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andreyandreev [35.5K]
3 years ago
7

Choose all the right answers.

Chemistry
1 answer:
vagabundo [1.1K]3 years ago
8 0
To change solar energy to chemical energy is the answer!
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The term “ average atomic mass “ is a ______average so is calculator different Lee from a normal average
hoa [83]

Average atomic mass of an element is a sum of the product of the isotope mass and its relative abundance.

For example: Chlorine has 2 isotopes with the following abundances

Cl(35): Atomic mass = 34.9688 amu; Abundance = 75.78%

Cl(37): Atomic mass = 36.9659 amu; Abundance = 24.22 %

Average atomic mass of Cl = 34.9688(0.7578) + 36.9659(0.2422) =

                                             = 26.4993 + 8.9531 = 35.4524 amu

Thus, the term “ average atomic mass “ is a <u>weighted</u> average so it is calculated differently from a normal average

3 0
3 years ago
Which is the smallest number?
My name is Ann [436]

Answer:

What I came up with was D equaling out to be 42

Explanation:

Im assuming you just do the math

A. 3.5 x 104 = 364

B. 2.4 x 103 = 247.2

C. 1.4 x 10^2_2 = 280

D. 4.9 x 10-7 = 42

I hope I helped.

5 0
3 years ago
This is true for the body being attracted as well. Bodies of water that are smaller than oceans, seas, and large lakes are _____
zhannawk [14.2K]

Answer:

:)  

:)

Explanation:

4 0
2 years ago
1 kg of water (specific heat = 4184 J/(kg K)) is heated from freezing (0°C) to boiling (100°C). What is the change in thermal en
Andrews [41]

Answer: 1560632 joules

Explanation:

The change in thermal energy (Q) required to heat ice depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = ?

Mass of frozen water (ice) = 1kg

C = 4184 J/(kg K)

Φ = (Final temperature - Initial temperature)

= 100°C - 0°C = 100°C

Convert 100°C to Kelvin

(100°C + 273) = 373K

Then, Q = MCΦ

Q = 1kg x 4184 J/(kg K) x 373K

Q = 1560632 joules

Thus, the change in thermal energy is 1560632 joules

5 0
3 years ago
If you want to prepare 80.0 mL of 4.00M acid ,How many mL of 12.4 M HCl are required ?
Oksi-84 [34.3K]

M_{A}V_{A}=M_{B}V_{B}\\(80.0)(4.00)=V_{B}(12.4)\\V_{B}=\frac{(80.0)(4.00)}{12.4} \approx \boxed{25.8 \text{ mL}}

5 0
2 years ago
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