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What is the mass in grams of 7.5 mol of C8H18?

Lana71 [14]

Answer:

$\boxed {\boxed {\sf 856.74 \ g \ C_8H _{18}}}$

Explanation:

To convert from grams to moles, the molar mass is used. These values tells us the grams in 1 mole of a substance. They can be found on the Periodic Table (they are equivalent to the atomic masses, but the units are grams per mole).

We are given the compound C₈H₁₈. Look up the molar masses of the individual elements.

Carbon (C): 12.011 g/mol
Hydrogen (H): 1.008 g/mol

Notice there are subscripts that tell us the number of atoms of each element. We must multiply the molar masses by the subscripts.

C₈: 8(12.011 g/mol)=96.088 g/mol
H₁₈: 18(1.008 g/mol)=18.144 g/mol

Add these 2 values together to find the molar mass of the whole compound.

C₈H₁₈: 96.088 g/mol +18.144 g/mol=114.232 g/mol

Use this number as a ratio.

$\frac{ 114.232 \ g \ C_8H_{18}}{1 \ mol \ C_8H_{18}}$

Multiply by the given number of moles: 7.5

$7.5 \ mol \ C_8H_{18}*\frac{ 114.232 \ g \ C_8H_{18}}{1 \ mol \ C_8H_{18}}$

The moles of C₈H₁₈ will cancel each other out.

$7.5 *\frac{ 114.232 \ g \ C_8H_{18}}{1}$

$7.5 *{ 114.232 \ g \ C_8H_{18}}$

$856.74 \ g \ C_8H _{18}$

7.5 moles of C₈H₁₈ is equal to <u>856.74 grams of C₈H₁₈</u>

8 0

2 years ago

What are the characteristics of a virtual image? Check all that apply.

tia_tia [17]

<h3>Answer:</h3>

Can only be seen while looking through a lens
May appear larger than the object
Appears to come from behind the lens

<h3>Explanation:</h3>

To answer this question we need to ask ourselves;

What is a virtual image?

A virtual image is an image that can not be formed on a screen.
It is formed by rays that appear to converge at a point.

What are the characteristics of a virtual image?

A virtual image can only be seen while looking through lenses
The image appear to come from behind the lens
It may be larger than the object
It can not be projected onto a screen.
It is always upright

6 0

3 years ago

Read 2 more answers

Calculate the rate of dissolution (dM/dt) of relatively hydrophobic drug particles with a surface area of 2.5×103 cm2 and satura

Crank

Answer:

$\large \boxed{\text{1.22 mg/s}}$

Explanation:

We can use the Noyes-Whitney equation to calculate the rate of dissolution.

$\dfrac{\text{d}M}{\text{d}t} = \dfrac{DA(C_{s} - C)}{d}$

Data:

D = 1.75 × 10⁻⁷ cm²s⁻¹

A = 2.5 × 10³ cm²

Cₛ = 0.35 mg/mL

C = 2.1 × 10⁻⁴ mg/mL

d = 1.25 µm

Calculations:

Cₛ - C = (0.35 - 2.1 × 10⁻⁴) mg·cm⁻³ = 0.350 mg·cm⁻³

d = 1.25 µm = 1.25 × 10⁻⁶ m = 1.25 × 10⁻⁴ cm

$\dfrac{\text{d}M}{\text{d}t} = \dfrac{(1.75 \times 10^{-7} \text{cm}^{2}\text{s}^{-1})(2.5 \times 10^{3} \text{ cm}^{2})(0.350\text{ mg$\cdot$cm$^{-3}$})}{1.25 \times 10^{-4} \text{ cm}} = \textbf{1.22 mg/s}\\\\\text{The rate of dissolution is $\large \boxed{\textbf{1.22 mg/s}}$}$

8 0

3 years ago

What is the molarity of a NaCl solution containing 9.0<br> moles of NaCl in 3.0 L of solution?

ss7ja [257]

Answer:

3 M

Explanation:

Molarity equation: M = n/v

n = moles of solute

v = liters of solution

9 moles of NaCl / 3 L

9/3 = 3 M

4 0

3 years ago

PLZ HELP IM GIVING 50 FRICKING POINTS. NO WRONG ANSWERS PLZ

Liula [17]

Answer:

I think it's B

Explanation:

I dont have much experience with the periodic table, but I just think its B.

3 0

3 years ago