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Greeley [361]
3 years ago
11

The specific heat of a substance is the energy required to produce a certain change in _____________.

Chemistry
2 answers:
kumpel [21]3 years ago
8 0

Answer:

d

Explanation:

it is the temperature change

34kurt3 years ago
8 0

Answer:

<em>The answer is D. </em>

Explanation:

<em>This is because it is temperature change.</em>

<em>Hope This Helps!</em>

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What kind of oxide is formed when a piece of sodium is dropped in the water
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Answer:

Sodium oxide is the product

Explanation:

4Na+O2->2Na2O

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Chemical A and Chemical B react in an exothermic reaction. What can be known about what will happen when Chemical A and Chemical
Fynjy0 [20]
In exothermic reactions, there is a release heat and the replacement of weak bonds with stronger ones.
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How do we think that Mars lost atmospheric gas?
grin007 [14]

the rocky planet was exposed to harsh and forceful solar winds, which blew the Martian atmosphere away.

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7 0
2 years ago
Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy
gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol    

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure ,  Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$  ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

  $Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

    $2P=PE^{6.4 \times 10^{-4}P}$

   $2=E^{6.4 \times 10^{-4}P}$

  $\ln 2 = 6.4 \times 10^{-4}P$

  ∴  P = 1083.04 atm

c). dG = Vdp -S dt  at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp $

            $=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

           $=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol  

7 0
3 years ago
1. How do metals and non-metals react with acids? Write and explain the chemical equation for the reaction of magnesium with sul
Nata [24]

Answer:

1. How do metals and non-metals react with acids?

Ans : Non metals does not react with acids while metals react with acids and produce hydrogen gas that burns with a 'pop'sound.

2. Write and explain the chemical equation for the reaction of magnesium with sulphuric acid and aluminium with hydrochloric acid.

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Mg(s) + H2SO4 (aq) MgSO 4(aq) +H2 (g)

Aluminium + Hydrochloric acid = Hydrogen + Aluminium chloride

2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)

6 0
3 years ago
Read 2 more answers
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