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During chemical reactions the bonds between atoms break and new bonds form. Energy must be absorbed to break a bond, so breaking

ss7ja [257]

Answer:

The equation is: CuCO3(s) → CuO(s) + CO2(g)

Explanation:

Copper carbonate decomposes at high temperatures, generating the products carbon dioxide (CO2) and copper oxide (II) CuO. In this type of decomposition reaction, a substance is broken generating two different compounds.

7 0

3 years ago

A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the

Sveta_85 [38]

Answer:

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

Explanation:

Using the formula

Ca Va = Cb Vb

Cb = 0.32 M

Vb = 50 mL = 50/1000 = 0.050L

Ca = 0.5 M

Va =?

Substituting for Va in the equation, we obtain:

Va = Cb Vb / Ca

Va = 0.32 * 0.05 / 0.5

Va = 0.016 / 0.5

Va = 0.032 L

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

6 0

3 years ago

Electrons is an excited state are more likely to enter into chemical reactions.

Alex17521 [72]

A. True would be the best answer

3 0

3 years ago

HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i

OleMash [197]

Answer: S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

$HNO_3+S\rightarrow H_2SO_4+NO$

On the reactant side:

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

On the product side:

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

3 0

3 years ago

A sample of iron absorbs 67.5 j of heat upon which the temperature of the sample increases from 21.5 °c to 28.5 °c. if the speci

asambeis [7]

Q = mcΔθ

67.5 = m x 0.45 x (28.5 - 21.5)

M = 67.5 / 3.15
= 21.4 g

3 0

3 years ago

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