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valkas [14]
4 years ago
9

Which sample is most likely to undergo the smallest change in temperature upon the absorption of 100 kj of heat?

Chemistry
1 answer:
Sonja [21]4 years ago
7 0
The sample that is most likely going to undergo the smallest temperature change is 50 GRAMS OF WATER.
This is because, water has a relatively higher specific heat capacity compared to lead and the higher the volume of the water, the higher will be the specific heat of the water. The specific heat capacity of a substance refers to the amount of energy needed to change the temperature of 1 kg of the substance by one degree Celsius. 
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Tracie measured 87.47 mg of cholesterol in 0.03 mL of blood. What is the density of this mixture in g/mL?
Lelu [443]

We can use a simple equation to solve this problem.


d = m/v


Where d is the density, m is the mass and v is the volume.


d = ?

m = 87.47 mg = 87.47 x 10⁻³ g

v = 0.03 mL


By applying the equation,

  d = 87.47 x 10⁻³ g / 0.03 mL

  d = 2.92 g/mL



Hence, the density of the mixture is 2.92 g/mL.


5 0
4 years ago
HURRY!! 30 points!!
AnnZ [28]

Answer:

The amount of each element that begins photosynthesis equals the amount of each element that is produced

Explanation:

3 0
3 years ago
Which of the following is true for a chemical reaction at equilibrium?
victus00 [196]

Answer:

youre gonna have to include the answers for me to help

Explanation:

7 0
3 years ago
How many moles of Aluminum are produced when34.2 g of alumina are broken down through the process of electrolysis
Elan Coil [88]

Answer:

0.670 mol

Explanation:

Step 1: Write the balanced reaction for the decomposition of alumina

2 Al₂O₃ ⇒ 4 Al + 3 O₂

Step 2: Calculate the moles corresponding to 34.2 g of Al₂O₃

The molar mass of Al₂O₃ is 101.96 g/mol.

34.2 g × 1 mol/101.96 g = 0.335 mol

Step 3: Calculate the moles of Al produced from 0.335 moles of Al₂O₃

The molar ratio of Al₂O₃ to Al is 2:4.

0.335 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃) = 0.670 mol Al

6 0
3 years ago
A gas has a volume of 45.0 mL and a pressure of .98 atm. If the pressure increased to 2.1 atm and the temperature remained the s
goblinko [34]

Answer: The new volume is 95.45 mL.

Explanation:

Given: V_{1} = 45.0 mL,         P_{1} = 0.98 atm

P_{2} = 2.1 atm,               V_{2} = ?

According to Boyle's law, at constant temperature the pressure of a gas is inversely proportional to its volume.

Therefore, formula used to calculate the new volume is as follows.

\frac{P_{1}}{V_{1}} = \frac{P_{2}}{V_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}}{V_{1}} = \frac{P_{2}}{V_{2}}\\\frac{0.98 atm}{45.0 mL} = \frac{2.1 atm}{V_{2}}\\V_{2} = 95.45 mL

Thus, we can conclude that the new volume is 95.45 mL.

7 0
4 years ago
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