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Paul [167]
3 years ago
6

What are the charges and relative masses of the 3 main subatomic particles

Chemistry
1 answer:
blondinia [14]3 years ago
5 0
Proton: positive- mass equal to neutron
Electron: negative- mass 1/1840 mass of proton 
Neutron: no charge- same mass as proton
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Does the space between the particles decrease or increase when matter changes it’s state to a solid?
bazaltina [42]

Answer:

i pretty sure it decreses

Explanation:

7 0
3 years ago
Read 2 more answers
A certain substance x decomposes. it is found that 50% of x remains after 100 minutes. how much x remains after 200 minutes if t
fiasKO [112]
The time needed for a substance becomes half of it mass is called half-time, after. In this case, the half time is 100 minutes. Then to know how much substance remain after 200 minutes the equation would be:

Substance remains= 1/2^(time/half-time) x substance initial mass
Substance remains = 1/2^(200 minutes/100minutes) x 1 times= 1/4 times
6 0
4 years ago
746mmHg, 115m^3, 21*C number of moles?
victus00 [196]

Answer:

n = 4678.13 mol

Explanation:

Given data:

Temperature = 21°C

Pressure = 746 mmHg

Volume = 115 m³

Number of moles = ?

Solution:

21+273 = 294 k

746 /760 = 0.982 atm

115×1000 = 115000 L

PV = nRT

n = PV/RT

n =   0.982 atm × 115000 L /0.0821 atm. L. K⁻¹. mol ⁻¹× 294 k

n = 112930 atm. L/24.14 atm. L. mol ⁻¹

n = 4678.13 mol

8 0
3 years ago
Starting with 250 mL of a 0.250 M solution of HBr; a) Calculate the initial pH of the solution.b) Calculate the pH after adding
maxonik [38]

Answer:

a.

pH =  0.602

b.

pH = 1.5

c.

pH = 7

d.

pH = 12.1

Explanation:

a ) To calculate the pH, use the following equation:

pH = -log [H+]

Hbr is a strong acid, so the [H+] concentration can be calculated as follow:

[HBr] = 0.250 M

As acid Hbr:

Hbr = H+ + Br-

As strong acid HBr dissociates at all, so

[HBr] = [H+] = 0.250 M

So the pH:

<u>pH = -log [0.250 M] = 0.602</u>

<u></u>

<u>b) Calculate the pH after adding 250 mL of 0.125M NaOH</u>

<u></u>

<u>I</u>n this point, the reactions starts:

<u></u>

HBr + NaOH = H2O + NaBr

- First, we gonna find the mol of each reactant:

HBr:

mol = [M] × L

mol = 0.250 M × 0.250 L

mol = 0.0625 mol HBr

NaOH:

mol = [M] × L

mol = 0.125 M × 0.250L

mol = 0.03125 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.03125 mol HBr = 0.03125 mol HBr

These are the moles free in the solution, and we going to use them to calculate the pH:

pH = -log [ H]

pH = - log [ 0.03125 ] = 1.5

<u>c) Calculate the pH after adding 500 mL of 0.125M NaOH</u>

NaOH:

mol = [M] × L

mol = 0.125 M × 0.500L

mol = 0.0625 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.0625 mol HBr = 0 HBr

pH = 7

<u>d) Calculate the pH after adding 600 mL of 0.125M NaOH.</u>

<u>NaOH:</u>

mol = [M] × L

mol = 0.125 M × 0.600L

mol = 0.075 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.075 mol NaOH – 0.0625 mol NaOH = 00125 NaOH

These are the moles free in the solution, and we going to use them to calculate the pH:

pOH = -log [ OH]

pOH = - log [ 0.0125 ] = 1.90

pH + pOH = 14

pH = 14- pOH = 14 – 1.90 = 12.1

4 0
3 years ago
How many joules are needed to change the temperate of 22g of water from 18°C to 33°C?
Ira Lisetskai [31]

Answer:

Q =  1379.4 J

Explanation:

Given data:

Mass of water = 22  g

Initial temperature = 18°C

Final temperature = 33°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g. °C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 33°C - 18 °C

ΔT =  15°C

Q = 522 g ×4.18 J/g.°C× 15°C

Q =  1379.4 J

5 0
3 years ago
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