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3 years ago

How many moles are in 558kg of dinitrogen dioxide?

1 answer:

4 0

How many grams N2O in 1 mol? The answer is 44.0128. We assume you are converting between grams N2O and mole. You can view more details on each measurement unit: molecular weight of N2O or mol This compound is also known as Nitrous Oxide.

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Help me asappp please

Basile [38]

Answer:

wait

Explanation:

wait

3 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

4 years ago

Diethyl ether is produced from ethanol according to the following equation:

juin [17]

The balanced chemical equation is given as:

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.

Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3

% yield = .775 = actual yield / 264.66

actual yield = 205.11 g CH3CH2OCH2CH3

6 0

3 years ago

¿Cuántos electrones conforman un enlace doble?

Aleks04 [339]

Answer:

Cada átomo aporta dos electrones al enlace, es decir, se comparten dos pares de electrones entre dos átomos. Un ejemplo es la molécula de Oxígeno (O2).

Explanation:

6 0

3 years ago

How do the given animal reproduce

Aliun [14]

Answer:

Which animals????????

3 0

3 years ago

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