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Orlov [11]
3 years ago
9

Use exponential (scientific) notation to express the value 385,500 to 1 s.f.

Chemistry
1 answer:
lara31 [8.8K]3 years ago
7 0

To express the value 385 500 in scientific notation with 1 significant figure we have to move the decimal point to the left until we're left with a number between 1 and 10:

385500.0

38550.00 (1 space)

3855.000 (2 spaces)

385.5000 (3 spaces)

38.55000 (4 spaces)

3.855000 (5 spaces)

The number of spaces we moved to the left, is going to be the exponent of the number 10 that we are going to multiply to that number

3.855000*10⁵

Finally, we round the number so we have only one number different from 0 and this is the number expressed with scientific notation to 1 significant figure:

<u>4.0 * 10⁵</u>

<u></u>

Have a nice day!

#LearnwithBrainly

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Answer: See below

Explanation:

1. To calculate the mass, you know you can convert by using molar mass. Since mass is in grams, we can use molar mass to convert moles to grams. This calls for the Ideal Gas Law.

Ideal Gas Law: PV=nRT

We manipulate the equation so that we are solving for moles, then convert moles to grams.

n=PV/RT

P= 100 kPa

V= 0.831 L

R= 8.31 kPa*L/mol*K

T= 27°C+273= 300 K

Now that we have our values listed, we can plug in to find moles.

n=\frac{(100kPa)(0.831L)}{(8.31kPa*l/molK)(300K)}

n=0.033mol

We use the molar mass of NO₂ to find grams.

0.033mol*\frac{46.005g}{1mol }=1.52 g

The mass is 1.52 g.

2. To calculate the temperature, we need to use the Ideal Gas Law.

Ideal Gas Law: PV=nRT

We can manipulate the equation so that we are solving for temperature.

T=PV/nR

P= 700.0 kPa

V= 33.2 L

R= 8.31 kPa*L/mol*K

n= 70 mol

Now that we have our values, we can plug in and solve for temperature.

T=\frac{(700kPa)(33.2L)}{(70mol)(8.31 kPa*L/molK)}

T=40K

The temperature is 40 K.

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3 years ago
What will happen when the rates of evaporation and condensation are equal​
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A sample of nitrogen gas contains 9.46x10^23 atoms of nitrogen. How much nitrogen is expressed in grams? round your answer to th
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  • No of. atoms=9.46×10^23

We know

\boxed{\star{\sf No\;of\:atoms=No\:of\: moles\times Avagadro's\: number}}

\\ \rm\Rrightarrow No\:of\;moles=\dfrac{No\:of\:atoms}{Avagadro's\:no}

\\ \rm\Rrightarrow No\:of\:moles =\dfrac{9.46\times 10^{23}}{6.022\times 10^{23}}

\\ \rm\Rrightarrow No\:of\:moles=1.6mol

Again

\boxed{\star{\sf Mass=Molar\:mass\times No\:of\:moles}}

\\ \rm\Rrightarrow Mass=1.6(14)

\\ \rm\Rrightarrow Mass=22.4g

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