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3 years ago

If 42 grams of carbon in 52 grams of oxygen are used how many grams of CO2 will be produced. I need to show work.

Chemistry

1 answer:

UNO [17]3 years ago

4 0

Answer:

How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide? ... be used to produce 1.99 grams of water. 1.99 mg H2O X. 1mol H2O. 18.0g X ... c. If the reaction produces 5.3 mg of carbon dioxide how many grams of water ... X. 25mol O2. 2mol C8H18. X. 32.0g O2. 1mol O2. = 4.80 x 103g O2. Answer ...

Explanation:

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Evaluate the following: 3.7 x 102m x 2.9 x 108 m

yulyashka [42]

Answer:

118201.68m6^2

Explanation: scientific notation

6 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.

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4 years ago

Calculate the mass (in g) of 2.1 x 1024 atoms of W.

vfiekz [6]

Set up your units to algebraically cancel out the original unit, and end up with the units of the answer in the final numerator. We'll first convert to moles by using Avogadro's number, then the atomic mass of tungsten (W) from the periodic table, since the mass, in grams, of one mole of any element is the atomic mass number:
2.1x10∧24 atoms W x (1 mole/6.022x10∧23 atoms) x (183.84g/1 mole) = 641.089 grams
Since your given number has only two significant figures (2.1), your answer should also only express in 2 sig figs: 6.4x10∧2 grams

6 0

3 years ago

At STP, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water?

Delvig [45]

Should be 1.8L.
2 moles of hydrogen react with 1 mole of oxygen. If 2 moles of hydrogen is 3.6L, 1 mole of oxygen should be 1.8L.

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3 years ago

if the gas pressure exerted by a gas at 45 degrees celsius in a volume of 1.944 ATM, how many moles of gas are present?

Firdavs [7]

If the.pressure exerted by a gas at [math]25^{\circ} \mathrm{C}[/math] in a volume of 0.044 L is 3.81 atm, how many moles of gas are present

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3 years ago