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Paladinen [302]
3 years ago
11

Bread rising in the oven chemical or physical change

Chemistry
1 answer:
Darya [45]3 years ago
8 0
Both because there is a chemical change where carbon dioxide is released (rising)
and also physical change from liquid to a solid.
You might be interested in
Match Term Definition
mart [117]

The correct answers are:

1. Lithium - C) Opaque solid with higher density

2. Lead - B) Malleable, soft, and shiny

3. Florine - D) Highly reactive gas

4. Krypton - A) Nonreactive gas

I hope that helps u!

:)


7 0
3 years ago
CHEM
castortr0y [4]
Molarity = moles / liter

a) M = 2/4 = 0.5 M

b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M

c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M
3 0
3 years ago
Read 2 more answers
Which of the following is the FIRST step in solving the SI conversion problems?
AVprozaik [17]

Answer:

A

Explanation:

7 0
3 years ago
Read 2 more answers
What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0
3 years ago
Calculate the theoretical yield of 1-bromobutane; base your calculations on using 1.0 g of 1-butanol (as the limiting reagent).
Tomtit [17]

C₄H₉OH + HBr = C₄H₉Br + H2O

Δmole of alcohol gives 1 mole of bromobutanol

HBr is in excess, so the yield of the product is limited by the alcohol

Wt. of 1 butanol = 18

Molar mass of the butanol = 74.12 g/mole

Moles of the alcohol = 1/74.12 = 0.01349 moles

So, moles of bromobutane = 0.01349 moles

Molar mass of C₄H₉Br = 137.018 g/moles

So, theoretical mass of bromobutane is = 0.01349 × 137.0.18

= 1.85 g


6 0
4 years ago
Read 2 more answers
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