Answer: The constraints could be lack of materials, funds, or time. Another constraint could be size or shape.
Explanation:
E :)
Using the given formula, the density of the material is 2.015 g/mL
<h3>Calculating Density </h3>
From the question, we are to determine the density of the material
From the given formula
Density = Mass / Volume
And from the given information,
Mass = 65.5 g
and volume = 32.5 mL
Putting the parameters into the equation,
Density = 65.5/32.5
Density = 2.015 g/mL
Hence, the density of the material is 2.015 g/mL.
Learn more on Calculating density here: brainly.com/question/24772401
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The answer for the following questions is explained below.
Explanation:
The two variables that affect kinetic energy are:
- mass and
- velocity
- velocity - The faster an object moves,the more the kinetic energy it has.
- mass - Kinetic energy increases as mass increases
The kinetic energy of an object depends on both its mass and its velocity
Kinetic energy increases as mass increases
For example,think about rolling a bowling ball and a golf ball down a bowling lane at same velocity
Here,the bowling ball has more mass than the golf ball
Therefore you use more energy to roll the bowling ball than to roll the golf ball
The bowling ball is more likely to knock down the pins because it has more kinetic energy than the golf ball
Answer:
See explanation
Explanation:
Using the formula
°C = (F-32) × 5/9
Where;
°C = temperature in degrees centigrade
F= temperature in Fahrenheit
F= (9/5 ×°C) +32
F= (9/5 × 110) + 32
F= 230°F
To convert -78°C to Kelvin
-78°C + 273 = 195 K
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
