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a_sh-v [17]
3 years ago
5

In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical ch

emist receives a sample of groundwater with a measured volume of 79.0 mL.
Calculate the maximum mass in micrograms of acetone which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Round your answer to significant digits.
Chemistry
1 answer:
siniylev [52]3 years ago
3 0

Answer:

m = 4.7 μg

Explanation:

Given data:

density of acetone = 60.0  μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0  μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7  μg

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A 7.83 g sample of hcn contains 0.290 g of h and 4.06 g of n. find the mass of carbon in a sample of hcn with a mass of 3.37 g.
xeze [42]
To obey the Law of Conservation of Mass, the sum of all individual elements of a compound is equal to the mass of the compound. So, if HCN has a mass of 7.83 grams, then

7.83 g = mass of H + mass of C + mass of N

We know the masses of H and N to be 0.290 g and 4.06 g, respectively. Hence, we can find for the mass of C:

7.83 = 0.29 + mass of C + 4.06
mass of C = 3.48 g

As an extension to the Law of Conservation of Mass, there is also a Law of Definite Proportions. According to Dalton's atomic theory, a compound is formed from a fixed ratio of its individual elements. From our previous calculations, we know that the mass ratio of H to C to N is 0.29 g: 3.48 g:4.06 grams. The ratio could also be expressed in percentages. Let's find the mass percentage of Carbon in HCN to be used later:

mass % of Carbon = (3.48 g/7.83 g)*100
mass % of Carbon = 44.44%

So, if you collect a different mass of HCN, say 3.37 g, the corresponding mass of Carbon is equal to:

Mass of Carbon = (3.37)(44.44%)
Mass of Carbon = 1.498 g
3 0
3 years ago
15.True or False:
Nadusha1986 [10]
True
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2 years ago
When 12 moles of o2 react with 1.1 mole of c10h8 what is the limiting reactant?
Trava [24]
1.1   moles   of C10H8  is the  limiting  reagent  in   the  reaction between   reaction  C10H8   and   O2
.
C10H8  +  12O2  ---->  10CO2   + 4H2O

C10H8  is  the  limiting   reagent   since   1.1  moles  of  C10H8  is  totally   consumed   during  the  reaction
6 0
3 years ago
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Gaseous ethane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . What is the theoretical yiel
Effectus [21]

Answer:

1) Write the balanced equation:

2C2H6 + 7O2 ---> 4CO2 + 6H2O

2) Determine limiting reagent:

C2H6 ⇒ 13.8 g / 30.0694 g/mol = 0.45894 mol

O2 ⇒ 45.8 g / 31.9988 g/mol = 1.4313 mol

C2H6 ⇒ 0.45894 / 2 = 0.22947

O2 ⇒ 1.4313 / 7 = 0.20447

Oxygen is limiting.

3) Determine theoretical yield of water:

The oxygen : water molar ratio is 7 : 6

7 is to 6 as 1.4313 mol is to x

x = 1.2268286 mol of water

4) Convert moles of water to grams:

1.2268286 mol times 18.015 g/mol = 22.1 g (to three sig figs)

Solution to (b):

14.2 g / 22.1 g = 64.2%

Explanation:

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3 years ago
Which of the following is an example of how a scientist might use a model?
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I do believe the answer is:D.

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