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katen-ka-za [31]
3 years ago
12

How many moles of potassium chloride (KCI) are in 252 mL of 0.33 mil/L KCI

Chemistry
1 answer:
Rom4ik [11]3 years ago
7 0

Answer:

0.083 moles KCl

Explanation:

We have 252 mL of 0.33 M KCl. "M" is molarity, and it's measured in mol/L. Since we know the molarity is 0.33 mol/L, in order to find how many moles 252 mL is, we need to multiply our given volume by the molarity.

However, our given volume is in mL, so let's convert to L first by dividing 252 by 1000 (because there are 1000 mL in 1 L):

252/1000 = 0.252 L

Now, we can multiply 0.252 L by 0.33 mol/L to get moles since the L units will cancel out:

0.252 L * 0.33 mol/L = 0.08316

We have two significant figures here, so round 0.08316 to 0.083.

The answer is 0.083 moles.

Hope this helps!

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Which of the following represents the correct ranking in terms of increasing boiling point? A. n-butane < 1-butanol < diet
bonufazy [111]

Answer:

The answer to your question is: letter E

Explanation:

Normally, the correct order of boiling points is:

                     Alcohols > Ketones > Ether > Alkane

Then

A. n-butane < 1-butanol < diethyl ether < 2-butanone

B. n-butane < 2-butanone < diethyl ether < 1-butanol

C. 2-butanone < n-butane < diethyl ether < 1-butanol

D. n-butane < diethyl ether < 1-butanol < 2-butanone

E. n-butane < diethyl ether < 2-butanone < 1-butanol

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3 years ago
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shepuryov [24]

Answer:2 is correct

Explanation:

6 0
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2 years ago
16.25 g of water at 54 C relaeases 402.7 J. What will be its final temp?
leonid [27]
Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?

Solving:

Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
\Delta\:T =  \frac{-402.7}{67.99}
\boxed{\Delta\:T \approx -5.92\:^0C}

If: ΔT (T final - T initial) = ?
-5.92^0 =  T_{final} -  54^0
T_{final} = 54^0 - 5.92^0
\boxed{\boxed{T_{final} = 48.08\:^0C}}\end{array}}\qquad\quad\checkmark

8 0
3 years ago
2. ATO kg bouting ball would require what force to accelerate down an alleyway at a rate of ons?
Mars2501 [29]

Answer:

  • 2. 30N
  • 3. 5,000N
  • 4. 15 kg
  • 5. 2,800 kg

Explanation:

<em>2. A 10 kg bowling  ball would require what force to accelerate down an alleyway at a rate of 3m/s² ?</em>

Notice that I completed the question with the garbled and missing values:

<u>Data:</u>

  • F = ?
  • m = 10 kg
  • a = 3m/s²

<u />

<u>Physical principles:</u>

  • Newton's second law: F=m\times a

<u>Solution:</u>

  • Substitute and compute

        F=10kg\times 3m/s^2=30N

<em></em>

<em>3. Salty has a car that accelerates at 5 m/s². If the car has a mass of 1000 kg, how much force does the car produce?</em>

Notice that I arranged the typos.

<u />

<u>Data:</u>

  • F = ?
  • m = ?
  • a = ?

<u>Physical principles:</u>

  • Newton's second law: F=m\times a

<u>Solution:</u>

  • Substitute and compute

       F=1,000kg\times5m/s^2=5,000N

<em>4. What is the mass of a falling rock if it produces a force of 147 N?</em>

<u>Data:</u>

  • F = 147N
  • m = ?
  • a = falling rock

<u>Physical principles:</u>

  • neglecting air resistance ⇒ a = g: gravitational acceleration: 9.8m/s²
  • Newton's second law: F=m\times a

<u>Solution:</u>

  • Clear m from Newton's second law

         m=\dfrac{F}{a}

  • Substitute with F = 147 N and a = g = 9.8m/s², and compute

      m=\dfrac{147N}{9.8m/s^2}=15Kg

<em></em>

<em>5. What is the mass of a truck if it produces a force of 14,000 N while accelerating at a rate of 5 m/s²?</em>

<u>Data:</u>

  • F= 14,000N
  • m = ?
  • a =​ 5m/s²

<u>Physical principles:</u>

  • Second Newton's law: F=m\times a

<u>Solution:</u>

  • Clear m from Newton's second law

         m=\dfrac{F}{a}

  • Substitute with F = 14,000 N and a = 5m/s², and compute

      m=\dfrac{14,000N}{5m/s^2}=2,800kg

6 0
3 years ago
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