Missing question: volume of <span>solution on the left is 10 mL.
V</span>₁(solution) = 10 Ml.
c₁(solution) = 0.2 M.<span>
V</span>₂(solution)
= ?.<span>
c</span>₂(solution)
= 0.04 M.<span>
c</span>₁ -
original concentration of the solution, before it gets diluted.<span>
c</span>₂
- final concentration of the solution, after dilution.<span>
V</span>₁
- <span>volume to
be diluted.
V</span>₂ - <span>final volume after
dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
</span>10 mL · 0.2 M = 0.04 M · V₂.
V₂(solution) = 10 mL · 0.2 M ÷ 0.04 M.
V₂(solution) = 50 mL.<span>
</span>
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
The energy transferred on the object is 1000 Joules.
Given:
An object on which a constant force of 100 N was applied to displace it over a distance of 10 meters.
To find:
The energy transfer occurs on the object.
Solution
The force applied on the object = F = 100 N
The displacement of the object = d = 10 m
The energy transferred on the object or work done is given by:
The energy transferred on the object is 1000 Joules.
Learn more about work done here:
brainly.com/question/8119756?referrer=searchResults
brainly.com/question/3951672?referrer=searchResults
Answer:
[Ag⁺] = 0.0666M
Explanation:
For the addition of Ag⁺ and CN⁻, the (Ag(CN)₂⁻ is produced, thus:
Ag⁺ + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = 1x10²¹ = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺]
As initial concentrations of Ag⁺ and CN⁻ are:
[Ag⁺] = 0.110L × (3.0x10⁻³mol / L) = 3.3x10⁻⁴mol / (0.110L + 0.230L) = 9.7x10⁻⁴M
[CN⁻] = 0.230L × (0.1mol / L) = 0.023mol / (0.110L + 0.230L) = 0.0676M
The equilibrium concentrations of each compound are:
[CN⁻] = 9.7x10⁻⁴M - x
[Ag⁺] = 0.0676M - x
[Ag(CN)₂⁻] = x
<em>Where x is reaction coordinate</em>
Replacing in Kf formula:
1x10²¹ = [x] / [9.7x10⁻⁴M - x]² [0.0676M - x]
1x10²¹ = [x] / 6.36048×10⁻⁸ - 0.000132085 x + 0.06954 x² - x³
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = x
-1x10²¹x³ + 6.954x10¹⁹x² - 1.32085x10¹⁷ x + 6.36x10¹³ = 0
Solving for x:
X = 9.614x10⁻⁴M
Thus, equilibrium concentration of Ag⁺ is:
[Ag⁺] = 0.0676M - 9.614x10⁻⁴M = <em>0.0666M</em>
Answer:
D
Explanation:
heterogeneous mixture you can see the different substances that go into it.
homogeneous mixture you can't see the difference, the mixture looks like one substance.