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3 years ago

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The chemical equation below shows the amount of the reactants used and the amount of only one of the products formed. How much w

ater (H2O) should be formed in this chemical reaction? CH4 mass 4 + 2O2 mass 16 CO2 11 + 2h2O mass ?

1 answer:

Karolina [17]3 years ago

6 0

Answer:

CH4 + 2O2 → CO2 + 2H2O

Explanation:

This is all i could come up with im sorry.

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Sodium is going in group 1, period 3. It contains A. One valence electron in its 3rd energy shell B three valence electron in it

topjm [15]

Answer:

The correct answer is option A, that is, one valence electron in its third energy shell and option C, that is, 11 electrons and 11 protons.

Explanation:

The outermost electrons and the ones that take part in the process of bonding are termed as valence electrons. The atomic number of sodium is 11, thus, it possesses 11 protons and the atoms are neutral so it suggests that sodium has 11 electrons. By electronic configuration, it can be seen that in sodium, two electrons are present in the first shell, 8 in the second, and only one electron in the third shell, that is, 2.8.1. The electron present in the third shell is the valence electron.

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3 years ago

How many H2O molecules are in 183.2 grams of H20 gas?

jek_recluse [69]

Answer: There are $61.24 \times 10^{23}$ molecules present in 183.2 grams of $H_{2}O$ gas.

Explanation:

Given: Mass = 183.2 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of water is 18 g/mol. Therefore, moles of $H_{2}O$ are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{183.2 g}{18 g/mol}\\= 10.17 mol$

According to the mole concept, there are $6.022 \times 10^{22}$ molecules present in one mole of a substance.

Hence, molecules present in 10.17 moles are calculated as follows.

$10.17 mol \times 6.022 \times 10^{23}\\= 61.24 \times 10^{23}$

Thus, we can conclude that there are $61.24 \times 10^{23}$ molecules present in 183.2 grams of $H_{2}O$ gas.

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3 years ago

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-

Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u> $Ni(s)\rightarrow Ni^{2+}+2e^-$

<u>Reduction half reaction (cathode):</u> $Ni^{2+}+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V

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3 years ago

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lawyer [7]

It receives heat energy.

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3 years ago