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Convert 16.8 L of nitrogen monoxide gas at STP to grams .

german

Answer:

See explanation

Explanation:

1 mole of a gas occupies 22.4 L

x moles occupies 16.8 L

x = 1 mole * 16.8 L/22.4 L

x = 0.75 moles

number of moles = mass/molar mass

mass = number of moles * molar mass

mass = 0.75 moles * 30.01 g/mol = 22.5075 g = 2.25 * 10^1 g

the coefficient of the scientific notation answer = 2.25

the exponent of the scientific notation answer = 1

significant figures are there in the answer = 6

the right most significant figure in the answer = 3

2.

number of moles = 12.5g/38g/mol = 0.3289 moles

1 mole occupies 22.4 L

0.3289 moles occupies 0.3289 moles * 22.4 L/1 mole

= 7.36736 L = 7.36736 * 10^0 L= 7.37 * 10^0 L

the coefficient of the scientific notation answer =7.37

the exponent of the scientific notation answer = 0

significant figures are there in the answer = 6

the right most significant figure in the answer= 3

8 0

3 years ago

What can occur in physical change ?

Alexandra [31]

Within a physical change, an element can change forms, such as going from solid to a liquid through melting. Color change can also occur during a physical change. Physical changes are very different from chemical changes. In a chemical change the element itself changes into something else within a reaction, such as combustion (burning).

Hope this helped

5 0

3 years ago

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$

$2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol$

$2\times \Delta H^o_{4}=470.3 kJ/mol$

$\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol$

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago

If there are 80 liters of argon gas in a piston and the piston expands the gas until its volume is 2000 liters with a pressure o

Mice21 [21]

Answer: The correct answer is A. 11.5 atm. The temperature is held constant at 293 K, therefore, we can use Boyle's Law to determine the initial pressure. Boyle's Law states that there is an inverse relationship between pressure and volume of gases. Therefore, as volume increases, the pressure will decrease and vice versa.

Further Explanation:

Boyle's Law can be mathematically expressed as:

$P_{initial}V_{initial}\ = P_{final}V_{final}$

In this problem, we are given the values:

P(initial) = ?

V(initial) = 80 L

P (final) = 0.46 atm

V (final) = 2000 L

Plugging in these values into the equation:

$P_{initial}(80 \ L) = \ (0.46 \ atm)(2000 \ L)\\P_{initial} \ = \frac{(0.46 \ atm) \ (2000 \ L)}{80 \ L}\\P_{initial}\ = 11.5 atm\\$

The initial pressure was 11.5 atm. Since the volume increased or expanded, the space where the gas particles move is bigger, so the frequency of collisions with the wall of the container and with other particles are effectively decreased. This, therefore, decreases the pressure from 11.5 to 0.46 atm.

Learn More

Learn about Charles' Law brainly.com/question/1421697
Learn about Ideal Gas Law brainly.com/question/6534668
Learn about Gay - Lusaac's Law brainly.com/question/1358307

Keywords: gas, Boyle's Law, Ideal Gas Law

7 0

3 years ago