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2 years ago

What variable represents specific heat in the equation Q = mcAT?

Chemistry

1 answer:

Elan Coil [88]2 years ago

4 0

Answer: A. The variable c

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Explanation:

Q = heat transferred

m = mass

c = specific heat

$\Delta T$ = delta T = change in temperature

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Why is it important to do further investigations to determine if a chemical change really occurred?

KatRina [158]

Answer:

Explanation:

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2 years ago

What two things combine with oxygen to form ground-level ozone?

STALIN [3.7K]

Answer:

c) NOx and VOC

oxides of nitrogen (NOx) and volatile organic compounds (VOC)

Explanation:

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2 years ago

The decomposition of hydrogen peroxide (H2O2(aq)) is proposed to follow a reaction mechanism of: Step 1: H2O2(aq) + Br–(aq)→H2O(

lys-0071 [83]

Answer:

Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without

Explanation:

Let us consider the equation below:

Step 1:

H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)

Step 2:

BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)

From the above equation, we can see that Br– is unchanged.

This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.

8 0

2 years ago

how do you know the state of a compound when writing out the equation. WILL MARK BRAINLIEST PLEASE HELP!!

tester [92]

Answer:

The state of matter of each compound or molecule is indicated in subscript next to the compound by an abbreviation in parentheses. For example, a compound in the gas state would be indicated by (g), solid (s), liquid (l), and aqueous (aq).

Explanation:

3 0

3 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

5 0

2 years ago