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Fofino [41]
3 years ago
13

Write the chemical symbol for the isotope of cobalt with 37 neutrons

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
4 0

Answer:

64Co27

Explanation:

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1<br> The process of a solid turning into a liquid is called
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Melting

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Once heat is added it will turn to a liquid

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The ionic mobility of alkali metal ions in aqueous<br>solution is maximum for​
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Rb+

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Chemical ____ are used to represent compounds ​
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Zinc and hydrochloric acid react to form zinc chloride and hydrogen gas according to
stealth61 [152]

Explanation:

Mole ratio of Zn to HCl = 1 : 2.

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6 0
3 years ago
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
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