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Maru [420]
2 years ago
5

Brian is making a cake for his friend's birthday. He combines the ingredients for the cake and then pours the batter into two pa

ns and bakes them for 45 minutes in the oven. While the pans are in the oven, he combines butter, sugar, and milk and mixes them together to make the icing. What type of changes created the cake and the icing?
A. Both the icing and the cake are formed by physical changes.
B. Both the icing and the cake are formed by chemical changes.
C. The icing is formed by a chemical change and the cake is formed by a physical change.
D. The cake is formed by a chemical change and the icing is formed by a physical change.
Chemistry
1 answer:
adell [148]2 years ago
5 0

Answer:

it’s D.The cake is formed by a chemical change and the icing is formed by a physical change

Explanation:

It’s because when the cake was baking there was nothing physical about it so it was a chemical change the icing is formed by a physical change because all you did was mix three ingredients together

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A substance has half the number of particles as 12 grams of carbon 12, how many moles are in the substance?
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A gas evolved during the fermentation of sugar was collected at 22.5°C and 702 mmHg. After purification its volume was found to
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Answer:

A) 0.95 mol

Explanation:

We will assume the gas given off in the fermentation is an ideal gas because that allows us to use the ideal gas equation.

PV  = nRT

First let's convert all measurements to units that we can use

P = 702 mmHg * 1 atm/760 mmHg = 0.92368 atm

V = 25.0 L

R = 0.08206 L-atm/mol-K

T = 22.5 °C +273.15 = 295.65 K

PV  = nRT

0.92368 atm * 25.0 L = n * 0.08206 L-atm/mol-K * 295.65 K

                                  n = 0.9518 mol

4 0
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An ideal gas at a given initial state expands to a fixed final volume. would the work be greater if the expansion occurs at cons
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Constant pressure

Explanation:

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w = -p\Delta V = -p(V_{f} - V_{i})

At constant temperature,

w = -RT \ln \left(\dfrac{V_{f}}{V_{i}} \right)

1 mol of an ideal gas at STP has a volume of 22.71 L.

Let's compare the work done as it expands under each condition from an initial volume of 22.71 L.

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A plot of -w vs V₂ gives a straight line (red) with a constant slope of 100 J/L as in the diagram below (Note that w is work done on the system, so -w is the work done by the system). \

Isothermal expansion

w= -8.314 \times 273.15 \ln \left(\dfrac{V_{f}}{22.71} \right)\\\\= -2271 \left( \ln V_{f} -\ln22.71 \right)\\= -2271 \left(\ln V_{f} - 3.123 \right)\\= 7092 - 2271\ln V_{f}

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Thus, the work done during an expansion at constant pressure is greater than if the system is at constant temperature.

4 0
3 years ago
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