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padilas [110]
2 years ago
13

How much of a 3.75 M KI solution would you need to prepare 252.5 mL of a 0.780 M KI solution?

Chemistry
1 answer:
saveliy_v [14]2 years ago
4 0
Answer: 52.5 mL

Hope this helps!
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What will increased temperature in a reaction cause
uranmaximum [27]
I think your answer is 2 because the temperature will rise so the particles will move faster
7 0
3 years ago
Please Help! It says fill in the blanks to complete the Punnett square below to show the results of a cross between a homozygous
makkiz [27]

Answer:

Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.

Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%

Explanation:

Hello,

The Punnett square for this cross turns into:

\left[\begin{array}{ccc}&G&g\\G&GG&Gg\\G&GG&Gg\end{array}\right]

It means that the genotypes and phenotypes are:

Genotypes: Homozygous (GG)=50%, Heterozygous (Gg)=50%.

Phenotypes: Homozygous gray (GG)=50%, Heterozygous gray (Gg)=50% or just Gray=100%

Best regards.

4 0
2 years ago
A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.
Alja [10]

Answer:

  • <u>Question 1: 0.2J/(gºC)</u>
  • <u>Question 2: 6,000J</u>
  • <u>Question 3: 300J</u>
  • <u>Question 4: 80g</u>
  • <u>Question 5: 74ºC</u>
  • <u>Question 6: 50g</u>

<u></u>

Explanation:

Question 1.<em> A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.</em>

<em />

The thermal energy equation is:

  • Q = m × C × ΔT

<em />

Substitute and solve for C:

  • 566J = 20g × C × (195ºC - 35ºC)
  • C = 566J / (20g × 160ºC)
  • C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

<em />

You must round to one significant figure because one factor has one significant figure).

<em />

<em />

Qustion 2.<em> 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)</em>

<em />

Use the thermal energy equation again:

  • Q = m × C × ΔT

<em />

Substitute and compute:

  • Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)
  • Q = 5,857.6J

Round to one significant figure: 6,000J

<em />

Question 3. <em>Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?</em>

  • Q = m × C × ΔT
  • Q = 50g × 0.420J/gºC × (35ºC - 50ºC)
  • Q = 315J

Round to one significant figure (because 50g has one significant figure)

  • Q = 300J

<em />

Question 4.<em> </em><em>Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?</em>

<em />

  • Q = m × C × ΔT

Solve for m:

  • m = Q / (C × ΔT)

Substitute and compute:

  • m = 3,000J / (0.712J/gºC × 50ºC)
  • m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. <em>If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.</em>

<em />

  • Q = m × C × ΔT

<em />

Q is negative, since it is released.

Substitute and solve for T:

  • - 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

  • T = - 7500J / 400g × 4.184J/gºC) + 70ºC

  • T = 74ºC

<em />

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

<em />

Question 6. <em>How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC</em>

  • Q = m × C × ΔT

Subsitute:

  • 9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

  • m = 9,500J / (4.184J/gºC × 50ºC)

  • m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

  • m = 50g.
8 0
3 years ago
A sample of a compound contains 32.0 g C and 8.0 g H. Its molar mass is 30.0 g/mol. What is the compound’s molecular formula
Grace [21]
The compound's molecular formula is C2H6. This is obtained by:
  
            mass             moles               divided by smallest moles
C          32g         32/12 = 2.67                                 1
H           8g           8/1.01 = 7.92                        approx. 3

Next, divide both terms by the smallest number of moles, 2.67. This gives 1 and 3. So the empirical formula is CH3 which has a molar mass of 15g/mol. Given the molar mass of the molecular formula as 30g/mole, we can calculate the factor by which to multiply the subscripts of CH3.

X = molar mass of molecular formula / molar mass of empirical formula = 30/15
X=2

So (CH3)2 is C2H6.
3 0
2 years ago
What are some examples of gases at room temperature
Lady_Fox [76]

Answer:

Nitrogen, Hydrogen, Oxygen, Chlorine, and Fluorine are all gases at room temperature.

Explanation:

6 0
2 years ago
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