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Andreyy89
3 years ago
9

Find the number of cm in 0.286 miles.

Chemistry
1 answer:
egoroff_w [7]3 years ago
4 0
You can use dimensional analysis: \frac{.286 mi}{} \frac{5280 ft}{1 mi} \frac{12 in}{1 ft} \frac{2.54 cm}{1 in}= 46027.2384 cm
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As the temperature of a fixed volume of a gas increases the pressure wil
Sati [7]

Answer:

As the temperature of a fixed volume of a gas increase the pressure will increase.

Explanation:

According to the Gay- Lussac's Law,

" The pressure of given amount of gas is directly proportional to the temperature at a constant volume"

Mathematical expression:

           P ∝ T

          P = CT

          P / T = C

As the temperature increase, the pressure also increase.

The initial and final expression of volume and pressure can be written as,

P₁ / T₁  = P₂ / T₂

3 0
3 years ago
The (triangle)Hrxn of formation of carbon dioxide is negative. Which statement is true? The reaction is endothermic. The reactio
ANEK [815]
The negative ion reactions that consist of the formation of carbon dioxide in the atmosphere is generally an exothermic reaction. By definition, an exothermic reaction takes place when the chemical process eventually releases heat as its by-product. It is in contrast in endothermic process wherein heat is absorbed.
7 0
3 years ago
Read 2 more answers
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
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3. Write a hypothesis that explains your inference about the acidity of the paper. How might you test your hypothesis by extendi
vredina [299]

Answer:

google

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go to the plus thing on your screen and search it. (please dont attack me im doing a dare  >:( )

8 0
3 years ago
If 50. 75 g of a gas occupies 10. 0 l at stp, 129. 3 g of the gas will occupy ________ l at stp.
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22.4L

of any gas contains 1 mol of that gas.

50.75g/10L*22.4L/1 mol= 113.68g/mol- this is the mole weight of your gas

1 mol/113.68g*129.3g=1.137403 mol

Set up a ratio

1.137403mol/x L=1 mol/22.4 L

X=25.477827L, or with sig figs, x=25.5L

8 0
2 years ago
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