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mars1129 [50]
3 years ago
11

What is the mass in grams of 85.32 ml of blood plasma with a density of 1.03 g/ml?

Chemistry
1 answer:
Westkost [7]3 years ago
4 0
Mass=density*volume
        =85.32*1.03
        =87.88 g
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siniylev [52]
Explain the question...
7 0
3 years ago
A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?
Whitepunk [10]

Let the total mass of compound is 100g

The mass of each element will be

Al = 22.10 g

P = 25.40 g

O = 52.50 g

In order to determine the molecular formula we will calculate the molar ratio of the given elements

Atomic weight of Al : 27 g/ mol

Atomic weight of P : 3 1g /mol

Atomic weight of O : 16 g /mol

Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819

Moles of P = mass / atomic mass = 25.40/ 31 = 0.819

Moles of O = mass / atomic mass = 52.50/ 16 = 3.28

Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present

moles of Al = 0.819 / 0.819 = 1

moles of P = 0.819 / 0.819 = 1

moles of O = 3.28 / 0.819 =  4

So the empirical formula will be  : AlPO4

8 0
3 years ago
How did technology impact the development of the cell theory?
lidiya [134]
It helped be able to look more closely at the the cells
8 0
3 years ago
A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 10
masya89 [10]

<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol

<u>Explanation:</u>

Rate of a gas is defined as the amount of gas displaced in a given amount of time.

\text{Rate}=\frac{V}{t}

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

So,

\left(\frac{\frac{V_{X}}{t_{X}}}{\frac{V_{O_2}}{t_{O_2}}}\right)=\sqrt{\frac{M_{O_2}}{M_{X}}}

We are given:

Volume of unknown gas (X) = 1.0 L

Volume of oxygen gas = 1.0 L

Time taken by unknown gas (X) = 105 seconds

Time taken by oxygen gas = 31 seconds

Molar mass of oxygen gas = 32 g/mol

Molar mass of unknown gas (X) = ? g/mol

Putting values in above equation, we get:

\left(\frac{\frac{1.0}{105}}{\frac{1.0}{31}}\right)=\sqrt{\frac{32}{M_X}}\\\\M_X=367.12g/mol

Hence, the molar mass of unknown gas is 367.12 g/mol

3 0
4 years ago
How many moles of NaOH are present in 27.5 mL of 0.270 M NaOH?
Naddik [55]
The solution for the question above is:

C = 0.270 
<span>V = 0.0275L </span>
<span>n = ? </span>

<span>Use the molar formula which is: C = n/V </span>
<span>Re-arrange it to: n = CV </span>
<span>n = (0.270)*(0.0275) </span>
<span>n = 0.007425 mols </span>
<span>(more precise) n = 7.425 x 10^-3 mols
</span>
7.425 x 10^-3 mols is the answer.
8 0
3 years ago
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