What is the molarity of a solution with 55 grams of CaCL2 dissolved in. 78 L of water?

Magnesium (average atomic mass = 24.3050 amu) consists of three isotopes with masses 23.9850 amu, 24.9858 amu, and 25.9826 amu.

iris [78.8K]

Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%

Explanation: Given that the average atomic mass(M) of magnesium

= 24.3050amu

Mass of first isotope (M1) = 23.9850amu

Mass of middle isotope (M2)=24.9858amu

Mass of last isotope(M3)= 25.9826amu

Total abundance = 1

Abundance of middle isotope = 0.10

Let abundance of first and last isotope be x and y respectively.

x+0.10+y =1

x = 0.90-y

M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope

24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y

Substitute x= 0.90-y

Then

y = 0.11

Since y=0.11, then

x= 0.90-0.11

x=0.79

Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%

5 0

3 years ago

What is a small-ball shaped organelle that helps the cell break down nutrients and old cell parts.

Gemiola [76]

Lysosome: only ones that break down old cell parts<span>
</span>

6 0

3 years ago

At equilibrium the partial pressures of N2O4 and NO2 are 0.35 atm and 4.3 atm. What is the Kp

ale4655 [162]

Answer:

Kp = 52.83

Explanation:

6 0

3 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

3 years ago

What is the mass in grams of 7.23 miles of dinitrogen trioxide?

nevsk [136]

Answer:

549.563868

Explanation:

1 mole is equal to 1 moles N2O3, or 76.0116 grams. so 76.0116 x 7.23 = 549.563868

3 0

3 years ago