Convert 1.25 ug into hg

Questions 13 and 14<br> look at the pic

stiv31 [10]

Answer:

13. D Hypothesis 14. A Analyze Matter, Solve Problems

Explanation:

7 0

3 years ago

In the laboratory, a quantity of I2 was reacted with excess H2 to give 1.26 moles of HI. It is also known that the percent yield

Anon25 [30]

Answer:

1.008moles of iodine

Explanation:

Hello,

This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.

Percent yield = (actual yield / estimated yield) × 100

Actual yield = 1.2moles

Estimated yield = ?

Percentage yield = 84%

84 / 100 = 1.2 / x

Cross multiply and solve for x

100x = 84 × 1.2

100x = 100.8

x = 100.8/100

x = 1.008moles

1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI

5 0

3 years ago

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho

timama [110]

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$ , the solution will occupy a larger volume than at $20^{o}C$ .

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$ , density of water=0.9982071 g/ml

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

= $\frac{C_{1}}{d_{1}}$

= $\frac{1.000 mol/L}{0.9982071 g/ml}$

= 1.0017961 mol/g

At $24^{o}C$ , density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

= $\frac{C_{2}}{0.9972995}$

Also, $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so, 1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

$C_{2} = 1.0017961 \times 0.9972995$

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

$C_{2}$ = $\frac{concentration}{volume}$

0.9990907 mol/L = $\frac{concentration}{0.5 L}$

concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where, $d_{air}$ = density of air = 0.0012 g/ml

$d_{weight}$ = density of callibration weights = 8.0g/ml

d = density of weighed object

Hence, the true mass will be calculated as follows.

True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

true mass(m) = 50.5268 g

= 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.

8 0

3 years ago

Calculate the atomic mass of silicon. The three silicon isotopes have atomic mass and relative abundances of 27.9769 amu (92.229

eduard

Nolur acil lütfen yalvarırım yalvarırım lütfen yalvarırım sana

8 0

3 years ago

A balloon with volume of 0.40 liters is at a temperature of 293 K. If the balloon is heated to 523 K, then what will the new vol

igor_vitrenko [27]

v=293+523

v=816

this is the answer

thank you

4 0

2 years ago