<span>Let x = amt of distilled water
:
A simple equation
.25(16) = .10(x+16)
4 = .10x + 1.6
4 - 1.6 = .1x
2.4 = .1x
x = 2.4/0.1
x = 24 oz of distilled water
:
:
Prove this by seeing the amt of antiseptic is the same (only the % changes)
.25(16) = .10(24+16)
4 = .1(40)
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Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
Answer: 1.
moles
2. 90 mg
Explanation:

According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus
moles of ozone is removed by =
moles of sodium iodide.
Thus
moles of sodium iodide are needed to remove
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide=
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of
.
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 53.3 mg
m (final mass after time T) = ? (in mg)
x (number of periods elapsed) = ?
P (Half-life) = 10.0 minutes
T (Elapsed time for sample reduction) = 25.9 minutes
Let's find the number of periods elapsed (x), let us see:






Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:




I Hope this helps, greetings ... DexteR! =)