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3 years ago

How do you write 145,000,000 in scientific notation?

Chemistry

1 answer:

belka [17]3 years ago

5 0

Answer:

$\boxed{1.45 \times 10^{8}}$

Explanation:

A number in scientific notation has the form N × 10ⁿ

where N is a decimal number called the mantissa and n is an integer called the exponent.

We must have 1 ≤ N < 10

Step 1. Determine the mantissa

Move the decimal place to the left to create a new number between 1 and 10.

145 000 000 ⟶ 1.450 000 00; N = 1.45

Step 2. Determine the exponent.

The exponent is the number of times you moved the decimal to get the mantissa.

You moved the decimal eight places to the left, so the exponent n = 8.

Step 3. Write the number in scientific notation

The number in scientific notation is $\boxed{1.45 \times 10^{8}}$ .

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katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of $0.80^oC$ and a freezing point depression constant $K_f=7.82^oC.kg/mol$ . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

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<u>Explanation:</u>

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The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

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Freezing point of solution = $?^oC$

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

$K_f$ = freezing point depression constant = $7.82^oC/m$

$m_{solute}$ = Given mass of solute (iron (III) chloride) = 81.1 g

$M_{solute}$ = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

$w_{solvent}$ = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

$0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC$

Hence, the freezing point of the solution is $-17.6^oC$

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