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2 years ago

Perform the calculation to the correct number of significant figures. (8.81 - 7.50)/0.0020

Chemistry

1 answer:

AleksAgata [21]2 years ago

4 0

Answer:

655

Explanation:

using bodmas, (bracket first), 8.81-7.50=1.31

1.31÷0.0020=655

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2 years ago

What would the products of a double-replacement reaction between KBr and CaO be? (Remember: In double-replacement reactions, the

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ANSWER

$\begin{gathered} \text{ 2KBr }+\text{ CaO }\rightarrow\text{ K}_2O\text{ }+\text{ CaBr}_2 \\ Option\text{ B} \end{gathered}$

EXPLANATION

Given that;

The two reactants are KBr and CaO

Double replacement reaction is a type of chemical reaction that occur when two reactants exchange cations and anions to yield new products.

$\text{ 2KBr + CaO }\rightarrow\text{ K}_2O\text{ + CaBr}_2$

Therefore, the resulting products of the given data are K2O + CaBr2

The correct answer is option B

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1 year ago

Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made

arlik [135]

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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3 years ago