States that the properties of elements are periodic or recurring and are correlated to their atomic number.
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:
360 g : 1 L = x g : 30 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000
x = 10.8 g
So, from 30 mL mixture, 10.8 g of NaCl could be extracted.
Let's calculate the same for 10 mL water instead of 30 mL.
360 g : 1 L = x g : 10 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3
Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
</span>
37.5 g/mL
37.5 0.001kg/0.001L
37.5 kg/L
Answer:
844.4cm³
Explanation:
Using Boyle's law equation;
P1V1 = P2V2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (cm³)
V2 = final volume (cm³)
According to the information in this question,
V1 = 650cm³
P1 = 760mmHg
P2 = ?
V2 = 10% reduction of V1
10% of 650 = 10/100 × 650
65
V2 = 650 - 65 = 585cm³
Using P1V1 = P2V2
P2 = P1V1 ÷ V2
P2 = (760 × 650) ÷ 585
P2 = 494000 ÷ 585
P2 = 844.44
final pressure (P2) = 844.4cm³