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3 years ago

I was wondering if I could get some help on this question..?

Chemistry

1 answer:

loris [4]3 years ago

5 0

1: Perform a control group experiment
2: Perform the same experiment with a different variable
3: Perform the same experiment with added variables
4: Collect results from each and revise hypothesis appropriately and repeat.

Hope this helped.

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Determine the approximate mass of the isotope of carbon in the image.

avanturin [10]

The answer is gonna be D I believe

7 0

3 years ago

Nitrogen gas at 300 k and 200 kpa is throttled adiabatically to a pressure of 100 kpa if the change in kinetic energy is negligb

erastovalidia [21]

Answer:

The temperature of the Nitrogen after throttling is $T_2 = 300 \ K$

Explanation:

From the question we are told that

The temperature is $T_1 = 300 \ K$

The pressure is $P = 200 \ kPa = 200 * 10^{3} \ Pa$

The pressure after being $P_1 = 100 \ kPa = 100 * 10^{3} \ Pa$

Generally from the first law of thermodynamics we have that

$Q - W = \Delta U + \Delta K$

Here $\Delta U$ is the change internal energy which is mathematically represented as

$\Delta U = C_p (T_2 - T_1)$

Here $C_p$ is the specific heat of the gas at constant pressure

$\Delta K$ is the change kinetic energy which is negligible

Q is the thermal energy which is Zero for an adiabatic process

W is the work done and the value is zero given that the gas was throttled adiabatically

$0= \Delta U +0$

=> $\Delta U = 0$

=> $(T_2 - 300) = 0$

=> $T_2 = 300 \ K$

3 0

3 years ago

Would you classify silicone as a element or compound

bearhunter [10]

The answer is element

5 0

3 years ago

During studies of the reaction below,

Leni [432]

Answer: The percent yield of the nitrogen gas is 11.53 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NO:

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol$

For $N_2O_4$ :

Given mass of $N_2O_4$ = 102.1 g

Molar mass of $N_2O_4$ = 92 g/mol

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol$

For the given chemical reactions:

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$ ......(2)

$N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)$ .......(3)

Calculating the experimental yield of nitrogen gas:

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of $N_2O_4$

So, 0.383 moles of NO will be produced from = $\frac{2}{6}\times 0.383=0.128mol$ of $N_2O_4$

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.128 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.128=0.384mol$ of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g$

Calculating the theoretical yield of nitrogen gas:

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 1.11 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 1.11=3.33mol$ of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g$

To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%$

Hence, the percent yield of the nitrogen gas is 11.53 %.

7 0

3 years ago

Please help me with this net ionic chem question. I’ll mark you brainiest if you know it.

rjkz [21]

Answer:

B: 3k+ + 3OH¯ + Fe3+ + 3KOH = Fe3(H) + + 3NO3¯

Explanation:

Potassium hydroxide (KOH) reacts with iron (III).

The normal balanced equation is;

Fe(NO3)3 + 3KOH = Fe(OH)3 + 3KNO3

Now the ionic equation will be;

3k+ + 3OH¯ + Fe3+ + 3KOH = Fe3(H) + + 3NO3¯

8 0

3 years ago