All categories

3 years ago

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

You might be interested in

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]

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

Explanation:

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,
n = 46.875%

Answer: 48.87%

5 0

3 years ago

Ions stick together with A . Covalent bonds B. Negative bonds 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]

-1602 kj

Sauce:me

Your welcome

8 0

3 years ago

Read 2 more answers