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kozerog [31]
3 years ago
7

Can someone PLEASE help me with this problem, I’ve tried it multiple times and keep getting it wrong :(

Chemistry
1 answer:
atroni [7]3 years ago
5 0

Answer:

Explanation:

Okay.

From convervation of mass in chemical processes, all the reactants combined have the same mass as all the products.

m react = m prod

m Hg + m O 2 = m HgO + m O 2

114.0g+12.8g=123.1g+ m O 2

Solving for mO2 = 3.70 g

This is what I know.

I do hope I helped you! :)

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In the periodic table, how are elements in the same group related to each other? A. They have the same atomic mass. B. They have
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Elements in the same group tend to have very similar properties (D). This is due to the number of valence electrons each group has.
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3 years ago
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5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure<br> water?
Margaret [11]

Answer:

\boxed {\boxed {\sf 2.8 \ m }}

Explanation:

The formula for molality is:

m=\frac{moles \ of \ solute}{kg \ of \ solvent}

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

moles= 0.210 \ mol \\kilograms = 0.075 \ kg

Substitute the values into the formula.

m= \frac{ 0.210 \ mol }{0.075 \ kg}

Divide.

m= 2.8 \ mol/kg= 2.8 \ m

The molality is <u>2.8 moles per kilogram</u>

5 0
3 years ago
Suggest two observations that would be made when rubidium is added to cold water.
tino4ka555 [31]

Answer:

1. Rubidium metal reacts very rapidly with water to form a colorless basic solution of rubidium hydroxide (RbOH) and hydrogen gas (H2).

2. Rubidium sinks because it is less dense than water. It reacts violently and immediately, with everything leaving the container. Rubidium hydroxide solution and hydrogen are formed.

6 0
2 years ago
A workman uses a moveable pulley to lift a heavy load of bricks. What is the input force, if the output force is 150 N?
NeTakaya

Answer:

the input force would be 75 N

Explanation:

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7 0
3 years ago
A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

4 0
3 years ago
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