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finlep [7]
2 years ago
8

When baking soda (NaHCO3) is heated, it decomposes, producing carbon dioxide, water, and sodium carbonate. The carbon dioxide is

responsible for the rising of the dough. How many grams of CO2 are produced from 1.0 g of NaHCO3?
Chemistry
1 answer:
11Alexandr11 [23.1K]2 years ago
7 0

Answer:

mass of carbon-dioxide = 0.262 g

Explanation:

Firstly, write the chemical equation of the reaction and balance the equation accordingly.

2NaHCO3(aq) → CO2(g) + H2O(l) + Na2CO3(aq)

Calculate the molecular mass of  baking soda

molar mass of baking soda = 23 + 1 + 12 + 48 = 84 g

2 moles of baking soda = 2 × 84 = 168 g

molar mass of CO2 = 12 + 32 = 44 g

if  168 g of baking soda produces 44 g of carbon-dioxide

1 g of baking soda will produce ? grams of carbon-dioxide

cross multiply

mass of carbon-dioxide = 44/168

mass of carbon-dioxide = 0.2619047619

mass of carbon-dioxide = 0.262 g

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The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup ca
geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

M=d\times V=1.00 g/mL\times 150 ml =150.0 g

Mass of solution = m = M = 150.0 g

NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol

Moles of NaOH = \frac{5.00 g}{40 g/mol}=0.125 mol

Energy released when 0.125 moles of NaOH added in water = Q

Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

Q'= 5,563.8 J

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = \Delta T

Q'=mc\times \Delta T

5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T

\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0
3 years ago
Michelle has a beaker containing NaCl. What is in Michelle's beaker
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<span>Sodium chloride is in her beaker </span>
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B) An element whose particles are molecules.<br>​
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Explanation:

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Calculate the amount of gold deposited when a current of 5A is passed through a solution of gold salt for 2hrs 15mins. If the sa
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Answer:

a. 82.68 g b. 9.8 min

Explanation:

a. The amount of gold deposited by 5 A current in 2 hrs 15 mins

Since charge Q = It where I = current = 5 A and time = 2 hrs 15 mins = 2 × 60 min + 15 min = 120 + 15 min = 135 min = 135 × 60 s = 8100 s

Q = 5 A × 8100 s

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Also, Q = nF where n = number of moles of gold deposited and F = Faraday's constant = 96500 C

n = Q/F = 40500 C/96500 C = 0.4195 moles ≅ 0.42 mole

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m = nM

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b. The time taken for 6g of gold to be deposited.

We first find the number of moles of gold in 6g of gold

Since n = m/M and m = 6 g

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= 2939.09 mol C

= 587.82 s

Changing t to minutes

587.82/60 s = 9.8 min

8 0
3 years ago
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