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WITCHER [35]
3 years ago
10

Claim on outer space

Chemistry
1 answer:
konstantin123 [22]3 years ago
6 0
Claim on outer space
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How many gallons of 80% antifreeze solution must be mixed with 60 gallons of 20% antifreeze to get a mixture that is 70% antifre
Anarel [89]
Let A be the 80% solution and B be the 20% solution and P be the produce solution of 70%. Va and Vb and Vp are the volumes of A and B and P respectively.
Va + 60 = Vp
0.7Vp = 0.8Va + 0.2(60)
Substituting the value of Vp from the first equation:
0.7(Va + 60) = 0.8Va + 12
30 = 0.1Va
Va = 300 gallons
6 0
3 years ago
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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
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Which Florida law requires owners and operators of motor vehicles to be financially responsible for damages and/or injuries they
s344n2d4d5 [400]

The answer in the given sentence above is the financial responsibility law as this law provides operators and owners to be held liable of the damages that they had provided or injuries that has proven to be their fault and they would likely provide financial support or payment to those affected.

3 0
3 years ago
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At 50 degrees celcius and standard pressure intermolecular forces of attraction are strongest in a sample of
Ivanshal [37]
At 50 degrees Celsius and standard pressure inter-molecular forces of attraction are strongest in a sample of ethanoic acid. 
Ethanoic acid has hydrogen atom bonded with a more electronegative atom; Oxygen. As a result, the molecule possesses strong intermolecular Hydrogen Bonds. Therefore; ethanoic acid, and all other carboxyllic acids have the tendency to form dimers.  
4 0
3 years ago
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Combustion analysis of toluene, a common organic solvent, gives 3.52 mg of co2 and 0.822 mg of h2o. if the compound contains onl
IRISSAK [1]
<span>C7H8 First, lookup the atomic weight of all involved elements Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Then calculate the molar masses of CO2 and H2O Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now calculate the number of moles of each product obtained Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK. moles CO2 = 3.52 / 44.0087 = 0.079984 moles H2O = 0.822 / 18.01488 = 0.045629 Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen. moles C = 0.079984 moles H = 0.045629 * 2 = 0.091258 So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting 1: 1.140953 The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like) 1 ( 1 : 1.140953) = 1 : 1.140953 2 ( 1 : 1.140953) = 2 : 2.281906 3 ( 1 : 1.140953) = 3 : 3.422859 4 ( 1 : 1.140953) = 4 : 4.563812 5 ( 1 : 1.140953) = 5 : 5.704765 6 ( 1 : 1.140953) = 6 : 6.845718 7 ( 1 : 1.140953) = 7 : 7.986671 8 ( 1 : 1.140953) = 8 : 9.127624 That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>
7 0
4 years ago
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