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LUCKY_DIMON [66]
3 years ago
9

If 60.0 grams of carbonic acid are sealed in a 2.00 L soda bottle at room temperature (298 K) and decompose completely via the e

quation below, what would be the final pressure of carbon dioxide assuming it had the full 2.00 L in which to expand? H₂CO₃(aq) → H₂O(l) + CO₂(g)
Chemistry
1 answer:
Tresset [83]3 years ago
6 0

Answer:

The final pressure of the carbon dioxide gas will 11.84 atm.

Explanation:

Moles of carbonic acid = \frac{60.0 g}{62 g/mol}=0.9677 mol

H_2CO_3(aq)\rightarrow H_2O(l) + CO_2(g)

According to reaction, 1 mol of carbonic acid gives 1 mole of carbon dioxide gas.

Then 0.9677 moles of carbonic acid will give :

\frac{1}{1}\times 0.9677 mol=0.09677 mol of carbon dioxide

Moles of carbon dioxide gas = n = 0.09677 mol

Volume of soda bottle = V_1=2 L

Pressure of the carbon dioxide gas = P

Temperature of the carbon dioxide gas = T = 298 K

PV=nRT (ideal gas law)

P=\frac{nRT}{V}=\frac{0.9677 mol\times 0.0821 atm L/mol K\times 298 K}{2 L}=11.84 atm

The final pressure of the carbon dioxide gas will 11.84 atm.

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CH3COOH + NaOH ---> CH3COONa + H2O

That is an acid-base reaction.

The speed of reaction is proportional to the concentrations of the reactants.

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Answer: decrease the rate of reaction.


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3 years ago
Which of the following elements would form the smallest ionic radius and why?
AnnyKZ [126]

Answer:

Sodium

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5 0
2 years ago
A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm
Ulleksa [173]

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100

Percent error = \frac{0.1}{63.2}\times 100

Percent error = 0.00158\times 100

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100

Percent error = \frac{0.2}{63.2}\times 100

Percent error = 0.00316\times 100

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100

Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100

Percent error = \frac{(0.5)}{63.2}\times 100

Percent error = 0.00791\times 100

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%




7 0
3 years ago
Assume that 12.0 g of oxygen are reacted with 20.0 g of magnesium to produce magnesium oxide.
FinnZ [79.3K]
Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.

next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.

use the mols of the limiting reagent to compare with the mols of the product.

take the mols of the product/mr of the product.

this will give u the mass.
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3 years ago
.................. are microorganism used to improve soil fertility.<br>plz ams it correct​
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Answer:

Some bacteria like <em><u>rhizobium</u></em> and <u><em>blue green algae</em></u> are able to fix nitrogen gas from the atmosphere to enrich the soil with nitrogen compounds and increase its fertility. The nitrogen-fixing bacteria and blue green algae are called <u><em>biological nitrogen fixers.</em></u>

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