The question is incomplete, here is the complete question:
The solubility of substance X in water at a certain temperature is 35.7 g /100. g. Suppose that you have 330.0 g of substance X. What is the minimum volume of water you would need to dissolve it all? (Assume that the density of water is 1.00 g/mL.)
<u>Answer:</u> The minimum volume of water that would be needed is 940.17 mL
<u>Explanation:</u>
We are given:
Solubility of substance X in water = 35.1 g/100 g
This means that 35.1 grams of substance X is dissolved in 100 grams of water
Applying unitary method:
If 35.1 grams of substance X is dissolved in 100 grams of water
So, 330.0 grams of substance X will be dissolved in = 
of water
To calculate the volume of water, we use the equation:
Density of water = 1 g/mL
Mass of water = 940.17 g
Putting values in above equation, we get:
Hence, the minimum volume of water that would be needed is 940.17 mL
The correct answer is : a. Salt water
Answer:
Besides electrical, chemical, and thermal energy, some other forms of energy include mechanical and sound energy. Any of these forms of energy can change into any other form. Often, one form of energy changes into two or more different forms.
The one with the largest atomic radius is barium
Hoped this helped:)
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
