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Inessa05 [86]
2 years ago
9

This is a change in an objects motion caused by an unbalanced force

Chemistry
2 answers:
OlgaM077 [116]2 years ago
4 0

Answer:

Pressure

Pressure is done when a force is applied to an object to course a motion

kati45 [8]2 years ago
3 0

Answer:

This is a change in an objects motion caused by an unbalanced force.

- Acceleration

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In a mixture of hydrogen and nitrogen gases, the mole fraction of nitrogen is 0.333. If the partial pressure of hydrogen in the
notka56 [123]

Answer:

P_T=112.4torr

Explanation:

Hello there!

In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:

P_{H_2}=x_{H_2}*P_T

And can be solved for the total pressure as follows:

P_T=\frac{P_{H_2}}{x_{H_2}}

However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:

x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667

Then, we can plug in to obtain the total pressure:

P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr

Regards!

4 0
2 years ago
One mole of ice at 0°C is added to two moles of water at 50°C under a constant external pressure of 1 atm. This process is carri
kotykmax [81]

Answer:

T2 = 29.79°C

Explanation:

Equliibrium signifies that heat loss = heat gained

Heat gained by Ice;

H = ML

Mass, M = Number of moles * Molar mass = 1 * 18 = 18g

l = 6.01 k J m o l = 334 J/g

C = 4.186 J/g

H =  18(334)

H = 6012

Heat lost by water

H = MCΔT

H = 18 * 4.186 * (50 - T2)

H = 3767.4 - 75.348T2

Since H = H, we have;

6012 = 3767.4 - 75.348T2

- 75.348T2 = 3767 - 6012

T2 = 2245 / 75.348

T2 = 29.79°C

8 0
2 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
2 years ago
1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to
Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0
2 years ago
Which term describes the information that a scientist gathers during an investigation? Data, hypothesis, observation, variable a
rosijanka [135]

Answer:hypothesis

Explanation:

4 0
3 years ago
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