Answer:
Mass in kg = 4.7*10^19 kg
Mass in tons = 5.2*10^16 tons
Explanation:
<u>Given:</u>
Total volume of sea water = 1.5*10^21 L
Mass % NaCl in seawater = 3.1%
Density of seawater = 1.03 g/ml
<u>To determine:</u>
Total mass of NaCl in kg and in tons
<u>Calculation:</u>
Unit conversion:
1 L = 1000 ml
The volume of seawater in ml is:
To convert mass from g to Kg:
1000 g = 1 kg
To convert mass from g to tons:
1 ton = 9.072*10^6 g
M ( HCl ) = ?
V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L
M ( NaOH ) = 0.113 M
V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L
number of moles NaOH:
n = M x V
n = 0.113 x <span> 0.0512 => 0.0057856 moles of NaOH
mole ratio:
</span><span>HCl + NaOH = NaCl + H2O
</span><span>
1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- </span><span> 0.0057856 moles NaOH
</span>
(moles HCl ) = <span> 0.0057856 x 1 / 1
</span>
= <span> 0.0057856 moles of HCl
</span>
M ( HCl ) = n / V
M = 0.0057856 / <span>0.0255
</span>
= 0.227 M
Answer A
hope this helps!
How to calcutate concentration of solution.
there is 12gram of solute in a 36 gram solution
then you take 12 divided by 24 because 36-12=24 which is youe solvent
Yasir wanted to study the relationship between sleep and room color. So he has done all the background study required on the topic and arrived at a hypothesis that people fall asleep more rapidly in a room painted in blue color than in a room painted yellow. Yasir inquired several people which color they like better -yellow or blue- and used their feedback to decide whether his hypothesis was correct. There was no experiment conducted to observe the effect of a room color on sleep and the variables to be tested in the experiment were not properly identified.
Therefore, the correct answer would be,
An experiment that directly tests the hypothesis
Variables to be tested by an experiment
The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are <span>C) carbon monoxide and carbon dioxide</span>
