If 2 mL of a liquid weighs 4 g, what is its density?

(4.1357 x 10^-15) x (5.4 x 10^2) Answer must be in scientific notation

nadya68 [22]

Answer:

22×10⁻¹³

Explanation:

Scientific notation:

Scientific notation is the way to express the large value in short form.

The number in scientific notation have two parts.

The digits (decimal point will place after first digit)

× 10 ( the power which put the decimal point where it should be)

for example the number 6324.4 in scientific notation will be written as = 6.3244 × 10³

When we multiply or divide the values the number of significant figures must be equal to the less number of significant figures in given value. Thus, in given value,

(4.1357× 10⁻¹⁵) × (5.4×10²)

22×10⁻¹³

5 0

3 years ago

If given 5.98 x 10^24 atoms of Copper, how many kilograms do you have?

Rufina [12.5K]

Explanation:

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7 0

3 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted 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 glucose solution

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

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$