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Margarita [4]
3 years ago
7

A compound X has the following percentage composition 66.7% carbon, 11.1% hydrogen and 22.2% oxygen .Calculate the empirical for

mula of X.The relative molecular mass of X is 72 calculate the molecular formula​
Chemistry
1 answer:
Digiron [165]3 years ago
4 0

Explanation:

c. h. o

66.7%. 11.1%. 22.2%

____. ____. ____

12. 1. 16

1.558. 11.1. 1.39. (divide by the smallest)

1. 8. 1

empirical formula=ch8o

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GuDViN [60]

Answer:

A phosphorus atom forms a P3- ion by gaining three electrons.

here

4 0
2 years ago
In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temp
asambeis [7]

Answer:

ΔH = 57 Kj/mole H₂O

Explanation:

6 0
3 years ago
Read 2 more answers
What kind of molecule is table sugar? a. Monosaccharide c. Trisaccharide b. Disaccharide d. Octosaccharide
Aleksandr-060686 [28]
Table sugar is a
B. Disaccharide
3 0
3 years ago
Consider the reaction directly below and answer parts a and b. C6H4(OH)2 (l) + H2O2 (l) à C6H4O2 (l) + 2 H2O (l)
dedylja [7]

Answer:

a. -206,4kJ

b. Surroundings will gain heat.

c. -115kJ are given off.

Explanation:

It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.

Using:

<em>(1) </em>C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

<em>(2) </em>H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ

<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ

It is possible to obtain:

C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)

From (1)-(2)+2×(3). That is:

<em>(1) </em>C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ

<em>-(2) </em>H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ

<em>2x(3) </em>2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ

The ΔH you obtain is:

+177,4kJ + 187,8kJ - 2×285.8 kJ =<em> -206,4kJ</em>

b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the <em>surroundings will gain this heat.</em>

c. 20,0g of H₂O are:

20,0g×\frac{1mol}{18,01g} = <em>1,11 mol H₂O</em>

As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:

1,11mol H₂O×\frac{-206,4kJ}{2mol} =<em> -115kJ</em>

I hope it helps!

8 0
3 years ago
Ag2O(s) → 2Ag(s) + ½ O2(g) ΔH° = 31.05 kJ Which statements concerning the reaction above are true? (1) heat is released (2) heat
Sidana [21]

Answer:

D) 2, 4, and 5

Explanation:

In order to fully comprehend the answer choices we must take a close look at the value of ΔH° = 31.05. The enthalpy change of the reaction is positive. A positive value of enthalpy of reaction implies that heat was absorbed in the course of the reaction.

If heat is absorbed in a reaction, that reaction is endothermic.

Since ∆Hreaction= ∆H products -∆H reactants, a positive value of ∆Hreaction implies that ∆Hproducts >∆Hreactants, hence the answer choice above.

5 0
3 years ago
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