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3 years ago

Describe the effects of heat on rocks

Chemistry

1 answer:

Monica [59]3 years ago

6 0

Answer:

rocks main constituent is calcium carbonate, which produce carbon dioxide and calcium oxide (quick lime) upon heating

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How many grams of water will form if 10.54 g H2 react with 95.10 g O2?

sergiy2304 [10]

Answer:

94.0 g H2O

Explanation:

5 0

3 years ago

Read 2 more answers

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c

gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= 135,583 J = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

Now that we have the heat of combustion, we need to calculate the molar heat.

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the conversion factor as 135.58 kJ/ 4.40 g.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

3 0

3 years ago

Please help: A 0.200 M NaOH solution was used to titrate a 18.25 mL HF

algol [13]

The molar concentration of the original HF solution : 0.342 M

Further explanation

Given

31.2 ml of 0.200 M NaOH

18.2 ml of HF

Required

The molar concentration of HF

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)

Titrant = NaOH(1)

Titrate = HF(2)

Input the value :

$\tt 0.2\times 31.2\times 1=M_2\times 18.25\times 1\\\\M_2=0.342$

7 0

3 years ago

Solid aluminum and gaseous oxygen read in a combination reaction to produce aluminum oxide. 4Al(s) + 3O_2(g) rightarrow 2Al_2O_3

Kryger [21]

Answer:

4.7 g. Option 5 is the right one.

Explanation:

4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)

We convert the mass of reactants to moles, in order to determine the limiting.

2.5 g Al / 26.98 g/mol → 0.092 moles of Al

2.5 g O₂ / 32g/mol → 0.078 moles of O₂

Ratio is 4:3. 4 moles of Al react with 3 moles of O₂

Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂

We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.

Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃

Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles

If we convert the moles to mass, we find the anwer:

0.046 mol . 101.96 g/mol = 4.69 g

5 0

3 years ago

The temperature of a 100.0 g sample of water is raised from 30degrees celsius to 100.0 degrees celsius. How much energy is requi

worty [1.4K]

Answer:

29260J

Explanation:

Given parameters:

Mass of water sample = 100g

Initial temperature = 30°C

Final temperature = 100°C

Unknown:

Energy required for the temperature change = ?

Solution:

The amount of heat required for this temperature change can be derived from the expression below;

H = m c (ΔT)

H is the amount of heat energy

m is the mass

c is the specific heat capacity of water = 4.18J/g°C

ΔT is the change in temperature

Now insert the parameters and solve;

H = 100 x 4.18 x (100 - 30)

H = 100 x 4.18 x 70 = 29260J

6 0

3 years ago