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sergejj [24]
2 years ago
9

The chemical equation below shows the reaction of sodium (Na) and chlorine (Cl) to form sodium chloride (NaCl).

Chemistry
1 answer:
Aleonysh [2.5K]2 years ago
5 0

The reactant in the equation give in 2na and cl2.


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The Earth's diameter at the equators is _____ its diameter at the poles. A:greater than B:equal to C:less than
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I think the answer is greater than
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2 years ago
Molecule of water contains hydrogen and oxygen in a 1:8 ratio by mass. This is a statement of __________.A) the law of multiple
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Answer:

The correct answer is B.

Explanation:

The molecule of water has 2 atoms of hydrogen and 1 atom of oxygen.

The ratio of masses are given as:

2\times 1 g/mol: 1\times 16 g/mol= 1 : 8

This illustrates the law of definite proportions which is also known as law of constant compositions .

The law states that 'the elements combining to form compound always combine in a fixed ratio by their mass.'

Whereas :

Law of multiple proportion states that when two elements combine with each other to form more than one compounds , the mass of one element with respect to the fixed mass of another element are in ratio of small whole numbers.

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7 0
3 years ago
In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0
sp2606 [1]

Answer:

The value of K_p for this reaction at 1200 K is 4.066.

Explanation:

Partial pressure of water vapor at equilibrium = p^o_{H_2O}=15.0 Torr

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Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:

P=p^o_{H_2O}+p^o_{H_2}

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3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)

The expression of K_p is given by:

K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}

K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066

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6 0
3 years ago
If 11.9 kJ are used to heat a sample of water the temperature increases from 20.0°C to
Kipish [7]

Answer:

m=4.51g

Explanation:

Hello!

In this case, since the energy involved during a heating process is shown below:

Q=mCp\Delta T

Whereas the specific heat of water is 4.184 J/(g°C), we can compute the heated mass of water by the addition of 11.9 kJ (11900 J) of heat as shown below:

m=\frac{Q}{Cp\Delta T}

Thus, by plugging in, we obtain:

m=\frac{11900J}{4.184\frac{J}{g\°C}(650\°C-20.0\°C)}\\\\m=4.51g

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