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stealth61 [152]
2 years ago
12

Plssssssssssssssssssssss helppppppppppppppppppppppppppppppppppppp

Chemistry
1 answer:
andriy [413]2 years ago
3 0

Answer:

protons and neutrons have different charges but they have approximately the same mass

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Balance the redox reaction equation (occurring in acidic solution) and choose the correct coefficients for each reactant and pro
algol13
Balanced chemical reaction: 
PbO₂<span>(s) + Sn(s)+ 4H</span>⁺(aq) → Pb²⁺(aq) + Sn²⁺(aq) + 2H₂O<span>(l).
Oxidation half-reaction: Sn </span>→ Sn²⁺ + 2e⁻.<span>
Reduction half-reaction: PbO</span>₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O.
Net reaction: Sn + PbO₂ + 4H⁺ + 2e⁻ → Sn²⁺ + 2e⁻ + Pb²⁺ + 2H₂O.
Oxidation is increase of oxidation number, reduction is decrease of oxidation number.
6 0
4 years ago
The reaction with hydrogen with oxygen produces water: 2 H2 + O2 ---&gt; 2H20. How many moles of O2 are required to react with 2
il63 [147K]

o2 is made and hydrogen help make it

4 0
3 years ago
H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles
abruzzese [7]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

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

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

For the given chemical reaction:

H_2(g)+F_2(g)\rightarrow 2HF(g)

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

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]

We are given:

\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = \frac{-14.1}{1}\times 2.20=-31.02J/K

Hence, the value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

7 0
3 years ago
Select the correct answer.
Kipish [7]
The answer should be a
8 0
3 years ago
Read 2 more answers
If I have 4.4 moles of nitrous oxide (laughing gas) that is kept at 32.9 °C in a container under 1.6 atm, what is the volume of
r-ruslan [8.4K]

Answer:

Explanation:

Using the ideal gas equation as follows:

PV = nRT

Where:

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Based on the information provided in this question;

P = 1.6atm

n = 4.4 moles

R = 0.0821 L*atm/mol*K

T = 32.9°C = 32.9 + 273 = 305.9K

V = ?

4 0
3 years ago
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