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3 years ago

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Write a word equation for the overall reaction that occurred in the baggie. The reactants are sodium bicarbonate, carbon dioxide

, and water. Write a balanced equation for the overall reaction that occurred in the baggie.

1 answer:

Fynjy0 [20]3 years ago

6 0

Answer:

Explanation:

In a baggie, sodium bicarbonate is decomposed into sodium carbonate and carbondioxie and water is given off as gases:

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Two samples of carbon tetrachloride were decomposed into their constituent elements. One sample produced 32.4 g of carbon and 37

Gnoma [55]

Answer:

The answer to your question is:

Explanation:

Law of constant composition states that a given compound always contains its component elements in a fixed ratio.

Data

Sample 1 32.4 g of C and 373 g of Cl

Sample 2 12.3 g of C and 112 g of Cl

Convert the grams to moles and compare the ratios

Sample 1 Sample 2

12 g of C --------------- 1 mol 12 g of C -------------- 1 mol

32.4 g ---------------- x 12.3 ------------- x

x = (32.4 x 1) / 12 x = (12.3 x 1) / 12

x = 2.7 moles of C x = 1.025 moles of C

35.5 g of Cl ---------- 1 mol 35.5 g of Cl ---------- 1 mol

373 g ------------------ x 112 g of Cl ----------- x

x = (373 x 1) / 35.5 x = (112 x 1) / 35.5

x = 10.5 moles of Cl x = 3.15 moles of Cl

Ratio

10.5 / 2.7 = 3.9 3.15/1.025 = 3.1

The ratio must be equal in both samples to say that the law is constant. So the answer is NO.

2.-

1) H3PO4 3 hydrogens

2) Na2HPO4 1 hydrogen

3) Ca(HCO3)2 2 hydrogens

4) Ba(C2H3O2)2 6 hydrogens

8 0

3 years ago

Methylhydrazine, CH6N2, is commonly used as a liquid rocket fuel. The heat of combustion of methylhydrazine is −1.30 × 103 kJ/mo

Step2247 [10]

Answer:

There is 5720 J of heat released.

Explanation:

Step 1: Data given

Molar mass of CH6N2 = 46.07 g/mol

The heat of combustion of methylhydrazine is −1.30 × 103 kJ/mol

Mass of methylhydrazine = 202.7 grams

Step 2: Calculate number of moles

Moles CH6N2 = mass of CH6N2 / Molar mass CH6N2

Moles CH6N2 = 202.7 grams / 46.07 g/mol

Moles CH6N2 = 4.4 moles

Step 3: Calculate heat released when 202.7 grams of CH6N2 is burned

q = 1300 J/mol * 4.4 moles

q = 5720 J

(Since there is heat released, ΔH = negative)

There is 5720 J released.

7 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Supposed 0.301 g of methane

Nina [5.8K]

Answer:

The maximum mass of CO2 that could be produced is 0.1925 grams

Explanation:

Step 1: Data given

Mass of methane = 0.301 grams

Mass of oxygen = 0.28 grams

Molar mass of ethane = 16.04 g/mol

Molar mass of oxygen = 32 g/mol

Step 2: The balanced equation

CH4 + 2O2 → CO2 + 2H2O

For 1 mole methane consumed, we need 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O

Step 3: Calculate moles of methane

Number of moles =Mass / Molar mass

Number of moles CH4 = 0.301 grams /16.04 g/mol

Moles CH4 = 0.0188 moles

Step 4: Calculate moles oxygen

Number of moles oxygen = 0.28 grams / 32 g/mol

Moles oxygen = 0.00875 moles

Step 5: the limiting reactant

For 1 mole methane consumed, we need 2 moles of O2

Oxygen is the limiting reactant. It will be completely consumed.

Methane is in excess. There will be consumed 0.00875 /2 = 0.004375 moles

There will remain 0.0188 - 0.004375 = 0.014425 moles

Step 6: Calculate moles of CO2

For 1 mole methane consumed, we need 2 moles of O2 to produce 1 mole of CO2

For 0.00875 moles oxygen , we have 0.004375 moles CO2

Step 7: Calculate mass of CO2

Mass CO2 = moles CO2 * Molar mass CO2

Mass CO2 =0.004375 * 44.01 g/mol = 0.1925 grams CO2

The maximum mass of CO2 that could be produced is 0.1925 grams

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VladimirAG [237]

Answer:

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I hope my answer has come to your help. God bless and have a nice day ahead!
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