Answer:
A. 0.90 L.
Explanation:
- NaOH solution will react with H₂SO₄ according to the balanced reaction:
<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>
<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>
- For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.
<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>
x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.
<em>∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH</em> = 2(0.6 L)(3.0 M)/(4.0 M) = <em>0.90 L.</em>
Answer:
I think it's 6 moles are produced
Input the atomic masses of Mg and P to give 134.84g/mol
Explanation:
The molar mass of a substance (atom or molecule or compound) is the mass in grams of one mole of the substance:
When dealing with an element the molar mass is the relative atomic mass expressed as g/mol.
For compounds, you add the atomic masses of the component atoms and you sum up.
You simply input the atomic mass of 3 atoms of Mg and 2 atoms of P
Atomic mass of Mg = 24.3g/mol
P = 30.97g/mole
Molar mass of Mg₃P₂ = 3(24.3) + 2(30.97) = 134.84g/mol
learn more:
Molar mass brainly.com/question/2861244
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The answer is 163.333993748 grams
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
