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Nookie1986 [14]
2 years ago
6

PLEASE HELP ME ANSWER

Chemistry
1 answer:
maks197457 [2]2 years ago
5 0

Answer: the second one multiplication

Explanation:

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Which compounds are classified as Arrhenius acids?
Olenka [21]

Answer:The correct answer is option 4.

Explanation:

Arrhenius acids are those compounds which gives H^+ ions when dissolved in their aqueous solution.

HA(aq)\rightarrow H^++A^-

Arrhenius bases are those compounds which gives OH^- ions when dissolved in their aqueous solution.

BOH(aq)\rightarrow OH^-+B^+

HBr \& H_2SO_4 are Arrhenius acids because they form H^+ions in their respective aqueous solution.

HBr(aq)\rightarrow H^++Br^-

H_2SO_4(aq)\rightarrow 2H^++SO_{4}^{2-}

Hence, the correct answer is option 4.

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3 years ago
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Which best describes the motion of molecules in the gaseous state?
artcher [175]

Answer:

They spread apart to fill the enclosed object. In other words they move freely...

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5 0
2 years ago
Combustion reactions are exothermic. The heat of reaction for the combustion of 2-methylheptane, C8H18, is 1.306×103 kcal/mol. W
WARRIOR [948]

Answer:

11.45kcal/g

2.612 × 10³ kcal

Explanation:

When a compound burns (combustion) it produces carbon dioxide and water. The combustion of 2-methylheptane can be represented by the following balanced equation:

2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O

It releases  1.306 × 10³ kcal every 1 mol of C₈H₁₈ that is burned.

<em>What is the heat of combustion for 2-methylheptane in kcal/gram?</em>

We know that the molar mass of C₈H₁₈ is 114.0g/mol. Then, using proportions:

\frac{1.306 \times 10^{3}Kcal}{1mol} .\frac{1mol}{114.0g} =11.45kcal/g

<em>How much heat will be given off if molar quantities of 2-methylheptane react according to the following equation? 2 C₈H₁₈ + 25 O₂ ⇄ 16 CO₂ + 18 H₂O</em>

In this equation we have 2 moles of C₈H₁₈. So,

2mol \times\frac{1.306 \times 10^{3}kcal }{1mol} =2.612\times 10^{3}kcal

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2 years ago
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<span>PV=nRT= a universal constant For any condition P1V1/n1T1=R and P2V2/n2T2=R i.e P1V1/n1T1=P2V2/n2T2 Becomes V1/n1=V2/n2 rearranging and solving V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>
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Copper and Aluminium..........
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