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dusya [7]
3 years ago
15

Please help! The answer has to have a decimal notation​

Chemistry
1 answer:
CaHeK987 [17]3 years ago
4 0

Answer:

.001 ? Try a notation calculator

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All radioactive nuclides undergo what
kotykmax [81]
They undergo nuclear fission.
6 0
3 years ago
In the combined gas law, if the volume is decreased, what happens to the temperature if pressure remains constant?
Flura [38]

The correct answer is A. In the combined gas law, if the volume is decreased and the pressure is constant, then the temperature decreases.

 <span>P1V1/ T1 = P2V2 / T2</span>

 <span>Assume the volume decrease by half; V2 = V1/2</span>

 <span>P1V1/ T1 = P2V1 /2 T2</span>

 <span>Cancelling terms,</span>

 <span>1/T1 = 1/2 T2</span>

T2 = T1/2

 <span>Thus, the temperature decreased.</span>

8 0
3 years ago
A gas system has an initial number of moles of 0.693 moles with the volume unknown. When the number of moles changes to 0.928 mo
lara [203]

Answer:

The initial volume in mL is 5959.2 mL

Explanation:

As the number of moles of a gas increases, the volume also increases.  Hence, number of moles and volumes are directly proportional i.e

n ∝ V

Where n is the number of moles and V is the volume

Then, n = cV

c is the proportionality constant

∴n/V = c

Hence n₁/V₁ = n₂/V₂

Where n₁ is the initial number of moles

V₁ is the initial volume

n₂ is the final number of moles

and V₂ is the final volume.

From the question,

n₁ = 0.693 moles

V₁ = ?

n₂ = 0.928 moles

V₂ = 7.98 L

Putting the values into the equation

n₁/V₁ = n₂/V₂

0.693 / V₁ = 0.928 / 7.98

Cross multiply

∴ 0.928V₁ = 0.693 × 7.98  

0.928V₁ = 5.53014

V₁ = 5.53014/0.928

V₁ = 5.9592 L

To convert to mL, multiply by 1000

∴ V₁ = 5.9592 × 1000 mL

V₁ = 5959.2 mL

Hence, the initial volume in mL is 5959.2 mL

5 0
3 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
3 years ago
Draw the R enantiomer of CFClBrI as a 3-D (wedge and dotted line) structure and as a Fischer projection (putting the I at the to
vazorg [7]
I don’t know This one is there a picture ?
4 0
3 years ago
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