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Magnesium oxide (MgO) forms when the metal burns in air. (a) If 1-25 9 of MgO contains 0.754 g of Mg, what is the mass ratio of

Vsevolod [243]

Answer:

Explanation:

a )

1.25 g MgO contains .754 g of Mg .Rest will be O

so oxide = 1.25 - .754 = 0.496 g

ratio of magnesium to oxide = .754/.496 = 1.52

b) 1.25 g of MgO contains .754 g of Mg

534 g of MgO contains .754 x 534 / 1.25 g = 322.11 g

5 0

3 years 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

3 years ago

β‑Galactosidase (β‑gal) is a hydrolase enzyme that catalyzes the hydrolysis of β‑galactosides into monosaccharides. A 0.387 g sa

gtnhenbr [62]

Answer:

The molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=icRT$

where,

$\pi$ = osmotic pressure of the solution = 0.602 mbar = 0.000602 bar

0.000602 bar = 0.000594 atm

(1 atm = 1.01325 bar)

i = Van't hoff factor = 1 (for non-electrolytes)

c = concentration of solute = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273.15 +25]=298.15 K$

Putting values in above equation, we get:

$0.000594 atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298.15 K\\\\c=2.4278\times 10^{-5} mol/L$

The concentration of solute is $2.4278\times 10^{-5} mol/L$

Volume of the solution = V =0.137 L

Moles of β‑Galactosidase = n

$C=\frac{n}{V(L)}$

$n=2.4278\times 10^{-5} mol/L\times 0.137 L$

$n=3.3261\times 10^{-6} mol$

To calculate the molecular mass of solute, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of β‑Galactosidase = $3.3261\times 10^{-6} mol$

Given mass of β‑Galactosidase= 0.387 g

Putting values in above equation, we get:

$3.3261\times 10^{-6} mol =\frac{0.387 g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=116,352.97 g/mol$

Hence, the molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

3 0

3 years ago

What is the uncertainty of 300.0ft

romanna [79]

Answer:

300.0+\-0.1

Explanation:

3 0

2 years ago

Gaseous methane ch4 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 1.44

Usimov [2.4K]

The balanced equation for the above reaction is;
CH₄ + 2O₂ ---> CO₂ + 2H₂O
Stoichiometry of CH₄ to O₂ is 1:2
The number of methane moles present - 1.44 g/ 16 g/mol = 0.090 mol
Number of oxygen moles present - 9.5 g/ 32 g/mol = 0.30 mol
If methane is the limiting reagent,
0.090 moles of methane react with 0.090x 2 = 0.180 mol
only 0.180 mol of O₂ is required but 0.30 mol of O₂ has been provided therefore O₂ is in excess and CH₄ is the limiting reactant.
Number of moles of water that can be produced - 0.180 mol
Therefore mass of water produced - 0.180 x 18 g/mol = 3.24 g
Therefore mass of 3.24 g of water can be produced

6 0

3 years ago

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