Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
As a solution is contained with 11% by mass of sodium
chloride, this will therefore conclude that the 100g of the solution is
composed of the 11g of the sodium chloride. In which, the correct answer would
be specified as letter c.
Answer:
It is the closest sir.
Have an amazing day and enjoy. P.S Good luck!
Explanation:
Answer: m-%(Ca) = 40.08 / 110.98
Explanation: molar mass of CaCl2 is 40.08+ 2·35.45 = 110.98
Think you have one mole substance. It contains 40.08 g Ca
Answer:
Answer is given below:
Explanation:
<em>Given Data:</em>
mass = 80kg
acceleration = 4 ms
force = 800N
<em>Find out:</em>
friction = ?
<em>Formula</em><em>:</em>
F-friction = weight - f-net
<em>Solution:</em>
weight = (80)(10)
= 800 N
F-net = ma =(80)(4) = 320N
F-friction = weight - F-net
=800 N - 320N
=480N
<em>Answer</em> :
Friction = 480 N
