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Anni [7]
3 years ago
12

Chase buys a helium balloon for a friend's birthday. As he walks out of the store, it starts snowing.

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

I think its A sry if im wrong

Explanation:

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For a particular isomer of C8H18, The following reaction produces 5093.7 KJ of heat per mole of C8H18(g) consumed under standere
Aleks04 [339]
02(g) = 0 kj/mol 
<span>CO2 (g) = -393.5 kj/mol </span>
<span>H20(g) = -241.8 kj/mol </span>

<span>H total = -5094 kJ
</span>5094kJ = [8(-393.5) + 9(-241.8)] - [X + 12.5(0)] 
<span>-5094 kJ = [-3148 + (-2176.2)] - [x + 0] </span>
<span>-5094 kJ = -5324.2 - x </span>

<span>add -5324.2 to -5094 </span>
<span>to get +230.2 = -x </span>

<span>move the negative to the other side </span>
<span>and you get -230 kj/mol</span>
7 0
3 years ago
How many atoms are in 1.50 moles of Hg?
Oxana [17]

Answer:

Atoms=9.033*10^23 atoms

Explanation:

Atoms=no.of miles*Avogadro's no.

Atoms=1.5*6.022*10^23

Atoms=9.033*10^23 atoms

3 0
3 years ago
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On which of the following factors does the amount of energy absorbed by an endothermic reaction depend?
jolli1 [7]

Answer: D!! ( difference in the potential energy of the reactants and products )

Explanation:

i have the same test

4 0
3 years ago
Figuring out how to make a better type of plastic is which type of research
quester [9]

Answer:applied chemistry

5 0
3 years ago
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How many moles of N2 in 57.1 g of N2?
SpyIntel [72]

We are given –

  • Mass of \bf N_2 is 57.1 g and we are asked to find number of moles present in 57.1 g of \bf N_2

\qquad\pink{\bf\longrightarrow  { Molar \:mass \:of \: N_2:-}  }

\qquad\bf  \twoheadrightarrow 14\times 2

\qquad\bf \twoheadrightarrow   28

\qquad____________________

Now,Let's calculate the number of moles present in 57.1 g of \bf N_2

\qquad\purple{\bf\longrightarrow  { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}

\qquad\bf   \twoheadrightarrow \dfrac{57.1}{28}

\qquad\bf  \twoheadrightarrow 2.04\: moles

__________________________________

7 0
2 years ago
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