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10 months ago

Which of the following can be mixed in solution with H₂CO3 to make a buffer?OA. Na₂CO3OB. HFOC. NH3OD. NaHCO3

1 answer:

6 0

A buffer system usually will need a weak acid, like H2CO3, which is carbonic acid, and a strong base, and the only compound that could figure a fairly strong base among the options given is Na2CO3. NH3 and NaHCO3 are weak bases and HF is a strong acid. Therefore the best answer will be letter A

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A sample of a compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. What is the empirical formula of this

Shtirlitz [24]

Answer:

The empirical formula is CH2O, and the molecular formula is some multiple of this

Explanation:

In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.

We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.

5 0

1 year ago

The equilibrium constant, Kp, for the following reaction is 0.497 at 500K.PCl5(g) PCl3(g) + Cl2(g)If an equilibrium mixture of t

Anestetic [448]

Answer: The equilibrium partial pressure of chlorine gas is 0.360 atm

Explanation:

For the given chemical equation:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl2(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{Cl_2}\times p_{PCl_3}}{p_{PCl_5}}$

We are given:

$K_p=0.497\\p_{PCl_3}=0.651atm\\p_{PCl_5}=0.471atm$

Putting values in above equation, we get:

$0.497=\frac{p_{Cl_2}\times 0.651}{0.471}\\\\p_{Cl_2}=0.360atm$

Hence, the equilibrium partial pressure of chlorine gas is 0.360 atm

5 0

2 years ago

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

Answer:

For a: The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Explanation:

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

For c:

For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

2 years ago

Check all statements below that are true.

docker41 [41]

Answer:

The answers are A,B,C.

Explanation: Just got it right on Edge 2020

3 0

2 years ago

How many moles are in 0.0688 g of silver chloride (AgCl)

Artist 52 [7]

Answer: 0.00048

Explanation:

Moles = Mass/Molar mass

0.0688

107.87

35.5

0.00048

6 0

2 years ago