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3 years ago

4. How many milligrams are in 5.25 x 10-13 kg? the “-13” is an exponent

Chemistry

1 answer:

rusak2 [61]3 years ago

8 0

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

Given mass = 5.25 x 10⁻¹³kg

The kilo- is a prefix that denotes 10³

therefore;

1000g = 1kilogram

the milli- is a prefix that denotes 10⁻⁻³

1000mg = 1g

Now that we know this, we can convert:

5.25 x 10⁻¹³kg x $\frac{1000g}{1kg} x \frac{1000mg}{1g}$ = 5. 25 x 10⁻¹³ x 10⁶mg

= 5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

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A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0°C. What is the new

dolphi86 [110]

Answer: The new pressure of the gas in Pa is 388462

Explanation:

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The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas at STP = $10^5Pa$

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 700.0 ml

$V_2$ = final volume of gas = 200.0 ml

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$T_2$ = final temperature of gas = $30^oC=273+30=303K$

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

$\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}$

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3 years ago

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

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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)

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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

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We are given:

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

Putting values in equation 1, we get:

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

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We are given:

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

Putting values in equation 1, we get:

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Change in temperature = 34.04°C

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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

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