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kykrilka [37]
3 years ago
6

How many dozen (dz) eggs are needed to make 12 muffins? What about 15.5

Chemistry
2 answers:
NARA [144]3 years ago
8 0

Answer:

I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer

romanna [79]3 years ago
6 0
Should be 1.2 I divided and got 1.2
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Select all that apply.
cupoosta [38]

Answer:

B and C

Explanation:

I think the answer correct is C because you never know in what temperature the block of ice is going to melt but if it says select all that apply its possible that B might be useful.

7 0
2 years ago
Please someone can help with this answer?
Annette [7]

Answer:

6,000kg/m3

Explanation:

6.00g/1cm3 x 1kg/1000g x 1cm3/0.000001m3

= 6.00kg/0.001m3

= 6,000kg/m3

5 0
3 years ago
The vapor pressure of chloroform is 173.11 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in chloroform is aspirin
Over [174]

Answer:

VP (solution) = 171.56 mmHg

Explanation:

Vapor pressure of pure solvent(P°) - Vapor pressure of solution (P') = P° . Xm

Let's replace the data:

173.11 mmHg - P' = 173.11 mmHg . Xm

Let's determine the Xm (mole fraction for solute)

Mole fraction for solute = Moles of solute / Total moles

Total moles = Moles of solute + moles of solvent.

Let's determine the moles

Moles of solvent → 623.4 g / 119.4 g/mol = 5.22 moles

Moles of solute → 9.322 g / 180.1 g/mol = 0.052 moles

Total moles = 0.052 + 5.22 = 5.272 moles

Xm = 0.052 moles / 5.272 moles = 0.009 → 9/1000

173.11 mmHg - P' = 173.11 mmHg . 9/1000

P' = - (173.11 mmHg . 9/1000 - 173.11 mmHg)

P' = 171.56 mmHg

4 0
2 years ago
Read 2 more answers
The chemical reaction in which compounds break up into simpler Constituents is a.
coldgirl [10]
Decomposition its the answer
8 0
3 years ago
Read 2 more answers
Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
WARRIOR [948]
Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
5 0
2 years ago
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