All categories

2 years ago

Helppppppppppppppppp

Chemistry

1 answer:

vovangra [49]2 years ago

7 0

Do not trust people that share links these people hack and put viruses. to your phone/computer’s

DO NOT TRUST THESE LINKS

You might be interested in

A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To what temperature must the gas in the

Ivahew [28]

Answer: The temperature to which the gas in the syringe must be heated is 720.5 K

Explanation:

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=1.88atm\\V_1=285mL\\T_1=355K\\P_2=2.50atm\\V_2=435mL\\T_2=?K$

Putting values in above equation, we get:

$\frac{1.88atm\times 285mL}{355K}=\frac{2.50atm\times 435mL}{T_2}\\\\T_2=\frac{2.50\times 435\times 355}{1.88\times 285}=720.5K$

Hence, the temperature to which the gas in the syringe must be heated is 720.5 K

8 0

3 years ago

How do core electrons relate to the ionization energy of the atom?

Archy [21]

Answer:

For any given element, ionization energy increases as subsequent electrons are removed. For example, the energy required to remove an electron from neutral chlorine is 1251 kJ/mol. ... An even sharper increase in ionization energy is witnessed when inner-shell, or core, electrons are removed.

Hope it helps :)

6 0

2 years ago

Sceince problem i need help

Cerrena [4.2K]

D. due to the the water it will bring sand with the water there for us is D.

3 0

2 years ago

Read 2 more answers

Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere

Colt1911 [192]

Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
    4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
  = 2x10^-8 M
c) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
  0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M

8 0

3 years ago

If atmospheric pressure suddenly changes from 1.00 atm to 0.892 atm at 298 k, how much oxygen will be released from 4.40 l of wa

garri49 [273]

Henry's law constant for oxygen is 0,0013 mol/L·atm. Air has 21,0% oxygen.
concentration of oxygen at 1 atm: 0,0013 mol/L·atm · 0,21 · 1 atm = 0,000273 mol/l.
concentration of oxygen at 1 atm: 0,0013 mol/L·atm · 0,21 · 0,892 atm = 0,000243 mol/l.
difference in concentration: 0,000273 - 0,000243 = 0,00003 mol/L.
n(oxygen) = 0,00003 mol/L · 4,40 L = 0,000132 mol.

4 0

3 years ago

Helppppppppppppppppp​

Helppppppppppppppppp