in mind Fick's law, which states:

Where:

As (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.

Thus the answer is substance 1.

Answer:

1.06 °C

Explanation:

From the question given above, the following data were obtained:

Mass of gold (M₉) = 10 g

Specific heat capacity of gold (C₉) = 0.129 J/gºC

Initial temperature of gold (T₉) = 24 °C

Mass of water (Mᵥᵥ) = 118 g

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Initial temperature of water (Tᵥᵥ) = 1 °C

Equilibrium temperature (Tₑ) =?

The equilibrium temperature of the system can be obtained as follow:

Heat loss by the gold = heat gained by the water

M₉C₉(T₉ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

10 × 0.129 (24 – Tₑ) = 118 × 4.184 (Tₑ – 1)

1.29(24 – Tₑ) = 493.712 (Tₑ – 1)

Clear bracket

30.96 – 1.29Tₑ = 493.712Tₑ – 493.712

Collect like terms

30.96 + 493.712 = 493.712Tₑ + 1.29Tₑ

524.672 = 495.002Tₑ

Divide both side by 495.002

Tₑ = 524.672 / 495.002

Tₑ = 1.06 °C

Therefore, the temperature of the system is 1.06 °C

The amount of heat of the system is measured by a device called a calorimeter. The final temperature of the system will be 1.06 degrees celsius.

The equilibrium temperature is the temperature that follows the law of thermodynamics and is said to be the system that has alike temperatures.  

Given,

Mass of Ag [tex]\rm (M_{g})[/tex] = 10g

Specific heat capacity of Ag [tex](\rm C_{g})[/tex] = [tex]\rm 0.129 J/g^{\circ}C[/tex]

The initial temperature of Ag [tex](\rm T_{g})[/tex] = [tex]24 ^{\circ}\;\rm C[/tex]

Mass of water [tex](\rm M_{w})[/tex] = 118 g

Specific heat capacity of water [tex](\rm C_{w})[/tex] = [tex]4.184 \rm \;J/g^{\circ}\;\rm C[/tex]

The initial temperature of water [tex](\rm T_{w})[/tex] = [tex]1 ^{\circ}\;\rm C[/tex]

Equilibrium temperature = [tex](\rm T_{e})[/tex]

The equilibrium temperature can be shown as, heat loss by the gold = heat gained by the water:

[tex]\rm \rm M_{g}C_{g}(T_{g} - T_{e}) = M_{w}C_{w}(T_{e}-C_{w})[/tex]

Substituting values in the equation:

[tex]\begin{aligned} 10 \times 0.129 (24 - \rm T_{e}) &= 118 \times 4.184 (\rm T_{e} - 1)\\\\\rm 1.29(24 - T_{e}) &= 493.712 (\rm T_{e} - 1)\\\\524.672 &= 495.002 \;\rm T_{e}\end{aligned}[/tex]

Now divide both the sides by 495.002:

[tex]\begin{aligned} \rm T_{e} &= \dfrac{524.672 }{495.002}\\\\\rm T_{e} &= 1.06 \;^{\circ}\rm C\end{aligned}[/tex]

Therefore, the final temperature of the system is 1.06 degrees celsius.

Learn more about equilibrium temperature here:

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Answer:

A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make 500 mL of solution.

Calculate the pH of this buffer.

Explanation:

The pH of a buffer solution can be calculated by using the Henderson-Hesselbalch equation:

[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]

The pH of the given buffer solution can be calculated as shown below:

The question is incomplete, the complete question is:

A solution of permanganate is standardized by titration with oxalic acid. To react completely with 0.0018 mol of oxalic acid required 28.18 mL of permanganate solution. The unbalanced chemical equation for the reaction in acidic solution is:

[tex]\mathrm{MnO}_{4}^{-}(aq)+\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(aq)\stackrel{\mathrm{Acidic}}{\longrightarrow}\mathrm{Mn}^{2+}(aq)+\mathrm{CO}_{2}(\mathrm{g})[/tex]

Determine the concentration of the permanganate solution in molarity.

Answer: The molarity of permanganate solution is 0.026 M

Explanation:

The balanced chemical equation follows:

[tex]2MnO_4^-(aq)+6H^+(aq)+5H_2C_2O_4(aq)\rightarrow 2Mn^{2+}+8H_2O(l)+10CO_2(g)[/tex]

Given values:

Moles of oxalic acid = 0.0018 moles

By the stoichiometry of the reaction:

If 5 moles of oxalic acid reacts with 2 moles of permanganate solution

So, 0.0018 moles of oxalic acid will react with = [tex]\frac{2}{5}\times 0.0018mol=0.00072mol[/tex] of permanganate solution

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)

Given values:

Moles of permanganate solution = 0.00072 moles

Volume of solution = 28.18 mL

Putting values in equation 1, we get:

[tex]\text{Molarity of permanganate solution}=\frac{0.00072\times 1000}{28.18}\\\\\text{Molarity of permanganate solution}=0.026M[/tex]

Hence, the molarity of permanganate solution is 0.026 M

Answer:

i didnt understand

Explanation:

Answer:

x = 2 (C₂H₆)

Explanation:

The general formula for alkanes is CₓH₂ₓ₊₂

2x + 2 = 6

Simply solve for n:

2x = 4

x = 2

Answer:

Silicon

Explanation:

Silicon is in group IV with strong Si-Si ( Silicon to Silicon ) bonds. Since it is a molecule, these bonds exert a strong molecular force hence adopting a giant covalent structure.