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4 years ago

Find the mistake Cu(NO3)2(s) + CuO(s) → 2NO(g) + O3(g)

1 answer:

3 0

there are 2 copper atoms on the left side of the arrow, but none on the other side. there are also 7 oxygen atoms on the left hand side, but only 5 on the right hand side of the arrow.

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The value of the entropy change for the process N₂ (g) + 3H₂ (g) --> 2NH₃ (g) is ________.

kodGreya [7K]

Answer:

negative

Explanation:

Entropy is a measure of the "disorder" in a system.

In this reaction, the amount of disorder decreases. This is because one gas molecule (NH₃) has more order than two gas molecules (N₂ and H₂). Therefore, the entropy change should be negative.

5 0

2 years ago

Which of these statements is true about the temperature of a substance?

Svetradugi [14.3K]

Answer:

(c)

Explanation:

4 0

3 years ago

Choose the correct simplification of x^12z^11/x^2z^4

PolarNik [594]

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▹ Answer

x¹⁰z¹⁵

▹ Step-by-Step Explanation

x¹²z¹¹/x²z⁴

Reduce

x¹⁰z¹¹ * z⁴

Calculate

x¹⁰z¹⁵

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

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5 0

4 years ago

Notice

mixer [17]

Answer:

gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata

Explanation:

vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs

3 0

3 years ago

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0

4 years ago