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VMariaS [17]
3 years ago
15

What is the GFM of H2O2

Chemistry
1 answer:
vichka [17]3 years ago
5 0

Answer:

Hydrogen peroxide

Explanation:

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A 2.000 g sample of CoCl2.xH2O is dried in an oven. When the anhydrous salt is removed from the oven, its mass is 1.565 g. What
Mashutka [201]

The value of X is 2

The total amount of sample taken is 2.00 g

The amount of sample left in the oven after drying is 1.565g

The amount of sample lost (mass of water driven out) = Total sample-Anhydrous salt left in the oven

                                 = 2.00 - 1.565

                                = 0.435 grams

The moles of anhydrous salt present in the hydrate = 1.565g/129.83g/mol = 0.01205

The moles of water present in the hydrate = 0.4350g/18.01g/mol = 0.02415

Therefore the ratio of these two are in 1:2 ratio

The complete chemical reaction is   CoCl2.2H20

To know more about the Similar calculation and explanation click here:

brainly.com/question/14967837

#SPJ4

6 0
1 year ago
5.40 g B determine the number of atoms
sdas [7]
Atomic mass Boron ( B ) = 10.811 u.m.a

10.811 g -------------- 6.02x10²³ atoms
5.40 g ----------------- ?? atoms

5.40 x ( 6.02x10²³) / 10.811 =

3.0069x10²³ atoms
4 0
3 years ago
Nitrogen gas consists of _____.
Anni [7]
Nitrogen gas consists of a diatomic nitrogen atom. So, the correct answer is B.
8 0
3 years ago
Read 2 more answers
How many grams of oxygen gas will be produced when 2.50 moles of potassium chlorate is decomposed?
yawa3891 [41]

Answer:

m_{O_2}=120gO_2

Explanation:

Hello!

In this case, since the decomposition of potassium chlorate is:

2KClO_3\rightarrow 2KCl+3O_2

We can see a 2:3 mole ratio between potassium chlorate and oxygen (molar mass 32.0 g/mol), thus, via stoichiometry, we compute the mass of oxygen that are produced by the decomposition of 2.50 moles of this reactant:

m_{O_2}=2.50molKClO_3*\frac{3molO_2}{2molKClO_3} *\frac{32.0gO_2}{1molO_2}\\\\m_{O_2}=120gO_2

Best regards!

8 0
3 years ago
Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
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