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docker41 [41]
3 years ago
5

Helium has a density of 1.79 x 10-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate t

he mass of helium that it would take to fill the balloon. Be sure to follow significant figure rules when calculating the answer.
Chemistry
2 answers:
iris [78.8K]3 years ago
7 0

 The mass    of  Helium that it would take  to fill the balloon is  1.1277 grams  of helium

<u><em>calculation</em></u>

<u> </u> mass  = density  x volume

Density  = 1.79  x 10 ⁻⁴ g/ml

Volume  = 6.3 L

  • convert  volume  into ml to make  the unit uniform

 that is  1 L = 1000 ml

             6.3 L =? L

by cross  multiplication

= ( 6.3 L x 1000 ml) / 1 L = 6300 ml


mass  is therefore  = 1.79 x 10⁻⁴ g/ml x 6300  ml= 1.1277 grams

zloy xaker [14]3 years ago
4 0
Density = 1.79x10⁻⁴ g/mL

1  L = 1000 mL

6.3 L = ?  

6.3 x 1000 => 6300 mL

Mass of helium:

m = D x V  

Therefore:

m = ( 1.79x10⁻⁴ ) x  6300

m = 1.1277 g of helium


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WARRIOR [948]
To answer the question above, let us a basis of the 1000 mL or 1 L. 
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Then, determine the mass of the alcohol by multiplying the total mass by the decimal equivalent of 5%. 
               mass of alcohol = 0.05(992.8 g) = 49.64 g
Then, determine the number of moles of ethyl alcohol by dividing the mass of alcohol by the molar mass (46 g/mol). 
                       n = 49.64 g/ (46 g/mol) = 1.08 mol
Then, divide the number of moles by the volume (our basis is 1 L)
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3 years ago
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So, the answer to 27.) would be <em>2x.</em> Both 6x and 2x can be divided by 2x, but they can't go any higher without the end-answer becoming a fraction. As such, 2x is the greatest common factor.

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3 0
3 years ago
Examine the photo of quartz below in which way does the quartz break
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3 years ago
A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
Leya [2.2K]

Answer:

See explanation

Explanation:

The molecular equation shows all the compounds involved in the reaction.

The molecular equation is as follows;

2NaF(aq) + Pb(NO3)2(aq) -------> PbF2(s) + 2NaNO3(aq)

The complete ionic equation shows all the ions involved in the reaction

The complete ionic equation;

2Na^+(aq) + 2F^-(aq) + Pb^2+(aq) + 2NO3^-(aq) -------->PbF(s) + 2Na^+(aq) +2NO3^-(aq)

The net Ionic equation shows the ions that actually participated in the reaction

The net ionic equation is;

2F^-(aq) + Pb^2+(aq)--------> PbF(s)

3 0
3 years ago
Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.
BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>

Now we can find percentage composition / percentage by mass of oxygen.

% composition = \frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound} × 100

% composition = \frac{144}{240.9} × 100 = <u>59.776%</u>

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0
2 years ago
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