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2 years ago

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Na2SO4(aq) +CaCl2(aq) — CaSO4(s) + 2NaCl(aq) Ionic Equation: 2Na+ (aq) + S02-(aq) + Ca2+(aq) + 2C1- (aq) CaSO4(s) + 2Na+ (aq) +

2Cl- (aq) Now, let's write the net ionic equation: [?] + []=[] A. 2Na+(aq) C. 2C1-(aq) B. Ca2+ (aq) D. Na+ (aq)
please help

1 answer:

devlian [24]2 years ago

5 0

Answer:

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Which is the IUPAC name for NO?

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Answer:

Nitrogen (ii) oxide

Explanation:

To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.

NOTE: The oxidation number of oxygen (O) is always – 2.

Thus the oxidation number of N in NO can be obtained as follow:

N + O = 0 (ground state)

N + (– 2) = 0

N – 2 = 0

Collect like terms

N = 0 + 2

N = +2

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Answer: The empirical formula is $C_3H_3O$ .

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 12.24 g

Mass of $H_2O$ = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, = $\frac{12}{44}\times 12.24=3.338g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, = $\frac{2}{18}\times 2.505=0.278g$ of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.278}{0.104}=3$

For H = $\frac{0.278}{0.104}=3$

For O = $\frac{0.104}{0.104}=1$

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is $C_3H_3O$ .

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