Does it require more energy to "vaporize" water at the boiling point or to melt water at the melting point? Explain.

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

butalik [34]

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

4 0

3 years ago

Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,

ludmilkaskok [199]

Answer: The empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

Explanation:

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = $\frac{5.80}{1.44}=4.03\approx 4$

For Oxygen = $\frac{1.44}{1.44}=1$

For Nitrogen = $\frac{1.44}{1.44}=1$

For Chlorine = $\frac{1.44}{1.44}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

3 0

3 years ago

Show All Work For Brainliest

sattari [20]

Answer:

1) 16.8 L CO2

2) 36.96 L NH3

3) 9.88 L CO2

4) 56.99 L H2O

Explanation:

How many liters of carbon dioxide gas will be produced when 75.0 g of calcium carbonate decomposes to form calcium oxide when at STP?

CaCO3 → CaO + CO2

Moles calcium carbonate = 75.0 grams / 100.09 g/mol

Moles calcium carbonate = 0.750 moles

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.750 moles CaCO3 we'll have 0.750 moles CO2

1 mol = 22.4 L

0.750 moles CO2 = 0.750 *22.4 L = 16.8 L CO2

2. Hydrogen gas reacts with 23.1 g of nitrogen gas to produce ammonia (NH3). What volume of ammonia will be produced at STP?

3H2 + N2 → 2NH3

Moles N2 = 23.1 grams / 28.0 g/mol

Moles N2 = 0.825 moles

For 3 moles H2 we need 1 mol N2 to produce 2 moles NH3

For 0.825 moles N2 we'll have 2*0.825 = 1.65 moles NH3

1 mol = 22.4 L

1.65 mol = 1.65 * 22.4 L = 36.96 L NH3

3. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide. How many liters of carbon dioxide will be produced from 23.5 g of iron (III) oxide when at STP?

Fe2O3 + 3CO → 2Fe + 3CO2

Moles Fe2O3 = 23.5 grams / 159.69 g/mol

Moles Fe2O3 = 0.147 moles

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.147 moles Fe2O3 we'll have 3*0.147 = 0.441 moles CO2

1 mol = 22.4 L

0.441 moles = 22.4 * 0.441 = 9.88 L CO2

4.How many liters of water vapor would be produced in the combustion of 12.5L of ethane, C2H6 at STP?

2C2H6 + 7O2 →4CO2 + 6H2O

22.4 L = 1 mol

12.5 L = 0.848 moles C2H6

For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.848 moles C2H6 we'll have 3*0.848 = 2.544 moles H2O

1 mol = 22.4 L

2.544 moles = 22.4 L * 2.544 = 56.99 L H2O

5 0

3 years ago

What is another name for the heat involved in a reaction?

wlad13 [49]

There is your answer

4 0

2 years ago

Which statement describes a system that consists of sugar crystals dissolving in water?

trapecia [35]

To know the answer, you either know what is really the nature and chemistry of a sugar solution. You can also know the answer by knowing the meaning of entropy. Entropy is often interpreted as the degree of disorder or randomness in the system. So the correct statement is that the system becomes more disordered and has an increase in entropy.

7 0

3 years ago

Read 2 more answers