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ivanzaharov [21]
2 years ago
14

The deepest hole on Earth is only about 12 km deep. So, you should know that most of the data you collected in this virtual lab

re actually scientific inferences, and factual data. Do you think any this information will change in your lifetime? Why or why not?
Chemistry
2 answers:
arsen [322]2 years ago
8 0
Sorry man i just need the points hope you find the answer soon
Ne4ueva [31]2 years ago
6 0

Answer:

                                       bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

Explanation:

You might be interested in
Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices
lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

  • P= 0.980 arm
  • V= 1.20 L
  • T= 287 K
  • R= 0.082 \frac{atmL}{molK}
  • n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082\frac{atmL}{molK}× 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082\frac{atmL}{molK}× 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

molar mass=\frac{0.458 grams}{0.04997 moles}

Solving:

<u><em>molar mass= 9.17 </em></u>\frac{g}{mol}

Finally, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

Learn more about

molar mass:

brainly.com/question/5216907

brainly.com/question/11209783

brainly.com/question/7132033

brainly.com/question/17249726

ideal gas law:

brainly.com/question/4147359

#SPJ1

6 0
2 years ago
Anthony's homework assignment is to demonstrate that an orange has already undergone a chemical change. Which of the following
wolverine [178]
B. Rotten orange is the correct answer. Hope this helps!
8 0
3 years ago
Read 2 more answers
What are the reagents for BaCO3(s)
Wittaler [7]

 The reagents for BaCO₃  is

    BaO and CO₂

      <em><u>Explanation</u></em>

Reagent is  a substance that bring  about a chemical reaction when added to a system.

Some  reagent  may be added to  see if a reaction has occurred.

 BaO  and  Co₂    are reagent  since they react  to produce BaCO₃ as below

BaO(s) + CO₂(g) → BaCO3(s)



6 0
3 years ago
4. Write the electronic configuration of first 20 elements in the periodic table.
DIA [1.3K]

Answer:

Name Atomic Number Electron Configuration Period 1 Hydrogen 1 1s1 Helium 2 1s2 Period 2 Lithium 3 1s2 2s1 Beryllium 4 1s2 2s2 Boron 5 1s2 2s22p1 Carbon 6 1s2 2s22p2 Nitrogen 7 1s2 2s22p3 Oxygen 8 1s2 2s22p4 Fluorine 9 1s2 2s22p5 Neon 10 1s2 2s22p6 Period 3 Sodium 11 1s2 2s22p63s1 Magnesium 12 1s2 2s22p63s2 Aluminum 13 

7 0
3 years ago
A wooden tool is found at an archaeological site. Estimate the age of the tool using the following information: A 100 gram sampl
lord [1]

Answer:

2578.99 years

Explanation:

Given that:

100 g of the wood is emitting 1120 β-particles per minute

Also,

1 g of the wood is emitting 11.20 β-particles per minute

Given, Decay rate = 15.3 % per minute per gram

So,

Concentration left can be calculated as:-

C left = [A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Also, Half life of carbon-14 = 5730 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{5730}\ years^{-1}

The rate constant, k = 0.000120968 year⁻¹

Time =?

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

So,  

\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}

\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}

0.7320=e^{-0.000120968\times t}

ln\ 0.7320=-0.000120968\times t

<u>t = 2578.99 years</u>

8 0
3 years ago
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