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
It depends on the substance,but for most substance it is in the gaseous state.
The specific heat capacity of the metal given the data from the question is 0.66 J/gºC
<h3>Data obtained from the question</h3>
- Mass of metal (M) = 76 g
- Temperature of metal (T) = 96 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 24.5 °C
- Equilibrium temperature (Tₑ) = 31 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
The specific heat capacity of the sample of the metal can be obtained as follow:
Heat loss = Heat gain
MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)
C × 4940 = 3263.52
Divide both side by 4940
C = 3263.52 / 4940
C = 0.66 J/gºC
Learn more about heat transfer:
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Answer:
The Law of Definite Proportions ensures that chemical compounds are always created using the same proportions, regardless of the amount of the compound which is being made
Answer:

Explanation:
Here, we wnat to get the most electronegative atom
On the periodic table, electronegativity increases across a period and decrease down the group
The elements listed are in the same group
Boron is the first member of the group
Boron is the most electronegative atom in the group