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9966 [12]
3 years ago
9

A gas can holds 2.9 gal of gasoline. what is the quantity in cubic centimeters

Chemistry
1 answer:
mash [69]3 years ago
3 0
1 gal ---------- 3785.41 cm³
2.9 gal --------- ??

2.9 x 3785.41 / 1 => 10977.689 cm³
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Element M has two isotopes,^104M and ^106M.
snow_lady [41]
To calculate atomic mass, you have to take to weighted average of the isotopes' masses. What that means is M = RA*106 + (1 – RA)*104, where RA is relative abundance expressed in decimal form. If you simplify the right side of that equation, you get M = 2*RA + 104. Doing a little more algebra yields RA = (M –104)/2 = (104.4 – 104)/2 = 0.4 / 2 = 0.2, which is 20%. So the answer is B.
3 0
3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0
3 years ago
A reaction produces 10.5 L if oxygen, but was supposed to produce 1 mol of oxygen. What is the percent yield ? (One mole of any
Rzqust [24]
To answer this item, we assume that oxygen behaves ideally such that it is able to fulfill the following equation,

 PV = nRT

If we are to retain constant the variable n and V. 

The percent yield can therefore be solved through the following calculation,

   n = (10.5 L)/(22.4 L)   x 100%

Simplifying,
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Answer: 48.87%
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3 years ago
Ions stick together with<br> A . Covalent bonds <br> B. Negative bonds<br> C. Ionic bonds
zysi [14]
The answer is C ionic bonds
8 0
3 years ago
How much heat (in kj) is released when 3.600 mol naoh(s) is dissolved in water? (the molar heat of solution of naoh is â445.1 kj
Fantom [35]
<span>-1602 kj

Sauce:me

Your welcome
</span>
8 0
3 years ago
Read 2 more answers
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