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aniked [119]
3 years ago
14

Choose all the answers that apply. Which of the following will decrease reaction rate? increase in pressure. decrease temperatur

e. increase concentration. decrease concentration
Chemistry
2 answers:
zloy xaker [14]3 years ago
4 0
Decrease temperature and decrease of concentration will decrease reaction rate, because less molecules will be able to react.
Oksana_A [137]3 years ago
4 0

Answer:

Decrease temperature and decrease of concentration will decrease reaction rate, because less molecules will be able to react.

Explanation:

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Chemistry what is stoichiometry and the different formulas
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Compared to a 0.1 M aqueous solution of NaCl, a 0.8 M aqueous solution of NaCl has a
Fudgin [204]
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3 0
4 years ago
All atoms that react
gregori [183]
Need more to the question
5 0
3 years ago
How does the brønsted-Lowry theory define acids and bases?
Leviafan [203]

They define acids as proton donors, and bases as proton acceptors

If you were to have:

HNO3 + H2O -> H3O+. + NO3-

You can see that the nitric acid (HNO3) gave a hydrogen ion which has 1 proton, 0 neutrons and 0 electrons to the water so we just say that it gave a proton.

Now let's see a base

NH3 + H2O -> NH4+ + OH-

Now, you can see that the ammonia (NH3) gained a hydrogen ion (proton) from the water to become ammonium(NH4). which means it accepted a proton

That's basically it. Feel free to ask if you have any further questions

8 0
3 years ago
Read 2 more answers
Two trials are run, using excess water. In the first trial, 7.8 g of Na2O2(s) (molar mass 78 g/mol) is mixed with 3.2 g of S(s).
Tema [17]

Answer:

The answer is "Option C".

Explanation:

Given equation:

2Na_20_2 (s)+S(s)+2H_2O \longrightarrow  4NaOH(aq)+SO_2(aq)

\to \Delta H^{\circ}_{rxn} (298\ K) = -610 \frac{kJ}{mol}

\to Na_2O_2  \ Mass = 7.8 \ g\\\\ \to  Na_2O_2 \ Molar \ mass = 78 \frac{g}{mol}

Na_2O_2 Has been the reactant which is limited since the two experiments are equal toNa_2O_2 for relationship between stress amounts.

Na_2O_2, n =\frac{Mass of Na_2O_2}{Molar mass of Na_2 O_2}=\frac{7.8 \ g}{78 \frac{g}{mol}} =0.1 \ mol \\\\q=\Delta H^{\circ}_{rxn} \times n = \frac{ -610 \ kJ}{ 2 \ mol \ Na_2 O_2} \times 0.1 \ mol  \ Na_2O_2= 30.5 \ KJ\\\\

Limiting reactant =Na_2O_2

q=30.5 \ kJ \approx 30 \ kJ

8 0
3 years ago
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