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2 years ago

Gasoline has a density of 0.740 g/ml. if you have 328 grams of gasoline, what is the volume in milliliters?

Chemistry

1 answer:

Ivahew [28]2 years ago

5 0

Answer: Volume of the gasoline measured is 443.24mL.

Explanation:

To calculate the density of the substance, we use the formula:

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

We are given:

Density of the gasoline = 0.740g/mL

Mass of the gasoline = 328 grams

Volume of the gasoline = ? mL

Putting values in above equation, we get:

$0.740g/mL=\frac{328g}{\text{volume}}\\\\V=443.24mL$

Hence, volume of the gasoline measured is 443.24mL.

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PLEASE ASSIST MEEEEEEE!!!!!!!!! 40 POINTS

lana66690 [7]

Answer:

Wavelength of this beam of light: $\rm 4.39\times 10^{-7}\; m$ .

Explanation:

The speed of light in vacuum is approximately $\rm 2.998\times 10^{8}\;m \cdot s^{-1}$ .

Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.

On the other hand, the frequency of a wave is the number of periods in unit time. $1\rm \; Hz$ means one oscillation per second. The frequency of this particular wave is $\rm 6.83\times 10^{14}\; Hz$ . In other words, there are $6.83\times 10^{14}$ oscillations in each second.

The period of oscillation will be equal to

$\displaystyle t = \frac{1}{f} = \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}}$ .

In that period of time, a beam of light in vacuum would have traveled

$\displaystyle \rm 2.998\times 10^{8}\; m\cdot s^{-1} \times \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}} = 4.39\times 10^{-7}\; m$ .

In other words, if this beam of light of frequency $\rm 6.83\times 10^{14}\; Hz$ is in vacuum, its wavelength will be equal to $\rm 4.39\times 10^{-7}\; m$ .

8 0

2 years ago

If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g)

Allisa [31]

Answer:

The correct answer would be : 33.8 g

Explanation:

Molar mass of ammonia,

Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)

= 1*14.01 + 3*1.008 = 17.034 g/mol

mass(NH3)= 25.0 g (given)

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(25.0 g)/(17.034 g/mol)

= 1.468 mol

Now,

Molar mass of O2

= 32 g/mol

mass(O2)= 45.0 g

similar as ammonia

n (O2)=(45.0 g)/(32 g/mol)

= 1.406 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

1.83456 mol of O2 is required for 1.46765 mol of NH3

by the calculation we have only 1.40625 mol of O2

Thus, the limiting agent will be - O2

now the Molar mass of NO,

= 1*14.01 + 1*16.0

= 30.01 g/mol (similar formula used for NH3)

Balanced equation :

mol of NO formed = (4/5)* moles of O2

= (4/5)×1.40625 (from above calculation)

= 1.125 mol

mass of NO = number of moles × molar mass

= 1.125*30.01

= 33.8 g

Thus, the correct answer would be : 33.8 g

5 0

2 years ago

How many lead (Pb) atoms will be generated when 5.38 moles of ammonia react according to the following equation: 3PbO+2NH3→3Pb+N

jasenka [17]

Answer:

4.86×10^23 molecule of Pb

Explanation:

Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.

So:

2 mol NH3/ 3 mol Pb

Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:

(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb

Then, we just need to use Avagadro's number to get the number of molecules.

(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb

4 0

2 years ago

Read 2 more answers

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition, the student measures out 5.8

ra1l [238]

Answer:

1. Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

2. Volume of 1.0M AgNO₃

41ml

Explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

(aq) for ions in aqueous solution
(s) for solid compounds or elements
(g) for gaseous compounds or elements

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s) ← answer

2. How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

number of moles = mass in grams / molar mass

molar mass of AgCl = 143.32g/mol

number of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

molarity = number of moles of solute / volume of solution in liters

1.0M = 0.040748mol / V

V = 0.040748mol / (1.0M) = 0.040748 liter

V = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer

4 0

2 years ago

A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall.How tall is the actual car?

Xelga [282]

let the actual height of car be x

now, according to question,

$\dfrac{1}{15} = \dfrac{8}{x}$

$x = 15 \times 8$

$x = 120 \: \: cm$

$height \: \: of \: \: car = 120 \: cm \: \: or \: \: 1.2 \: \: m$

5 0

1 year ago