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3 years ago

5

Seawater is mostly a solution of sodium chloride in water. The concentration varies, but marine biologists took a sample with an

amount concentration of 0.56 mol/L. Calculate the mass of sodium chloride in the biologists 5.0L

1 answer:

sineoko [7]3 years ago

3 0

Answer:

1.0.16

2.a.355mg

b. 8.07mmol/L

3.7.83%

Explanation:

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Determine the molarity for each of the following solution solutions:

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Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution</u>

Formula used :

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

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

<u>(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution</u>

Formula used :

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

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

<u>(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution</u>

Formula used :

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

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

<u>(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution</u>

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution</u>

Formula used :

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

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

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HELP ME PLEASE!!!

krok68 [10]

Answer:

Br (Bromine)

H (Hydrogen)

Explanation:

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Which of the following radiations consists of a helium nucleus emitted by some radioactive substances? Alpha radiation Beta radi

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Answer: Alpha radiation

Explanation: Alpha decay : When a larger radioactive nuclei decays into smaller nuclei by releasing alpha radiation, the mass number and atomic number is reduced by 4 and 2 units respectively.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}Y+_2^4\textrm{He}$

Beta decay : When a larger radioactive nuclei decays into smaller nuclei by releasing beta radiation, the atomic number is increased by 1 unit.

$_Z^A\textrm{X}\rightarrow _{Z+1}^{A}Y+_{-1}^0e$

Gamma decay : When a larger radioactive nuclei decays into smaller nuclei by releasing gamma radiation, the mass number remains same.

$_Z^A\textrm{X}^*\rightarrow _Z^A\textrm{X}+_0^0\gamma$

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A thermometer is in a beaker of water. Which statement best explains why the thermometer reading initially increases when LiBr(s

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