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mash [69]
3 years ago
10

Which of these would most likely be considered a pure substance?

Chemistry
2 answers:
Karolina [17]3 years ago
5 0

Answer:

c. Water Vapour

Hope it helps.

aliya0001 [1]3 years ago
4 0

the answer is C- water vapour

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Which structure is not an isomer of pentane and state why?
Musya8 [376]

Answer:

2,2-dimethylpropane : C5H12

becoz formula of propane is

C3H8

5 0
3 years ago
Which pH value would a very acidic solution have?<br> a. 0<br> b. 6<br> c. 9<br> d. 14
Artist 52 [7]
The answer is A. 0 because 9 and 14 would be a pH of basic solutions because they are greater than 7; 0 and 6 are acidic because they are less than 7 but zero is smaller than 6 so it is the most acidic
6 0
3 years ago
Read 2 more answers
1. A rock with a mass of 66.5g<br>occupies 23.0 cm. Density?​
telo118 [61]

Answer:

2.89 g/cm^3

Explanation:

Since density equals mass over volume (or also seen as d=\frac{m}{v} ), simply divide 66.5 grams by 23.0 cm. This will output an answer of 2.89 g/cm^3.

7 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
The growth rate is the death rate minus the birth rate these change because birth rates and
zysi [14]

Answer:

Hey there!

False. The growth rate is actually the birth rate minus the death rates.

Let me know if this helps :)

4 0
3 years ago
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