Help with all of these please!

The air in a balloon is heated up. Give two effects this has on the air particles.

Lisa [10]

Density and mass increases

4 0

3 years ago

A sample of limestone (calcium carbonate, CaCO3) is heated at 950 K until it is completely converted to calcium oxide (CaO) and

pav-90 [236]

Answer:

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Explanation:

Equations of reactions:

A: CaCO₃(s) ---> CaO(s) + CO₂(g)

B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)

Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g

From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂

This means that 56 g of CaO produces 74 g of Ca(OH)₂

mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO

Therefore, 6409.73 g of CaO were produced in reaction A

From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂

Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles

Therefore, 114.46 moles of CO₂ were produces as well.

Molar volume of gas at STP = 22.4 litres

Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L

However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation

Using P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?

V₂ = P₁V₁T₂/P₂T₁

V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)

V₂ = 9141.404 L

Therefore, volume of CO₂ produced in the first step is 9141.404 L

3 0

3 years ago

The combustion of 0.590 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

Alenkinab [10]

Answer:

2.943 °C temperature change from the combustion of the glucose has been taken place.

Explanation:

Heat released on combustion of Benzoic acid; :

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.590 g

Moles of benzoic acid = $\frac{0.590 g}{122.12 g/mol}=0.004831 mol$

Energy released by 0.004831 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004831 mol=15.5955 kJ=15,595.5 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.125°C

$Q=C\times \Delta T$

$15,595.5 J=C\times 2.125^oC$

$C=7,339.05 J/^oC$

Heat released on combustion of Glucose: :

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=1.400 g

Moles of glucose = $\frac{1.400 g}{180.16 g/mol}=0.007771 mol$

Energy released by the 0.007771 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.007771 mol=21.6030 kJ=21,603.01 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$21,603.01 J=7,339.05 J/^oC\times \Delta T'$

$\Delta T'=2.943^oC$

2.943 °C temperature change from the combustion of the glucose has been taken place.

8 0

3 years ago

HELP ME OUT PLS!!!

alina1380 [7]

D Direction of motion I hope this helps

4 0

3 years ago

roundup an herbicide manufactured by monsanto has the formula c3h8no5p. how many moles of molecules are there in a 500g sample o

uranmaximum [27]

N= m/Mr Find the Mr of C3H8NO5P and then divide the mass by the Mr. Sorry I don't have periodic table with me so I can't calculate the Mr for you

3 0

4 years ago