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allochka39001 [22]
3 years ago
14

Which electrons participate in chemical bonding?

Chemistry
2 answers:
OverLord2011 [107]3 years ago
7 0
B.) Valence Electrons. The nucleus of the valence electrons attracts and pulls atoms together.
VMariaS [17]3 years ago
4 0

Answer:

B

Explanation:

I took the test

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A sample of sulfur hexafluoride gas occupies a volume of 5.10 L at 198 ºC. Assuming that the pressure remains constant, what tem
Ludmilka [50]

Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

Temperature = 198 °C = 471 K

The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L

⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K

⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

5.10 L / 471 K = 2.50 L / T2

T2 = 2.50 L / (5.10 L / 471 K)

T2 = 230.9 K = -42.1

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

8 0
3 years ago
The rate constant for a certain reaction is k = 4.70×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the
Lynna [10]

Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is  a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

-\frac{d[A]}{dt} =kt

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

-\frac{d[A]}{dt} =kt

\Rightarrow -\frac{d[A]}{[A]}=kdt

Integrating both sides

\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀]  = k.0 + c

⇒c= -ln[A₀]  

Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]=  0.375

⇒[A] = 0.686

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

5 0
3 years ago
Logan combines hydrogen chloride (hcl) and calcium hydroxide ca(oh)2. as a result, a reaction occurs and new substances are form
scoundrel [369]
B. Calcium chloride and water.
8 0
3 years ago
Read 2 more answers
A 100.00 ml of volume of 0.500 M HCl was mixed with 100.00 ml of 0.500 M KOH in a constant pressure calorimeter. The initial tem
Virty [35]

Answer:

The heat of the reaction = -1985 J = -1.985 kJ

The enthalpy of the reaction is -39.7 kJ/ mol

Explanation:

<u>Step 1: </u>Data given

Volume of HCl = 100 mL the heat of the reaction = 0.1 L

Molarity of HCl solution = 0.500 M

Volume of KOH = 100 mL = 0.1 L

Molarity of KOH solution = 0.500 M

Initial temperature = 23.0 °C

Final temperature = 25.5 °C

Specific heat of the solution = 3.97 J/°C *g

Density of the solution = 1g/ mL

<u>Step 2: </u>Calculate heat

Q = m*c*ΔT

with m = the mass of both solution : 100g + 100 g ( since density = 1g/mL) = 200 g

c = the specific heat = 3.97 J/°C*g

ΔT  = T2 -T1 = 25.5 = 23 = 2.5 °C

Q = 200g *3.97 J/°C*g * 2.5°C = 1985 J  (= -1985 J because it's exothermic)

<u />

<u>Step 3:</u> Calculate the number of moles

Number of moles = Molarity * Volume

Number of moles = 0.5 * 0.1 L = 0.05 moles

(Moles of the acid are equal to the moles of water produced.

Moles of solution = 0.05 moles)

<u>Step 4: </u>Calculate the enthalpy of the reaction

ΔH = heat change /Number of moles

    = -1.985 kJ/ 0.05 moles

   =- 39700 J/mol = -39.7 kJ/ mol

The enthalpy of the reaction is -39.7 kJ/ mol

The enthalpy of the reaction is negative, this means the reaction is exothermic ( which means the final temperature is higher than the initial temperature.)

7 0
3 years ago
Help please i am very dumb
UkoKoshka [18]

Answer:

1. 25%

2. 1.25

3. 1

Explanation:

Be sure to look at the x and y axis to answer these questions. All you need to do is look at the graph.

- Hope that helps! Please let me know if you need further explanation.

7 0
3 years ago
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