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andriy [413]
3 years ago
12

what volume of carbon 4 oxide measured at STP will produce when 21 gram of sodium hydrogen carbonate for is completely decompose

d accor. ding to the equation. (Na=23. H=1 C=12 O=16) molar vol =22.4dmcube. pleeeeeeeeeeese help​
Chemistry
1 answer:
IgorLugansk [536]3 years ago
5 0

The volume of CO₂ at STP = 2.8 L

<h3> Further explanation </h3>

Given

21 gram of sodium hydrogen carbonate-NaHCO₃

Required

Volume of CO₂

Solution

The decomposition of sodium bicarbonate into sodium carbonate, carbon dioxide, and water :

2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)

mol of NaHCO₃ :

= mass : MW NaHCO₃

= 21 g : 84 g/mol

= 0.25

From the equation, mol ratio of NaHCO₃(s) :CO₂(g) = 2 : 1, so mol CO₂ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.25

= 0.125

At STP, 1 mol gas = 22.4 L, so for 0.125 mol :

= 0.125 x 22.4 L

= 2.8 L

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oksano4ka [1.4K]

The organism would no longer grow.

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When an animal eats corn or peanuts, will it also eat the seeds?
nasty-shy [4]

Answer:yes

Explanation:it doesnt know

3 0
3 years ago
A 59.1g sample of aluminum is put into a calorimeter (see sketch at right) that contains 250.0g of water. The aluminum sample st
Rainbow [258]

Answer:

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

Explanation:

Step 1: Data given

Mass of aluminium = 59.1 grams

Mass of water = 250.0 grams

Initial temperature of aluminium = 91.3 °C

Initial temperature of water = 16.0 °C

Final temperature = 19.5 °C

Pressure remains constant

Specific heat capacity of water = 4.186 J/g°C

Step 2: Calculate specific heat of aluminium

Heat lost = heat gained

Qlost = -Q heat

Q = m*c*ΔT

heat aluminium = - heat water

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒m(aluminium) = mass of aluminium = 59.1 grams

⇒c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 91.3 = -71.8 °C

⇒ m(water) = 250.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 16.0 = 3.5 °C

59.1 * c(aluminium) * -71.8 °C = 250.0 * 4.186 J/g°C * 3.5 °C

c(aluminium) = 0.863 J/g°C

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

3 0
3 years ago
mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0
3 years ago
What is the molar mass of Na2CO3? 60.0 g/mol 106.0 g/mol 118.0 g/mol 141.0 g/mol
Dmitry [639]

Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.

M(Na₂CO₃) = 2 · Ar(Na) + Ar(C) + 3 · Ar(O).

M(Na₂CO₃) = 2 · 23 + 12 + 3 · 16 · g/mol.

M(Na₂CO₃) = 46 + 12 + 48 · g/mol.

M(Na₂CO₃) = 106 g/mol; molar mass of sodium carbonate.

Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.

8 0
3 years ago
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