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3 years ago

By the end of the 19th century, more than half of the population believed in the presence of atoms.

Physics

1 answer:

suter [353]3 years ago

6 0

Answer:

hello your question is vague hence I will provide a general answer about what was known about atoms in the 19th Century

answer :

By the end of the 19th century more than half of the population believed in the presence of atoms and also in the 19th Century is was a known fact that atoms determine the chemical properties of an element

Explanation:

By the end of the 19th century more than half of the population believed in the presence of atoms and also by the 19th Century is was a known fact that atoms determine the chemical properties of an element.

John Dalton reintroduced the presence of atoms in 1800 with evidence which he used to develop atomic theory. with this theory people believed in the presence of atoms in nature and also that atoms determine the chemical properties of an element .

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Ray Of Light [21]

Answer:

The answer is C.

Explanation:

4 0

3 years ago

A tank contains 0.568 mol of molecular nitrogen (N2). Determine the mass (in grams) of nitrogen that must be removed from the ta

Marizza181 [45]

Answer:

$\Delta M = 7.812 gm$

Explanation:

we know that for ideal gag we have

pV =nRT

Since volume, gas constant R and T are constant, so we have

$\frac{p}{n] = constant$

$\frac{44.7}{0.568} =\frac{22.8}{n}$

n = 0.289 mole

hence mass removed $\Delta M =( 0.568 - 0.289)*( molecular\ weight\ of \ nitrogen)$

$\Delta M = (0.568 - 0.289) *28 gm$

$\Delta M = 7.812 gm$

8 0

4 years ago

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic

olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma

$a = \frac{F}{m}$ ...... (1)

acceleration (a) $= \frac{dv}{dt}$ ......(2)

substituting (2) into (1)

Hence, F $= \frac{mdv}{dt}$

$\frac{dv}{dt} = \frac{F}{m}$

$dv = \frac{F}{m} dt$

$dv = \frac{1}{m}Fdt$

Integrating both sides

$\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt$

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

$v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt$ ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

$v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt$