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Aneli [31]
3 years ago
15

Can someone please help me please?

Chemistry
2 answers:
pishuonlain [190]3 years ago
7 0

Answer:

the weighted average of the masses of the isotopes of the element.

Explanation:

every element has siblings, which are called isotopes. Isotopes are an element that has the same amount of protons and electrons, but it's mass is different because it has a different amount of neutrons.

the atomic mass is the average of all of those isotopes.

madam [21]3 years ago
5 0

Answer:

option no B

Explanation:

hope helps you

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The gas inside a balloon exerts a pressure of 3atm at 298 K. What is the temperature required to increase the pressure to 6 atm?
podryga [215]

Answer: 596 atm

Explanation:

Given that,

Original pressure of balloon P1 = 3atm

Original temperature of balloon T1 = 298K

New pressure of balloon P2 = 6atm

New temperature of balloon T2 = ?

Since pressure and temperature are given while volume is constant, apply the formula for Pressure law

P1/T1 = P2/T2

3 atm / 298K = 6 atm / T2

To get the value of T2, cross multiply

3 atm x T2 = 6 atm x 298K

3 atmT2 = 1788 atmK

Divide both sides by 3 atm

3 atmT2 / 3 atm = 1788 atmK / 3 atm

T2 = 596 atm

Thus, a temperature of 596 atmospheres is required to increase the pressure to 6 atm.

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3 years ago
The ocean stores carbon from its interactions with the atmosphere, so the ocean serves as (a) _________ for the carbon cycle.
anyanavicka [17]

\huge{\underline{ \pink{Answer}}}

\tt \bf \: carbon \:  \: sink

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.

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3 years ago
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Irina-Kira [14]
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3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 129 GPa and an original cross-sectional diameter of 4.4
miss Akunina [59]

Explanation:

The cross-sectional area of the specimen is calculated as follows.

         A_{o} = \frac{pi}{4} d^{2}

                     = \frac{3.14}{4} \times (\frac{4.4}{1000})^{2}

                     = 1.5197 \times 10^{-5} m^{2}

Equation of stress is as follows.

              \sigma = \frac{F}{A_{o}}

And, the equation of strain is as follows.

             \epsilon = \frac{\Delta l}{l_{o}}

Hence, the Hook's law is as follows.

              E = \frac{\sigma}{\epsilon}

       E = \frac{\frac{F}{A_{o}}}{\frac{\Delta l}{l_{o}}}

          = \frac{F \times l_{o}}{A_{o} \times \Delta l}

or,    l_{o} = \frac{E \times \Delta l \times A_{o}}{F}          

                   = \frac{129 \times 10^{9} \times \frac{0.48}{1000} \times 1.662 \times 10^{-5}}{1570}

                  = 0.6554 m

or,         l_{o} = 655.4 mm

Thus, we can conclude that the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm is 655.4 mm.

5 0
4 years ago
Correct answer. Correct.Multiple-Concept Example 3 reviews the concepts necessary to solve this problem. Radiation with a wavele
Makovka662 [10]

Answer:

answer is gold (4.82 eV)

Explanation:

the answer is explained in the image below

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