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2 years ago

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Arrange the following foods in order of increasing acidity : beets. PH=5.9;Camembert cheese,PH=7.4;egg white,pH=8.0;sauerkraut,P

H=3.5;yogurt,PH=4.5

1 answer:

fiasKO [112]2 years ago

8 0

Egg white, Camembert cheese, beets, yogurt, sauerkraut

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Naturally occurring indium has two isotopes, indium-113(112.9040580 amu) and Indium-115 (114.9038780 amu. The atomic mass of ind

Kruka [31]

Answer:

The answer to your question is: letter c (96%)

Explanation:

Indium -113 (112-9040580 amu) ₁₁₃In

Indium-115 (114.9038780 amu) ₁₁₅In

Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In

Formula

Atomic mass = m₁(%₁) +m₂(%₂) / 100

%₁ = x I established this is an equation

%₂ = 100 - x

Substituting values

114.82 = 112.8040x + 114.9039(100-x) /100 and know we expand and simplify

114.82 = 112.8040x + 11490.39 - 114.9039x /100

11482 = 112.8040x -114.9039x +11490.39

11482 - 11490.39 = 112.8040x -114.9039x

-8.39 = -2.099x

x = 3.99

Then % of Indium-115 = 100 - 3.99 = 96

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3 years ago

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Electrons move or cannot move from one object to another when rubbed together

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Which sequence is an example of radioactive decay?

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3 years ago

At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide:

gayaneshka [121]

<u>Answer:</u> The moles of bromine gas at equilibrium is 0.324 moles.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .......(1)

Calculating the initial moles of hydrogen and bromine gas:

<u>For hydrogen gas:</u>

Moles of hydrogen gas = 0.682 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$\text{Molarity of solution}=\frac{0.682mol}{2.00L}=0.341M$

<u>For bromine gas:</u>

Moles of bromine gas = 0.440 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$\text{Molarity of solution}=\frac{0.440mol}{2.00L}=0.220M$

Now, calculating the molarity of hydrogen gas at equilibrium by using equation 1:

Equilibrium moles of hydrogen gas = 0.566 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$\text{Molarity of solution}=\frac{0.566mol}{2.00L}=0.283M$

Change in concentration of hydrogen gas = 0.341 - 0.283 = 0.058 M

This change will be same for bromine gas.

Equilibrium concentration of bromine gas = $(\text{Initial concentration}-\text{Change in concentration})=0.220-0.058=0.162M$

Now, calculating the moles of bromine gas at equilibrium by using equation 1:

Molarity of bromine gas = 0.162 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$0.162M=\frac{\text{Moles of bromine gas}}{2.00L}\\\\\text{Moles of bromine gas}=(0.162mol/L\times 2.00L)=0.324mol$

Hence, the moles of bromine gas at equilibrium is 0.324 moles.

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3 years ago