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Ne4ueva [31]
3 years ago
14

Which statement is false

Chemistry
1 answer:
ella [17]3 years ago
8 0
Answer is D.

Hope it helped you.
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1.name the gas produced by the reaction between calcium carbonate and dilute hydrochloric acid?
GuDViN [60]
I dont know anything about this since i am not in college or high school i am in Middle school but I hope I will learn it
4 0
4 years ago
B. The equilibrium constant for the reaction 2H2(g) + S2(g) 2H2S(g) is Keq = . (3 points)
eimsori [14]

Answer:

i. Keq=4157.99.

ii. More hydrogen sulfide will be produced.

Explanation:

Hello,

i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:

Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99

ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.

Best regards.

4 0
3 years ago
The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc
jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br_2(g

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}

Given: -\frac{d[HBr]}{dt}]=0.301

Putting in the values we get:

\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}

+\frac{1d[Br_2]}{dt}=0.151

Thus the rate of appearance of Br_2 is 0.151

8 0
3 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
¿How do the products of the reaction to the phenol red test and the splint test? Please help me it's for today.! :((​
laila [671]

Answer:

The answer in this question is show you made Sodium Hydroxide and Hydrogen Gas.In order to do the products of the reaction relate to the phenol red test and the splint test you need to show that you made Sodium Hydroxide and Hydrogen Gas. Show that you made Sodium Hydroxide and Hydrogen Gas so that the products of the reaction relate to the phenol red test and the splint test.

5 0
3 years ago
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