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2 years ago

A 12.5-g marble display 5.0 mL of water. What is its density?

Chemistry

2 answers:

Irina-Kira [14]2 years ago

8 0

In g/mL, 12.5/5, so 2.5 g/mL

anastassius [24]2 years ago

5 0

12.5 g/ 5.0 mL= 2.5 g/L

The final answer is 2.5 g/L~

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2 years ago

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A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

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3 years ago

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Explanation:

Atomic numbers for neutral atoms are the same as the number of electrons. The number of electrons are used in writing the subshell notation where we can see the highest princpal quantum number(n).

In order to solve this problem, a good knowledge of the periodic table and shell notation is required.

For example;

Given a number 5;

We know the number is made of both the atomic number of the element and the highest quantum number.

It is easier for us to know the highest quantum number from the shell notation:

shell notation 1s²2s²2p¹

highest principal quantum number is 2

We are left with atomic number 5-2, which is 3

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3 years ago

How many parts per million of fluoride in a solution that is 500 grams of fluoride and 500,000 liters water

erik [133]

0.000001‬ppm

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Mass of fluoride = 500g

Volume of water = 500000liters

Unknown:

Parts per million of fluoride = ?

Solution:

The parts per million is the amount of solute in milligram dissolved in a liter of water or milligram per kilogram of solvent

It is a unit used to express very small concentration.

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500g = 500 x 10⁻³mg = 0.5mg

Concentration in parts per million = $\frac{mass in mg}{Volume in liters}$

Concentration in parts per million = $\frac{0.5}{500000}$ = 0.000001‬ppm

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3 years ago

There are 154,000 mg of sugar in a

Kipish [7]

Explanation:

1g = 1000mg

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No of days she can drink = 154÷ 11 = 14days

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3 years ago