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3 years ago

What does it mean when your teacher wants you to round 143 to two significant figures

Chemistry

1 answer:

zhenek [66]3 years ago

5 0

Answer:The answer is 3...I think

Explanation:

143

Sig Figs

Decimals

Scientific Notation

1.43 × 102

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What is the result of chemical digestion of carbohydrates proteins and fats?

worty [1.4K]

Protein=Amino acid
Carbohydrates=glucose
FATS=LIPIDS

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3 years ago

A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol

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.

Answer: The freezing point of the solution is $-17.6^oC$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

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)

where,

Freezing point of pure solvent = $0.80^oC$

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 $-17.6^oC$

6 0

3 years ago

How many ml of concentrated hydrochloric acid would be required to make 1 L of a 0.2 M solution? Assume that concentrated hydroc

fgiga [73]

Answer: The volume of concentrated hydrochloric acid required is 16.53 mL

Explanation:

To calculate the volume of concentrated solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

Conversion factor: 1 L = 1000 mL

$M_1=12.1M\\V_1=?mL\\M_2=0.2M\\V_2=1L=1000mL$

Putting values in above equation, we get:

$12.1\times V_1=0.2\times 1000\\\\V_1=16.53mL$

Hence, the volume of concentrated hydrochloric acid required is 16.53 mL

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3 years ago

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navik [9.2K]

Answer:sodium

Explanation: because its not hard to look on the PERIODIC TABLE

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3 years ago

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almond37 [142]

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3 years ago