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r-ruslan [8.4K]

2 years ago

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When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3) is heated, it releases carbon dioxide gas, which is r

esponsible for the rising of cookies, donuts, and bread. (a) Write a balanced equation for the decomposition of the compound (one of the products is Na2CO3). (b) Calculate the mass of NaHCO3 required to produce 20.5 g of CO2. Chang, Raymond. Chemistry (p. 113). McGraw-Hill Higher Education. Kindle Edition.

1 answer:

butalik [34]2 years ago

4 0

Answer:

There is 78.25g NaHCO3 required

Explanation:

Step 1: Balance the equation

2 NaHCO3 → Na2CO3 + CO2 + H20

For 2 moles of NaHCO3 consumed, there is produced 1 mole of Na2CO3, 1 mole of CO2 and 1 mole of H2O.

Step 2: calculating moles of CO2

mass of Co2 = 20.5g

Molar mass of CO2 = 44.01 g/mole

moles of CO2 = 20.5 / 44.01 = 0.4658 moles

Step 3: Calculating moles of NaHCO3

Since we have for 1 mole CO2 produced, there is 2 moles of NaHCO3 consumed.

To calculate number of moles of NaHCO3, we have to multiply the number of moles of CO2, by 2.

⇒ 0.4658 x2 = 0.9316 moles

Step 4: Calculating the mass of NaHCO3

mass of NaHCO3 = moles of NaHCO3 x Molar mass of NaHCO3

mass = 0.9316 moles x 84g/ mole = 78.25g NaHCO3

There is 78.25g NaHCO3 required

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How many liters of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa)?

OLga [1]

Using ideal gas equation,

P\times V=n\times R\times T

Here,

P denotes pressure

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R denotes gas constant

T denotes temperature

The values at STP will be:

P=100 kPa

T=293 K

R=8.314472 L kPa K⁻¹ mol⁻¹

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2 years ago

Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample

Dimas [21]

Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

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We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

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2 years ago

In a particular mass of kau(cn)2, there are 6.66 × 1020 atoms of gold. What is the total number of atoms in this sample?

Vinvika [58]

Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.

Number of gold, $Au$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms (given)

From the formula of compound that is $KAu(CN)_2$ it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.

So, the number of atoms of each element is:

Number of potassium, $K$ atoms in $KAu(CN)_2$ = $6.66\times 10^{20}$ atoms

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Total number of atoms in $KAu(CN)_2$ = Number of gold, $Au$ atoms+Number of potassium, $K$ atoms +Number of carbon, $C$ atoms + Number of nitrogen, $N$ atoms

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2 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

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ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

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ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

2 years ago