All categories

1 year ago

Calculate the mass of olive oil in a 750mL bottle if the density of olive oil is 0.92 g/mL. *Use the GRASP method.

Chemistry

1 answer:

kap26 [50]1 year ago

8 0

The mass of olive oil : 690 g

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

Given

volume of olive oil = V=750 ml

density of olive oil = ρ = 0.92 g/ml

Required

the mass of olive oil

Analysis

the formula :

$\tt m=\rho\times V$

Solution

$\tt m=0.92\times 750\\\\m=690~g$

Paraphrase

the mass of olive oil : 690 g

You might be interested in

A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.

solong [7]

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al + 3Cl₂ → 2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

Cl₂ : AlCl₃

3 : 2

0.06 : 2/3×0.06 = 0.04

Al : AlCl₃

2 : 2

0.10 : 0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

4 0

1 year ago

The temperature on a distant, undiscovered planet is expressed in degrees B. For example, water boils at 180 ∘ B and freezes at

marin [14]

Answer:

40.3∘C

Explanation:

At planet B;

Water boils = 180∘C

Water freezes = 50∘C

In this planet the temperature difference = 180 - 50 = 130 compared to earth where the temperature difference is; 100 - 0 = 100

This means;

130 ∘C = 100 ∘C

x ∘C = 31 ∘C

x = 31 * 130 / 100

x = 40.3∘C

5 0

2 years ago

Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 <em> initial partial pressure</em>

-x -x +x +x <em> react</em>

105.1-x 7.96-x x x <em>equilibrium</em>

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

Read 2 more answers

Match each situation below to a letter on the illustration.

Mnenie [13.5K]

Answer:a

Explanation:

4 0

1 year ago

Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0

ExtremeBDS [4]

Answer:

The freezing point will be $2.9^{0}C$

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

$Depressioninfreezing point=K_{f}Xmolality$

Where

Kf= $20\frac{^{0}C}{m}$

the molality is calculated as:

$molality=\frac{moles_{solute}}{mass_{solvent}}$

$moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol$

$massofcyclohexane=25g=0.025Kg$

$molarity=\frac{0.0045}{0.025}=0.18m$

Depression in freezing point = $20X0.18=3.6^{0}C$

The new freezing point = $6.5^{0}C-3.6^{0}C=2.9^{0}C$

5 0

2 years ago