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2 years ago

Please help with 2 chem questions thank you!

Chemistry

2 answers:

xeze [42]2 years ago

8 0

The correct answers are A and C.

andrew11 [14]2 years ago

6 0

The 1st one is A

The second one is I am 90% sure is A as well

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Is gravity a matter??

algol13

Answer:

No, gravity isn't matter

Explanation:

Gravity is a <u>force</u> that attracts matter towards the center of a physical body with mass.

3 0

2 years ago

In which case would recalibrating a thermometer be an important next step in an experiment dealing with boiling points?

Sonbull [250]

I believe the answer would be D

7 0

2 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

2 years ago

How many grams of hydrogen are needed to produce 1.80 g of water

JulsSmile [24]

Your answer is going to be 200g

7 0

2 years ago

Estimate the molar mass of a gas that effuses at 1.80 times the effusion rate of carbon dioxide. answer in units of g/mol.

Effectus [21]

From Grahams Law the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure.
Therefore; R1/R2 = √mm2/√mm1
The molecular mass of Carbon dioxide is 44 g
Hence; 1.8 = √(44/x
3.24 = 44/x
x = 44/3.24
= 13.58
Therefore, the molar mass of the other gas is 13.58 g/mol

3 0

2 years ago