Answer:
H2Br + 2KOH ----- K2Br + 2H2O
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:
The answer is D
Explanation:
Although natural processes continually form fossil fuels, such fuels are generally classified as non-renewable resources because they take millions of years to form and the known viable reserves are being depleted much faster than new ones are being made. ... The use of fossil fuels raises serious environmental concerns.
BRAINLIEST PLZ
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
Answer:
3) NaCl.
Explanation:
<em>∵ ΔTf = iKf.m</em>
where, <em>i</em> is the van 't Hoff factor.
<em>Kf </em>is the molal depression freezing constant.
<em>m</em> is the molality of the solute.
<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>
<em></em>
- For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
<em>So, for sugar: i = 1.</em>
<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>
<em></em>
- For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.
<em>So, i for NaCl = 2.</em>
<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>
<em></em>
<em>So, the right choice is: 3) NaCl.</em>
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