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3 years ago

how many oxygen atoms are there in six dinitrogen monoxide molecules? express your answer as an integer

Chemistry

1 answer:

stepan [7]3 years ago

7 0

Answer:

6 oxygen atoms

Explanation:

Step 1: Given data

Number of dinitrogen monoxide molecules (N₂O): 6

Number of oxygen atoms (O): ?

Step 2: Calculate the appropriate ratio

The ratio of dinitrogen monoxide molecules to oxygen atoms is 1:1.

Step 3: Use the ratio to calculate the number of oxygen atoms

6 molecule N₂O × (1 atom O/1 molecule N₂O): 6 atom O

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Explanation:

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3 years ago

A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af

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Explanation:

The given reaction is as follows.

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

n = $0.05 L \times 0.1 M$

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After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

n = $0.025 L \times 0.1 M$

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$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial: 0.005 mol 0.0025 mol 0 0

Change: -0.0025 mol -0.0025 mol +0.0025 mol

Equibm: 0.0025 mol 0 0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

[HA] = $\frac{0.0025 mol}{V}$

[NaA] = $\frac{0.0025 mol}{V}$

$[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

pH = $pK_{a} + log \frac{A^{-}}{HA}$

$pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

= $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

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3 years ago

Aluminum metal and manganese (IV) oxide react to produce aluminum oxide and manganese metal. What mass of manganese (IV) oxide r

gayaneshka [121]

Answer: 1.77 kg of manganese (IV) oxide reacts to produce 1.12kg of manganese metal.

Explanation:

The balanced chemical equation is:

$4Al+3MnO_2\rightarrow 2Al_2O_3+3Mn$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of manganese = $\frac{1.12\times 1000g}{55g/mol}=20.4mol$

According to stoichiometry :

3 moles of $Mn$ is produced by = 3 moles of $MnO_2$

Thus 20.4 moles of $Mn$ is produced by = $\frac{3}{3}\times 20.4=20.4moles$ of $MnO_2$

Mass of $MnO_2=moles\times {\text {Molar mass}}=20.4moles\times 86.9g/mol=1773g=1.773kg$ (1kg=1000g)

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3 years ago

How many orbitals are in the n=3 level

solniwko [45]

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3 years ago

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