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3 years ago

A student says that since the atomic theory is just a theory, it should not be considered useful. Which

Chemistry

1 answer:

Shkiper50 [21]3 years ago

8 0

Answer:

O Scientific theories are the results of many experiments and observations.

Explanation:

Going through the options;

O Scientific theories change over time.

This is true and supports the student's opinion. So wrong option!

O Scientists often do not agree about specific details of scientific theories.

This is true. However, it supports the student's opinion. So wrong option!

O Scientists often propose competing theories.

This is true. However, it supports the student's opinion. So wrong option!

O Scientific theories do not become Scientific Laws.

This is False. Theories that have withstood the test of time and countless experiments become laws. However it support's the student's opinion. So wrong option!

O Scientific theories are the results of many experiments and observations.

This is true and is against the student's opinion. This option is correct!

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The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be

Stels [109]

Explanation:

1) Initial mass of the Cesium-137= $N_o$ = 180 mg

Mass of Cesium after time t = N

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

Half life of the cesium-137 = $t_{1/2}=30 years[p/tex]where,
[tex]N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$N=180mg\times e^{-\frac{0.693}{30 years}\times t}$

Mass that remains after t years.

$N=180 mg\times e^{0.0231 year^{-1}\times t}$

Therefore, the parent isotope remain after one half life will be, 100 grams.

t = 70 years

$N_o=180 mg$

$t_{1/2}= 30 yeras$

$N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}$

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

$N_o=180 mg$

$t_{1/2}= 30 yeras$

N = 1 mg

t = ?

$1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}$

$\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t$

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

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3 years ago

Wharis the mass in grams of 2.000 mol of oxygen atoms?

Alika [10]

Answer:

32.00 g. Hope this helps! PLEASE GIVE ME BRAINLIEST!!!!! =)

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3 years ago

Complete this equation for the dissociation of na2co3(aq). omit water from the equation because it is understood to be present.

Savatey [412]

Net overall dissociation:
Na2CO3 ---> 2Na(-) + CO3(2-)

*The ion charge is in parenthesis

6 0

3 years ago

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3 A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state.

Masja [62]

When a specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state, this emitted energy can
<span>be used to determine the "identity of the element".</span>

5 0

3 years ago

A 32.4 L gas sample at STP is compressed to a volume of 28.4 L, and the temperature is increased to 352 K. What is the new press

Sindrei [870]

Answer:

1.47 atm

Explanation:

Step 1: Given data

Initial volume (V₁): 32.4 L

Initial pressure (P₁): 1 atm (standard pressure)

Initial temperature (T₁): 273 K (standard temperature)

Final volume (V₂): 28.4 L

Final pressure (P₂): ?

Final temperature (T₂): 352 K

Step 2: Calculate the final pressure of the gas

We can calculate the final pressure of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

P₂ = P₁ × V₁ × T₂ / T₁ × V₂

P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm

7 0

3 years ago