Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
I belive it is synaptic cleft
Answer:
Option d. 0.10 m Cr₂(SO₄)₃
Explanation:
Formula for the osmotic pressure is determined as:
π = M . R . T . i
So you have to take account the i (Van't Hoff factor, numbers of ions dissolved)
Urea is an organic compound, so the i value is 1
Zync acetate can be dissociated:
Zn(CH₃COO)₂ → 1Zn²⁺ + 2CH₃COO⁻
In this case, the i is 3. (you see, the stoichiometry of ions)
Cr₂(SO₄)₃ → 0.10 m
Chromium sulfate is dissociated:
Cr₂(SO₄)₃ → 2Cr³⁺ + 3SO₄⁻²
i = 5
BaI₂ → 0.16 m
BaI₂ → 1Ba²⁺ + 2I⁻
i = 3
Answer:
There is 541.6 grams of calcium carbonate produced from 5.11 moles
Explanation:
Step 1: Data given
Number of moles of sodium carbonate ( Na2CO3) = 5.11 moles
Molar mass of sodium Carbonate:
⇒ Molar mass of Sodium = 22.99 g/mole
⇒ Molar mass of Carbon = 12.01 g/mole
⇒ Molar mass of Oxygen = 16 g/mole
Molar mass of Na2CO3 = 2* 22.99 + 12.01 + 3*16 = 105.99 g/mole
Step 2: Calculate mass of 5.11 moles of sodium carbonate
mass = number of moles of sodium carbonate * Molar mass of sodium carbonate
mass = 5.11 moles * 105.99 g/moles = 541.6 grams ≈ 542 grams
There is 541.6 grams of calcium carbonate produced from 5.11 moles Na2CO3
The original radioisotope underwent double-alpha decay, where the original nucleus lost a total of 4 protons and 4 neutrons.
The original radioisotope would have an atomic number of 86+4 (protons), and would have an atomic mass of 222+8 (protons + neutrons). This would make the element Thorium-230, or 90Th230.