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Eduardwww [97]
2 years ago
7

BRAINLIEST!

Chemistry
2 answers:
vodomira [7]2 years ago
8 0

Answer:

The ions in metal behavior by repelled by electro

Harrizon [31]2 years ago
3 0

Answer:

attracted towards each other

Explanation:

hope u like the ans

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The structure of 3-propylpentane is best described as
victus00 [196]
The correct answer is b
7 0
3 years ago
Read 2 more answers
200.0 mL OF A 5.10 M solution are diluted to 500.0 mL. What is the concentration of the resulting solution? *
erastova [34]

Answer:

Concentration of the resulting solution = 2.04 M

Explanation:

<u>Data:</u>

M1 = 5.10 M

V1 = 200.0 mL

V2 = 500.0 mL

M2 = ?

By modifying the volume of solution, keeping the amount of solute constant, the concentration changes. To perform the calculations, the equation will be:

M1*V1=M2*V2

Where M1 is the initial concentration of the solution, M2 the final concentration and V is the value of the volumes of the initial and final solution.

Clearing the value of M2 from the equation and replacing the values ​​we have:

M2=\frac{M1*V1}{V2} =\frac{5.10M*200mL}{500mL} =2.04M

3 0
3 years ago
Read 2 more answers
Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

8 0
3 years ago
The pressure on a 200 milliliter sample of CO2 (g) at constant temperature is increased from
balu736 [363]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 100 ml.</em></u>

Explanation:

 Given:

Initial pressure (P_{1}) = 600 mm of Hg

Final pressure (P_{2}) = 1200 mm of Hg

Initial volume (V_{1}) = 200 ml      

To find:

Final volume (V_{2})

We know;

According to the ideal gas equation,

    P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

 From the above mentioned equation,

        P × V = constant

\frac{P_{1} }{P_{2} } = \frac{V_{1} }{V_{2} }

Where,

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1})  represents the initial volume of the gas

(V_{2})  represents the final volume of the gas

So;

\frac{600}{1200} = \frac{V_{2} }{200}    

V_{2} = 100 ml

<u><em>Therefore the final volume of the gas is 100 ml.</em></u>                                                                                                                                                                              

5 0
3 years ago
Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a
NISA [10]

Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is  ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0
3 years ago
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