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4 years ago

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What happens to the acidity of solutions A and B after dry ice is added? The pH of solution A decreases, and the pH of solution

B increases. The pH of solution A increases, and the pH of solution B decreases. The pH of solution A increases, and the pH of solution B increases. The pH of solution A decreases, and the pH of solution B decreases.

2 answers:

Dafna11 [192]4 years ago

6 0

Answer:

its D on Ed

Explanation:

Dennis_Churaev [7]4 years ago

4 0

answer is d.............

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3 years ago

A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o

Mumz [18]

<u>Answer:</u>

<u>For a:</u> The empirical formula for the given compound is $CH$

<u>For b:</u> The empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

<u>Explanation:</u>

<u>For a:</u>

The chemical equation for the combustion of hydrocarbon follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

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

We are given:

Conversion factor used: 1 g = 1000 mg

Mass of $CO_2=5.86mg=5.86\times 10^{-3}g$

Mass of $H_2O=1.37mg=1.37\times 10^{-3}g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in $5.86\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in $1.37\times 10^{-3}g$ of water, $\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g$ of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.133\times 10^{-3}$ moles.

For Carbon = $\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1$

For Hydrogen = $\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is $CH$

<u>For b:</u>

The chemical equation for the combustion of menthol 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 $CO_2$ = 0.2829 g

Mass of $H_2O$ = 0.1159 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.013g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00065 moles.

For Carbon = $\frac{0.0064}{0.00065}=9.84\approx 10$

For Hydrogen = $\frac{0.013}{0.00065}=20$

For Oxygen = $\frac{0.00065}{0.00065}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is $C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Hence, the empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

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3 years ago

What is the SI unit of pressure and from what units is it derived

irga5000 [103]

The Pascal (symbol: Pa) is the SI derived unit of pressure. It is used to quantify internal pressure, stress, Young's modulus and ultimate tensile strength. It is defined as one Newton per square meter. It is named after the French polymath Blaise Pascal.

The Pascal (Pa) or kilo pascal (kPa) as a unit of pressure measurement is widely used throughout the world and has largely replaced the pounds per square inch.

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4 years ago

The formula for a compound of Li+ ions and Br- ions is written LiBr. Why can’t it be written Li2Br? Why isn’t it written BrLi?

maksim [4K]

Lithium has charge of +1 and bromide has charge of - 1. So they combine to form the compound lithium bromide which is expressed as LiBr.

<h3><u>Explanation:</u></h3>

Lithium is an alkali metal placed in group 1 or periodic table. It has a valency of 1 which is achieved as lithium loses an electron to achieve a charge of +1.

Bromine is a halogen which is placed in group 17 of periodic table. It has a valency of 1 which is achieved as bromine looses an election to achieve a charge of - 1.

Lithium is the cation and bromide is the anion. So lithium is written in front and bromine following the cation. And as both of their valencies are 1, so they form the compound LiBr.

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3 years ago