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3 years ago

13

The ______ properties of a substance can be observed only when it undergoes a change to become an entirely different kind of sub

stance with different properties.
Enter the answer

1 answer:

Margaret [11]3 years ago

4 0

Answer:chemical

Explanation:

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What would happen if the cell cycle stopped in a multicellular organism

oksano4ka [1.4K]

The organism would no longer grow.

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3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

<u>The reaction </u>

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) <u>Limiting reagent</u>: $NO$

3) <u>Ammount of excess reagent</u>:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

How many control(s) are in an experiment

weqwewe [10]

You can have as many controls as necessary, But they must remain equal at all times in order to get the most accurate results

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3 years ago

The incredible catalytic power of enzymes can perhaps best be appreciated by imagining how challenging life would be without jus

ioda

Answer:

t = 7.58 * 10¹⁹ seconds

Explanation:

First order rate constant is given as,

k = (2.303 /t) log [A₀] /[Aₙ]

where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>

[A₀] = 615 calories;

[Aₙ] = 615 - 480 = 135 calories

k = 2.00 * 10⁻²⁰ sec⁻¹

substituting the values in the equation of the rate constant;

2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)

(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)

t = 2.303 / 3.037 * 10⁻²⁰

t = 7.58 * 10¹⁹ seconds

8 0

3 years ago

At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va

Bas_tet [7]

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

$P=X_{A}P_{A}+X_{B}P_{B}$

Here, $X_{A}$ is mole fraction of A, $X_{B}$ is mole fraction of B, $P_{A}$ is partial pressure of A and $P_{B}$ is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

$X_{A}+X_{B}=1$

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

$X_{octane}=1-X_{hexane}=1-0.58=0.42$

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

$P=X_{hexane}P_{hexane}+X_{octane}P_{octane}$

Putting the values,

$P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg$

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

4 0

3 years ago