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Tema [17]
3 years ago
11

what is the pressure inside a tank when it contains 87.3 moles of Helium if it holds 52.0 miles at 9.63 atm

Chemistry
1 answer:
Vladimir [108]3 years ago
3 0
We are given
n = 87.3 moles He

no = 52.0 moles
Po = 9.63 atm

Using the relationship between Pressure and number of moles
Po/no = P/n
Substituting
9.63 atm / 52.0 mol = P / 87.3 mol
Solve for P<span />
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What is the percent of nitrogen in dinitrogen pentoxide?
Lostsunrise [7]

Answer: D.) 25.9%

Explanation:

Dinitrogen pentoxide chemical formular : N2O5

Calculating the molar mass of N2O5

Atomic mass of nitrogen(N) = 14

Atomic mass of oxygen(O) = 16

Therefore molar mass :

N2O5 = (2 × 14) + (5 × 16) = 28 + 80 = 108g/mol

Percentage amount of elements in N205:

NITROGEN (N) :

(Mass of nitrogen / molar mass of N2O5) × 100%

MASS OF NITROGEN = (N2) = 2 × 14 = 28

PERCENT OF NITROGEN : (28/108) × 100%

0.259259 × 100%

= 25.925%

= 25.9%

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3 years ago
What part is the independent variable and what part is the dependent variable in the scenario: The blood pressure of a soldier i
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What part is the independent variable and what part is the dependent variable of the seminary the blood pressure of a soldier is measured while he’s resting soldiers and exposed to a stressful environment and his blood pressure is measured in
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What is the Relative Formula Mass of Potassium bromide?
Aleonysh [2.5K]
The relative formua for Potassium bromide is Kbr
6 0
3 years ago
Need some help, please. Explain why anions are always larger than the atoms from which they are derived, while cations are alway
ANTONII [103]

The question requires us to explain the differences in radii of neutral atoms, cations and anions.

To answer this question, we need to keep in mind that a neutral atom presents the same number of protons (positive particles) and electrons (negative particles). Another important information is that the protons are located in the nucleus of the atom, while the electrons are around the nucleus. Also, there is an electrostatic force between protons and electrons, which means that they the protons tend to attract the electrons to the nucleus.

While a neutral atom presents the same number of protons and electrons, a cation is an ion with positive charge, which means it has lost one or more electrons. In a cation, the balance between protons and electrons doesn't exist anymore: now, there is more positive than negative charge (more protons than electrons), and the overall attractive force that the protons have for the electrons is increased. As a result, the electrons stay closer to the nucleus and the radius of a cation is smaller than the neutral atom from which it was derived.

On the other side, anions present negative charge, which means they have received electrons. Similarly to cations, the balance between protons and electrons doesn't exist anymore, but in this case, there are more electrons than protons. In an anion, the overall attractive force that the protons have for the electrons is decreased. As a result, the electrons are "more free" to move and, as they are not so attracted to the nucleus, they tend to stay farther from the positive nucleus compared to the neutral atom - because of this, the radius of an anion is larger than the neutral atom from which it was derived.

3 0
1 year ago
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
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