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3 years ago

Q8.what happened to the mixture?

Chemistry

1 answer:

Natali [406]3 years ago

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What is sodium bicarbonate an example of? a buffer an acid a liquid a base

ludmilkaskok [199]

Answer:

base

Explanation:

carbonic acid - H2CO3 - is a weak acid. Therefore, HCO3 itself is its conjugate base. The Na(sodium) ion is neutral.

This means that NaHCO3 is a base. (a weak one)

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irakobra [83]

B: produces energy for the cell

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What are the transition metals

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Theyre the big bunched up group in the middle of the periodic table

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3 years ago

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If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?

Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

3H2 + N2 ------> 2NH3

from reaction 3 mol 1 mol

given 11.8 mol 1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

3H2 + N2 ------> 2NH3

from reaction 1 mol 2 mol

given 1.01 mol x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0

3 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

3 years ago