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natka813
9 days ago
13

Which statement best describes why specific heat capacity is often more useful than heat capacity for scientists when comparing

two materials?
Specific heat capacity is an intensive property and does not depend on sample size.
Specific heat capacity is an extensive property and does not depend on sample size.
Specific heat capacity is an intensive property and depends on the sample size.
Specific heat capacity is an extensive property and depends on the sample size.
Chemistry
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One hour of bicycle riding can require 500-900 kcal of energy, depending on the speed, the terrain, and the weight of the racer.
alisha [2963]
145 hours. Explanation: Riding a bicycle for one hour expends 505 kcal of energy. Given that one gram of body fat equals 7.70 kcal, and 1 pound of body fat is equivalent to 454 grams: 1 lb = 454 g; thus, 21 lb = 21 × 454/1 = 9534 g. Moreover, converting 9534 g of body fat gives us 9534 × 7.70 kcal/1 = 73411.8 kcal. If riding for one hour burns 505 kcal, then to lose 73411.8 kcal, it would require 73411.8 kcal x 1 hour/505 kcal = 145 hours.
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3 months ago
For the following equilibrium: A+2B⇋C+3D If the change in concentration for B is 0.44 M, what will be the change in concentratio
Anarel [2989]
0.22 M
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3 months ago
What is the density (in g/L) of a gas with a molar mass of 16.01 g/mol at 1.75 ATM and 337 K?
Anarel [2989]
To solve for density, you can use the formula--> Density= PM/ RT, where P stands for pressure, M for molar mass, R represents the gas constant, and T is temperature. 

P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k

Thus, the density calculation becomes: density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L
8 0
2 months ago
Read 2 more answers
Sodium only has one naturally occuring isotope, 23 Na , with a relative atomic mass of 22.9898 u . A synthetic, radioactive isot
KiRa [2933]

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

6 0
4 months ago
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