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wolverine [178]
3 years ago
11

Find molar mass of a gas when the density is given as 1.25g / L

Chemistry
1 answer:
adelina 88 [10]3 years ago
7 0
1.25 g x 22.4 l / 1 mol = 28 g/mol
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Drag each label to the correct location on the image identify the parts of the energy diagram
diamong [38]
If you would’ve attached a picture I’m sure it would’ve been a lot easier.
4 0
2 years ago
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What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 mole of c
yuradex [85]

Answer:

Theoretical moles of V are 1.6 moles

Explanation:

The theoretical yield of a reaction is defined as the amount of product you would make if all of the limiting reactant was converted into product.

In the reaction:

V2O5(s) + 5Ca(i) → 2V(i) + 5CaO(s)

Based on the reaction, 1 mol of V2O5 needs 5 moles of Ca for a complete reaction. As there are just 4 moles, <em>limiting reactant is Ca.</em> As there are produced 2 moles of V per 5mol of Ca, Theoretical moles of V are:

4 moles of Ca × (2mol V / 5Ca) = <em>1.6 moles of V</em>

<em></em>

I hope it helps!

5 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
Determine the oxidation number of sodium in Na202
Olegator [25]

Answer:

+1

Explanation:

Na₂O₂

NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.

Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:

Na₂O₂ = 0 (oxidation number of ground state compound is zero)

2Na + 2O = 0

O = –1

2Na + 2(–1) = 0

2Na – 2 = 0

Collect like terms

2Na = 0 + 2

2Na = 2

Divide both side by 2

Na = 2/2

Na = +1

Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1

5 0
3 years ago
An infectious disease is _____.
e-lub [12.9K]

Answer: caused by organismis

Explanation:

3 0
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