Balance the equation Mg(s) + O2(g) -> MgO(s)

Marina CMI [18]

Answer:

9.1 mol

Explanation:

The balanced chemical equation of the reaction is:

CO (g) + 2H2 (g) → CH3OH (l)

According to the above balanced equation, 2 moles of hydrogen gas (H2) are needed to produce 1 mole of methanol (CH3OH).

To convert 36.7 g of hydrogen gas to moles, we use the formula;

mole = mass/molar mass

Molar mass of H2 = 2.02g/mol

mole = 36.7/2.02

mole = 18.17mol

This means that if;

2 moles of H2 reacts to produce 1 mole of CH3OH

18.17mol of H2 will react to produce;

18.17 × 1 / 2

= 18.17/2

= 9.085

Approximately to 1 d.p = 9.1 mol of methanol (CH3OH).

6 0

3 years ago

Which property of water makes it effective in protecting firefighters?

bezimeni [28]

Answer:

water has a high specific heat

Explanation:

this should work as an answer

5 0

2 years ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

3 years ago

Why do you think large differences in temperature between the wet bulb and dry bulb thermometers are possible only at higher tem

olganol [36]

Answer:

This is the temperature indicated by a moistened thermometer bulb exposed to the air flow. The evaporation is reduced when the air contains more water vapor. The wet bulb temperature is always lower than the dry bulb temperature but will be identical with 100% relative humidity.

Explanation:

7 0

3 years ago

Question 21

Mumz [18]

There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.

Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.

5 0

2 years ago

Balance the equation Mg(s) + O2(g) -> MgO(s)​

Balance the equation Mg(s) + O2(g) -> MgO(s)