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3 years ago

NEED ANSWER, ASP! WILL MARK BRAINLIEST!!

Chemistry

1 answer:

Phoenix [80]3 years ago

7 0

Answer:

Answers are below

Explanation:

1. A

2. B

3. A

4. A

5. A

6. B

7. A

8. B

9. B

10. B

11. A, B

12. A

13. A

14. A,B

15. A

16. Not sure

17. Maybe both not sure

18. A

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A student increases the temperature of a 417cm³ balloon from 278k to 231k. Assuming constant pressure, what should the new volum

viva [34]

Charles law states that volume of gas is directly proportional to temperature at constant pressure
V/T = k
where V - volume , T - temperature and k - constant
$\frac{V1}{T1} = \frac{V2}{T2}$
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation

in the question it states that the temperature has been increased from 278 K to 231 K but it should actually be temperature is decreased from 278 K to 308 K
substituting the values in the equation
$\frac{417cm^{3} }{278K} = \frac{V}{308 K}$
V = 462 cm³
the answer should be D. 462 cm³

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3 years ago

How many protons dose p have

Softa [21]

Answer:

i have to see the question

Explanation:

5 0

3 years ago

Read 2 more answers

Please explain to me how to answer no.2 questions

bulgar [2K]

Answer: -

IE 1 for X = 801

Here X is told to be in the third period.

So n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a) Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b) So the atomic radius of X is lesser than that of Y.

c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.

Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.

4 0

3 years ago

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

= 25kg

Mass ratio of oil to solution $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75= E₁ × N₁

75 = E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

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3 years ago

The region in which an electron is likely to be found is known as a(n) ____________.

Zinaida [17]

Answer:

orbital

Explanation:

electrons are found in an orbital

3 0

3 years ago