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Ira Lisetskai [31]
3 years ago
10

The area of the 48 contiguous states is 3.02 x 106 miles2. Assume that these states are completely flat, no mountains and no val

leys. What mass of water would cover these states with a rainfall of two inches?
Chemistry
1 answer:
valentinak56 [21]3 years ago
7 0

Answer:

1 mile = 1,609 meters

3.02E6 mi^2 x (1609^2m^2/mi^2) = 7.818E12 m^2

1 in= 2.54 cm => 2 in = 5.08cm = .0508m

7.818E12 m^2 x .0508m = 3.97E11 m^3

1 cubic meter = 1,000,000 cubic centimeters

3.97E11 m^3 x (1,000,000 cm^3 / m^3) = 3.97E17 cm^3

1000 cm^3 = 1L

3.97E17 cm^3 x (1L / 1000cm^3) = 3.97E14 L  

or 3.97 x 10^14 L

convert to grams since 1L is 1000g

(3.97x10^14) x1000g which is 3.974x10^ 17g which is the final ans

Explanation:

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Which of these periods contain elements with electrons in s, p, d, and forbitals?
vodka [1.7K]

Answer:

I learnt k,l,m

Explanation:

so 1-3periods have klm and after that klmn

5 0
3 years ago
Read 2 more answers
Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.
zubka84 [21]

<u>Answer:</u> The value of \Delta G^o of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)

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

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

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

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]

We are given:

\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol

  • The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]

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

\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]

We are given:

\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}

where,

\Delta H^o_{rxn} = standard enthalpy change of the reaction =-67200 J/mol

\Delta S^o_{rxn} = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol

Hence, the value of \Delta G^o of the reaction is 28.38 kJ/mol

7 0
3 years ago
A sample of argon gas has a volume of 73 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the fina
NeX [460]

Answer:

V₂ → 106.6 mL

Explanation:

We apply the Ideal Gases Law to solve the problem. For the two situations:

P . V = n . R . T

Moles are still the same so → P. V / R. T = n

As R is a constant, the formula to solve this is: P . V / T

P₁ . V₁ / T₁ = P₂ .V₂ / T₂   Let's replace data:

(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C

((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂

58.66 mL.atm = 0.55 atm . V₂

58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL

3 0
3 years ago
Consider the cell notation Cd(s) | Cd(NO3)2(aq) || AgNO3(aq) | Ag(s) with standard reduction potentials for Cd2+ and Ag+ as -0.4
MakcuM [25]

The value of the Gibbs free energy shows us that the reaction is spontaneous.

<h3>What is the Gibbs free energy?</h3>

The Gibbs free energy is a quantity that helps us to be able to determine the spontaneity of a reaction.

In order to obtain the Gibbs free energy, we must obtain the Ecell as follows; 0.799V - (-0.402V) = 1.201 V

Now;

△G = -nFEcell

△G = -(2 * 96500 * 1.201)

△G = -232kJ/mol

Learn more about free energy:brainly.com/question/15319033?

#SPJ1

8 0
1 year ago
The rate constant for this first‑order reaction is 0.550 s−10.550 s−1 at 400 ∘C.400 ∘C. A⟶products A⟶products How long, in secon
Alisiya [41]

Answer:

t=2.08s

Explanation:

Hello,

In this case, for first order reactions, we can use the following integrated rate law:

ln(\frac{[A]}{[A]_0} )=kt

Thus, we compute the time as shown below:

t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s

Best regards.

5 0
3 years ago
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