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Luden [163]
3 years ago
10

The reaction taking place in the sun's core _______.

Chemistry
1 answer:
kogti [31]3 years ago
5 0
The answer is B

Explanation:The core of the sun is known as nuclear fusion and involves hydrogen nuclei combining together to form helium.
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Need help!!!!!!!!!!!!!
natulia [17]

Answer:

density

Explanation:

3 0
3 years ago
Discuss electrochemical principle about rusting of iron​
Levart [38]

Answer:

The electrochemical phenomenon of rusting of iron can be described as : At Anode: Fe (s) undergoes oxidation to releases electrons. Electrons released at anode move to another metal and reduce oxygen in presence of H+. It is available from H2CO3 formed from the dissolution of CO2 from air into water.

7 0
3 years ago
Describe what happens when a liquid is heated to its boiling point
kolezko [41]
Boiling<span> is the process by which a </span>liquid<span> turns into a vapor when it is </span>heated to its boiling point<span>. The change from a </span>liquid<span> phase to a gaseous phase occurs when the vapor pressure of the </span>liquid<span> is equal to the atmospheric pressure exerted on the</span><span>liquid</span>
7 0
3 years ago
Benzene contains only carbon and hydrogen and has a molar mass of 78.1 g/mol. Analysis shows the compound to be 7.44% hydrogen b
Sindrei [870]

Hydrogen - 7.44%

Carbon - (100-7.44)% = 92.56%


Lets take 100 g of benzene, then we have

Hydrogen - 7.44 g

Carbon - 92.56 g


n - number of moles

n(H) = 7.44g *1 mol/1.0g = 7.44 mol

n(C) = 92.56 g* 1mol/12.0 g ≈ 7.713 mol


n(C) : n(H) = 7.713 mol : 7.44 mol = 1:1

Empirical formula is CH.


M(CH) = (12.0+ 1.00) g/mol = 13.0 g/mol

M (benzene) = 78.1 g/mol

M (benzene)/M(CH)= 78.1 g/mol/13.0 g/mol = 6

So, molecular formula of benzene is C6H6.


6 0
3 years ago
The heat of vaporization of water at the normal boiling point, 373.2 K, is 40.66 kJ/mol. The molar heat capacity of liquid water
Tatiana [17]

Answer:

\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}

Explanation:

Hello!

In this case, according to the Kirchhoff's law for the enthalpy change, it is possible to compute the heat of vaporization at 300.2 K by considering the following thermodynamic route:

\Delta _{vap}H(300.2K)=Cp_{liq}(T_b-T\°)+\Delta _{vap}H\°+Cp_{vap}(T-T_B)

Whereas the first term stands for the effect of taking the liquid from 298.15 K to 373.15 K, the second term stands for the standard enthalpy of vaporization and the last term that of the vapor from the boiling point to 300.2 K; thus we plug in to obtain:

\Delta _{vap}H(300.2K)=75.37\frac{J}{mol*K} (373.2K-298.15K)+40,660\frac{J}{mol} +36.4\frac{J}{mol*K}(300.2K-373.2K)\\\\\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}

Best regards!

6 0
3 years ago
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