Answer:
Isotopes – caused by varying numbers of neutrons in an element – have many practical uses in our society. ... In geology and archaeology, radioactive isotopes are used to determine the age of a sample while hydrologists can use isotope signatures to distinguish between different groundwater types.
Explanation:
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Answer:
70 mL of 5% HCl and 30 mL of 15% HCl
Explanation:
We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:
5x + 15y = 8
Since x and y are fractions of a total, they must equal one:
x + y = 1
This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:
y = 1 - x
This expression is substituted into the first equation and we solve for x.
5x + 15(1 - x) = 8
5x+ 15 - 15x = 8
-10x = -7
x = 7/10 = 0.7
We then calculate the value of y:
y = 1 - x = 1 - 0.7 = 0.3
Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:
(100 mL)(0.7) = 70 mL
Similarly, the volume of 15% HCl we need is:
(100 mL)(0.3) = 30 mL
Answer:
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Answer:
10043.225 J
Explanation:
We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:
Mass (m) = 15.5g
Latent heat of fussion of water (L) = 334J/g
Heat (Q1) =..?
Q1 = mL
Q1 = 15.5 x 334
Q1 = 5177 J
Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.
This is illustrated below:
Mass = 15.5g
Initial temperature (T1) = 0°C
Final temperature (T2) = 75°C
Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C
Specific heat capacity (C) of water = 4.186J/g°C
Heat (Q2) =?
Q2 = MCΔT
Q2 = 15.5 x 4.186 x 75
Q2 = 4866.225 J
The overall heat energy needed is given by:
QT = Q1 + Q2
QT = 5177 + 4866.225
QT = 10043.225 J
Therefore, the amount of energy required is 10043.225 J
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