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likoan [24]
2 years ago
10

1. What is the density, in g/cm², of a substance weighing 22.2 cg with dimensions of 0.333

Chemistry
1 answer:
RUDIKE [14]2 years ago
7 0
  • Mass=22.2cg=2.22g

Volume:-

\\ \tt\longmapsto 0.333(0.444)(0.555)=0.08cm^3

So

\\ \tt\longmapsto Density=\dfrac{Mass}{Volume}

\\ \tt\longmapsto Density=\dfrac{2.22}{0.08}=27.75g/cm^3

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Answer:

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How does a liquid substance change into gas? ​
Afina-wow [57]

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3 years ago
What is the mass of 80.0mL of ethanol at 20 degress celcius
Oxana [17]

Answer: What is the mass, in grams, of 135 mL of ethanol? d=0.789 g/mL - the ethanol density. V=135 mL - the volume of ethanol. m=0.789g/mL*135mL=106.515g ~ 106.5g- the mass of ethanol.

Explanation:

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4 0
3 years ago
A 53. 5 mL ample of an 5. 4 % (m / v) KBr olution i diluted with water o that the final volume i 205. 0 mL Expre your anwer to t
gogolik [260]

The concentration after dilution is 1.4%.

We are aware that concentration and volume are related to each other by the formula -

C_{1} V_{1} = C_{2} V_{2}, where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side.

Keep the values to calculate final concentration.

C_{2} = (53.5 × 5.4)/205.0

Performing multiplication on Right and Side

C_{2} = 288.9/205.0

Performing division on Right Hand Side

C_{2} = 1.4%

Hence, the final concentration is 1.4%.

Learn more about concentration -

brainly.com/question/17206790

#SPJ4

The complete question is -

A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.

Calculate the final concentration and express your answer to two significant figures and include the appropriate units.

3 0
1 year ago
The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?
LuckyWell [14K]
Empirical formula is the simplest ratio of whole numbers of components in a compound. 
Assuming for 100 g of the compound 
                                Cu                                 As                             S
mass                      48.41 g                          19.02 g                      32.57 g
number of moles    48.41 / 63.5 g/mol      19.02 / 75 g/mol        32.57 / 32 g/mol 
                                = 0.762 mol                = 0.2536 mol            = 1.018 mol 
divide by the least number of moles 
                               0.762 / 0.2536             0.2536 / 0.2536         1.018 / 0.2536
                               = 3.00                          = 1.00                         = 4.01
once they are rounded off 
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
8 0
3 years ago
Read 2 more answers
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