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lara31 [8.8K]
3 years ago
15

Which of the following represents an endothermic reaction?

Chemistry
1 answer:
Softa [21]3 years ago
8 0

Answer:

<h2><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em></h2>

The second one would be one of the application of an endothermic reaction as because endothermic reaction always have the heat in the process to mix the reactants and form the products as here in a burning wood when the wood is supplied with heat then it change into products as ashes and smoke fumes So here due to the heat is involved to form the products thats the reaction is endothermic reaction.

Hope it helps

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How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
Which of the following would be considered an Exothermic Reaction?*
Digiron [165]

Answer:

Burning wood

Explanation:

the fire releases heat into the air from the burning wood

6 0
3 years ago
What information does a phase diagram give
Finger [1]
The amount of energy required to change the temperature or phase of a reactant
7 0
3 years ago
ABSOLOTLY NO ROBOTS I AM TIERD OF THEM.
NikAS [45]

Answer:

You have a trait (plant color) and you have two possible outcomes for it (purple and white). You know that some alleles for white were present in one of the parent plants, so you know they must be present in the offspring too. But they're expression is covered up by the purple alleles from the other plant, so you know that purple is dominant to white.

8 0
3 years ago
Pyridinium is a weak acid having a pKa of 5.2. How much pyridine (the conjugate base of pyridinium) must be added to an aqueous
Whitepunk [10]

Answer:

Amount of pyridine required = 0.0316 M

Explanation:

pH of a buffer solution is calculated by using Henderson - Hasselbalch equation.

pH=pK_a+log\frac{[Conjugate\ base]}{[weak\ acid]}

Pyridinium is a weak acid and in the presence of its conjugate base, it acts as buffer.

Henderson - Hasselbalch equation for pyridine/pyridinium buffer is as follows:

pH=pK_a+log\frac{[Py]}{PyH^+]}

pH = 4.7

pK_a=5.2

PyH^+ (Pyridinium)=0.100 M

Substitute the values in the formula

pH=pK_a+log\frac{[Py]}{PyH^+]}\\4.7=5.2 log\frac{[Py]}{0.100}

4.7-5.2=log\frac{[Py]}{0.100} \\-0.5=log\frac{[Py]}{0.100}\\\frac{[Py]}{0.100}=antilog -0.5\\\frac{[Py]}{0.100}=0.316

\frac{[Py]}{0.100} =0.316

[Py]=0.0316\ M

Amount of pyridine required = 0.0316 M

7 0
3 years ago
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