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3 years ago

Someone please helpppp!!!!!!Question 7

Chemistry

1 answer:

Snezhnost [94]3 years ago

5 0

answer:

612.5

Explanation:

from the equation, no. of mole of iron oxide is obtaibed from mass over molar mass equals to 5.5 moles, taking mole ratio as 1:2 to get moles of iron as 10.94 moles, then the mass is obtained by moles of iron times molar mass

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How many molecules are in 15.0 g of oxygen gas (O2)?

frosja888 [35]

Answer:

2.82*10^23

Explanation:

please see attached for work!

3 0

3 years ago

At 850°C, CaCO3 undergoes substantial decomposition to yield CaO and CO2. Assuming that the ΔH o f values of the reactant and pr

Xelga [282]

Answer:

The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ

Explanation:

Step 1: Data given

temperature = 850 °C

Mass of 68.10 grams of CO2

ΔH°f (CaO) = -635.6 kJ/mol

ΔH°f (CO2) = -693.5 kJ/mol

ΔH°f (CaCO3) =-1206.9 kJ/mol

Step 2: The balanced equation

CaCO3(s) → CaO(s) + CO2(g)

Step 3: Calculate ΔH°reaction

ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)

ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) - ΔH°f (CaCO3)

ΔH°reaction = (-635.6 kJ/mol + -693.5 kJ/mol) + 1206.9 kJ/mol

ΔH°reaction = -122.2 kJ /mol

Step 4: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 68.10 grams / 44.01 g/mol

Moles CO2 = 1.547 moles

Step 5: Calculate the enthalpy change for 68.10 grams of CO2

-122.2 kJ/mol * 1.547 moles = -189.04 kJ

The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ

7 0

3 years ago

Gold and hydrochloric acid

Sever21 [200]

Answer:

hydrochloric acid can be used to clean gold

hydrochloric acid does not gold in any way

3 0

3 years ago

__Ba3N2 + __ H2O —-> __Ba(OH)2 + __ NH3

sergeinik [125]

Answer:

is there a pic?

Explanation:

6 0

2 years ago

If, as is typical, each of them breathes about 500 cm3 of air with each breath, approximately what volume of air (in cubic meter

Allisa [31]

Volume of brethe consumed in a normal breathe is known to be tidal volume. The tidal volume of a person is 500 mL. There are 12 times a person can breathe in a minute.

Number of breathes in a minute $= 12 breathe$

Consumption of air per brethe $= 500 mL$

Consumption of air per minute = $= 12 * 500 = 6 L$

Consumption of air per year $= 6 * 60 * 24 * 365 = 31.53 * 10^5 L$

Thus, 31.53 * 10^5 L of air consumed in a year by an astronaut.

4 0

3 years ago