All categories

3 years ago

How many grams of potassium nitrate (KNO3) are required to prepare 0.250L of a 0.700 M

Chemistry

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer:

17.78g

Explanation:

m/M = C×V

m/101.102= 0.7× 0.25

m= 101.102×0.7×0.25= 17.78g

You might be interested in

How many kilograms are 1.21 x 10^-6 moles of zinc ?

Svetradugi [14.3K]

Answer: The answer is 2.3 x 10 4's.

7 0

3 years ago

Acid rain can be destructive to both the natural environment and human-made structures. The equation below shows a

guapka [62]

Answer:

Explanation:

What the equation is tell you is that for every 3 mols of NO2 you get 2 mol of HNO3

3 mol NO2 / 2 mol HNO2 ===> 300.00 mol NO2 / x Cross multiply

3x = 2 * 300

3x = 600 Divide by 3

3x/3 = 600/3 Do the division

x = 200.00

3 0

3 years ago

Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at

kozerog [31]

Answer:

20 g/mol

Explanation:

We can use Graham’s Law of diffusion:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

Solve for M₂:

$M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}$

$M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}$

$= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}$

7 0

3 years ago

Which of the following is a half-reaction? A. Zn+CuSO4−> B. 2Cl−−>Cl2+2e− C. H2+1/2O2−>H2O D. −>Cu+ZnSO4

malfutka [58]

Answer:

2Cl——>Cl2+2e-

Explanation:

It shows an electron loss or gain

8 0

2 years ago

What is the molar mass of glucose, C3H603?

ziro4ka [17]

Answer:

The mass percentage of carbon can be found easily using the molar mass of C6H12O6, 180.1559 g/mol. We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass. C6H12O6 = CO2 + C3H6O3 + CH3OCH3 Take fructose for example. Compound.

Explanation: I looked it up

4 0

2 years ago