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makkiz [27]
3 years ago
10

Consider this reaction: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g).

Chemistry
1 answer:
iragen [17]3 years ago
5 0
Option are as follow,

A. temperature, concentration and surface area 
<span>B. temperature, and concentration only </span>
<span>C. concentration and surface area only </span>
<span>D. temperature and surface area only
</span>
Answer:
           Option-<span>A. Temperature, Concentration and Surface area
</span>
Explanation:

1) Increasing Temperature:
                                              Increase in temperature increases the Kinetic energy of molecules. This results in increase in the velocity and rate of collisions between reactants. Hence, greater the number of collisions between reactants per time greater will be the probability of formation of product per unit time.

2) Increasing Concentration
                                            Increase in concentration results in increase in number of particles of reactants per unit area, hence collision rate increases resulting in rate of reaction.

3) Increasing Surface Area
                                             Grinding of Zn results in the increase of surface area of Zinc. So greater the surface area greater is the exposure of Zinc metal to HCl molecules, hence the rate of formation of product increases.
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I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
3 years ago
For which of the following processes would you expect there to be an increase in entropy? Ag+(aq) + Cl-(aq) AgCl(s) H2O(g) H2O(l
Lyrx [107]
CaBr₂(s) → Ca⁺²(aq) + 2Br⁻ (aq)     ΔS>0
6 0
3 years ago
Read 2 more answers
139%
GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}

\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}

\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}

\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}

Based on no. of iron reacted,  

\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)

n = m/M

\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}

% composition ofFeCl_3  

=  (9.75 / 10.39)^{*} 100

= 94%

6 0
3 years ago
12-How many elements are in the mixture pictured below?<br> A-7<br> O B-4<br> C-3<br> D-2
vovikov84 [41]

Answer:

C

Explanation:

3

Because you don't count the dots. You count how many colors there are. Which are Blue, Red, and Black.

Hope this helped!

4 0
3 years ago
Read 2 more answers
**ASAP PLEASE**
Anna007 [38]

The answer to your question is C.)


8 0
3 years ago
Read 2 more answers
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