1. Which of the following facts did Dalton realize?

Chemistry

1 answer:

Ne4ueva [31]3 years ago

8 0

He discovered choice C! this is because atoms of the same element can have different numbers of neutrons

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a steel cylinder is left out in the sun the pressure at 2:00 pm was 1014 kpa at 33.0 c at 7:00 am at a temperature of 21.4c what

e-lub [12.9K]

The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.

Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:

p/T is constant ==> p_1 / T_1 = p_2/T_2

From which you obtain:

p_2 = [p_1 / T_1] * T_2

T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K

p_1 = 1014 kPa

p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa

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3 years ago

(02.01 MC)

Rina8888 [55]

Answer:

C. Particle size

Explanation:

The sand, which has smaller particles, will go through the sieve, while the rice (with a larger particle size) will not

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3 years ago

What is the best definition

Ann [662]

Answer:

Explanation:

thermal expansion♡

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2 years ago

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What are some uses for graphite

erma4kov [3.2K]

Because it is a solid material, it is known as a dry lubricant. This is useful inapplications where “wet” lubricants, such as oil, cannot be used. Graphite is the only non-metal element that is a good conductor of electricity. Natural graphite is used mostly in what are called refractory applications

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3 years ago

The radioactive element carbon-14 has a half-life of 5750 years. a scientist determined that the bones from a mastodon had los

Monica [59]

Radioactive elements obey 1st order kinetics,

For 1st order reaction, relation between rate constant (k) and half life [t(1/2)] is,
k = $\frac{0.693}{t(1/2)} = \frac{0.693}{5750} = 1.205 X 10^-^4 hr^-^1$

Also, for 1st order reaction, we have
t = $\frac{2.303}{k} log \frac{\text{initial conc.}}{\text{final conc.}}$

Given that: the bones from a mastodon had lost 78.5% of their C14,
∴ initial conc. of C14 = 100%, conc. of C14 left after time 't' = 21.5%

∴t = $\frac{2.303}{1.205 X 10^(-4)} log \frac{\text{100}}{\text{21.5}}$ = 1.2758 X 10^4 hours

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3 years ago