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3 years ago

How do scientists write very large numbers

Chemistry

2 answers:

Tatiana [17]3 years ago

8 0

Answer:Explained

Explanation:

Scientists uses Scientific notations to write big number for ex.

50000000 can be written in the form of $5\times 10^7$

789400 can be written as $7.894\times 10^6$

0.000462 can be written as $4.62\times 10^4$

Likurg_2 [28]3 years ago

4 0

PLZ Brainly
By using the scientific-notation, is a mathematical abbreviation for writing very large (or small) numbers
EXMP: 150,000,000
1.5 X 10^8

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2 years ago

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Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

Option d: 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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3 years ago

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3 years ago

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