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Ann [662]
2 years ago
9

Which characteristic of planets depends upon its distance from the Sun?

Chemistry
2 answers:
Svetradugi [14.3K]2 years ago
8 0

Answer:

I think its the length of its year

Artemon [7]2 years ago
3 0

Answer:

the length of its year

Explanation:

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A.
Andru [333]

Answer:

Where is the question?

7 0
2 years ago
Calculate the standard cell potential at 25 ∘c for the reaction x(s)+2y+(aq)→x2+(aq)+2y(s) where δh∘ = -793 kj and δs∘ = -319 j/
QveST [7]
First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
        = - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts   
 
5 0
3 years ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
Which statement is true? Energy cannot be created or destroyed. Energy can be destroyed but not created. Energy can be created b
vaieri [72.5K]
Energy can not be created and cannot be destroyed
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6 0
3 years ago
Read 2 more answers
Calculate the maximum wavelength of light that will cause the photoelectric
Drupady [299]

Answer:

Explanation:

Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J

So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .

energy of photon of wavelength λ = hc / λ

where h = 6.67  x 10⁻³⁴

c = 3 x 10⁸

Putting the values in the equation above

6.67  x 10⁻³⁴  x  3 x 10⁸ / λ =  3.67 X 10⁻¹⁹

λ  = 6.67  x 10⁻³⁴  x  3 x 10⁸ /  3.67 X 10⁻¹⁹

= 5.452 x 10⁻⁷

= 5452 x 10⁻¹⁰ m

= 5452 A .

6 0
3 years ago
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