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PolarNik [594]
2 years ago
8

Observe: click reset. select the gas collection setup. chemists use this apparatus to collect any gases produced in the reaction

. from the reaction flask, gases travel through a long tube and into a cylinder of water. as gases bubble into the cylinder, the water is displaced (removed) until the cylinder is filled with gas. click play and observe the cylinder. was any gas produced in the reaction? [ ]
Chemistry
1 answer:
Oxana [17]2 years ago
6 0

Hello. Although it is not possible to see the video to which the question refers, we can see, in the question, that water is removed from the cylinder as a gas is added to the same cylinder. So we can see that a gas was produced in the reaction, through the displacement of water, that is, if the water starts to be removed from the cylinder, it is because the reaction is producing a gas. If the gas is not produced, water will not come out of the cylinder.

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A sample of gas contains 3.0L of nitrogen at 320kPa. What volume would be necessary to decrease the pressure at 110kPa
Phoenix [80]

Answer:

V_{2} = 8.92 L

Explanation:

We have the equation for ideal gas expressed as:

PV=nRT

Being:

P = Pressure

V = Volume

n = molar number

R = Universal gas constant

T = Temperature

From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:

P_{1} V_{1} =nRT

320Kpa*0.003m^{3} =nRT

1000L = 1m^{3}

So

nRT= 0.96

Now, as for the final condition:

P_{2}V_{2}=nRT

P_{2} V_{2} =0.96

clearingV_{2}

V_{2} =\frac{0.96}{P_{2} }

V_{2} =0.00872m_{3}

V_{2} = 8.92 L

6 0
3 years ago
Please help :( this is for science.
elena55 [62]
Solar eclipse only i believe
4 0
2 years ago
Balance the following reactions using either the oxidation number method or the half-reaction method. Identify which element has
Marina86 [1]

6Ce^{4+} + I^- + 6OH^- → 6Ce^{3+} + IO_3^- + 3H_2O is the balanced chemical equation.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical reaction is an equation that has equal numbers of each type of atom on both sides of the arrow.

Half-reaction method:

Unbalanced chemical equation:

Ce^{4+} + I^-→ Ce^{3+} + IO^{3-}

Oxidation half-reaction:

I^-+ 6OH^- - 6e- → IO^{3-} + 3H_2O

Reduction half-reaction:

Ce4^+ + e^- → Ce^{3+}

Balanced chemical equation:

6Ce^{4+} + I^- + 6OH^-→ 6Ce^{3+} + IO^{3-} + 3H_2O

Oxidation number method:

Unbalanced chemical equation:

Ce^{4+} + I^-→ Ce^{3+} + IO^{3-}

I^{-1} -6e^-→ I^{+5}

Ce^{4+} + e^- → Ce^{3+}

Balanced chemical equation:

6Ce^{4+} + I^{-1} → 6Ce^{3+} + I^{+5}

or

6Ce^{4+} + I^- + 6OH^-→ 6Ce^{3+} + IO_3^- + 3H_2O

Hence, 6Ce^{4+} + I^- + 6OH^-→ 6Ce^{3+} + IO_3^- + 3H_2O is the balanced chemical equation.

Learn more about the balanced chemical equation here:

https://brainly.in/question/46754758

#SPJ1

6 0
2 years ago
The reaction 2NO(g)+Cl2(g)→2NOCl(g) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate o
Luba_88 [7]

Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.

Explanation : Given,

\frac{d[NO]}{dt} =21 torr/min

The balanced chemical reaction is,

2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)

The rate of disappearance of NO = -\frac{1}{2}\frac{d[NO]}{dt}

The rate of disappearance of Cl_2 = -\frac{d[Cl_2]}{dt}

The rate of formation of NOCl = \frac{1}{2}\frac{d[NOCl]}{dt}

As we know that,

\frac{d[NO]}{dt} =21 torr/min

So,

-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}

\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min

And,

\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}

\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min

Now we have to calculate the rate change.

Rate change = Reactant rate - Product rate

Rate change = (21 + 10.5) - 21 = 10.5 torr/min

Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.

8 0
3 years ago
What is the reaction energy q of this reaction? use c2=931.5mev/u. express your answer in millions of electron volts to three si
Ann [662]
₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹
Mass of reactants = 235.04393 + 1.008665 = 236.052595 amu
Mass of products = 139.92144 + 93.91523 + 2* (1.008665) = 235.854000 amu
Mass defect Δ m = 236.052595 - 235.854000 = 0.198 amu
Reaction energy released Q = Δ m * 931.5
                                              = 0.198 * 931.5 = 185 MeV
5 0
3 years ago
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