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Vinil7 [7]
3 years ago
13

Predict what must happen when a beaker containing a solution of lithium chloride is heated. check all that apply. view available

hint(s) check all that apply. energy will be transferred into the system. the δh3 value will increase. the δhsoln value will decrease. the reaction will become endothermic.
Chemistry
2 answers:
weqwewe [10]3 years ago
6 0

Answer:  energy will be transferred into the system and the reaction will become endothermic.

Explanation:   When the solution of Lithium chloride is heated that means the system is solution of lithium chloride and by heating, we are providing the sufficient amount of energy to the system. Thus in other words, energy will be transferred into the system.

Moreover, the system now will represent an example of endothermic reaction as the energy is being provided to the system and the process of dissolution is basically an endothermic reaction in which the heat is being transferred to the system.

slava [35]3 years ago
5 0

The correct answer is that the energy will be transferred into the system and as well as the solubility of lithium chloride will likely increase. It Is because LiCl is soluble when is dispersed in hot water that is why the solubility of lithium chloride will increase and that the heating solution will likely transfer the heat energy in the system in which the energy will be transferred in the system is correct.

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An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At
anygoal [31]

Answer: [N2]₀ = 10M and [H2]₀ = 11M

Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:

I is initial amount;

C is change in concentration;

E is for equilibrium concentration;

For the mixture,

       N2                       3H2                2NH3

I      [N2]₀                     [H2]₀                  0

C     - x                          -3x                 +2x

E     [N2]₀ - x =8      [H2]₀ - 3x =5       2x =4

With the product, we can find "x":

2x=4

x=2M

With x=2, find the concentrations:

[N2]₀ - x = 8

[N2]₀ = 10M

[H2]₀ - 3x = 5

[H2]₀ = 11M

The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.

8 0
3 years ago
Read 2 more answers
An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is
Juli2301 [7.4K]

Answer:

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}  

Explanation:

The gas ideal law is  

PV= nRT (equation 1)

Where:

P = pressure  

R = gas constant  

T = temperature  

n= moles of substance  

V = volume  

Working with equation 1 we can get  

n =\frac{PV}{RT}

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

\frac{m}{mw} =\frac{PV}{RT}  or  

m =\frac{P*V*mw}{R*T}   (equation 2)

The cylindrical container has a constant pressure p  

The volume is the volume of a cylinder this is

V =(pi)*r^{2}*h

Where:

r = radius  

h = height  

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is  V =(pi)*r^{2}*h

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:  

T =\frac{T_{1} + T_{O}}{2}  

Replacing these values in the equation 2 we get:

m =\frac{P*V*mw}{R*T}   (equation 2)

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}    

8 0
3 years ago
A gardener waters the plants in a garden using a garden hose during a drought. The gardener observes that most of the water give
Anna71 [15]

Answer:

Break up the soil to increase the number of pores should be your answer.

Explanation:

If you do that it will increase the amount of water in the plant.

8 0
3 years ago
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Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.
Sholpan [36]

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. n_{CuO}= \frac{4}{80}= 0,05 (mol)

⇒n_{CuCl2}= n_{CuO}=0,05 mol

⇒m_{CuCl2}= 0,05×135=6,75 (g)

c. n_{HCl}=2× n_{CuO}=0,1 (mol)

⇒m_{HCl}= 0,1×36,5= 3,65 (g)

⇒m_{dd HCl}= \frac{m_{HCl}}{10}×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2=\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }×100=\frac{6,75}{36,5+4} ×100≈ 16,67%

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3 years ago
Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
Vladimir79 [104]

Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

7 0
3 years ago
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