All categories

2 years ago

4) Ethanol has a density of 0.8 g/cm3. What is the mass of 225 cm3 of ethanol

Chemistry

1 answer:

AveGali [126]2 years ago

7 0

Answer:

$\huge\boxed{\sf m = 180\ g}$

Explanation:

Given:

Density = ρ = 0.8 g/cm³

Volume = V = 225 cm³

Required:

Mass = m = ?

Formula:

m = ρV

Solution:

m = (0.8)(225)

m = 180 g

$\rule[225]{225}{2}$

Hope this helped!

<h2>~AnonymousHelper1807</h2>

You might be interested in

If 1.20 moles of an ideal gas occupy a volume of 18.2 l at a pressure of 1.80 atm, what is the temperature of the gas, in degree

scoundrel [369]

We can calculate for temperature by assuming the equation for ideal gas law:

P V = n R T

Where,

P = pressure = 1.80 atm

V = volume = 18.2 L

n = number of moles = 1.20 moles

R = gas constant = 0.08205746 L atm / mol K

Substituting to the given equation:

T = P V / n R

T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm / mol K)

T = 332.70 K

We can convert K unit to ˚C unit by subtracting 273.15 to Kelvin, therefore

T = 59.55 ˚C

6 0

2 years ago

Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).

mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

CuSO₄.5H₂O → CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

Number of moles = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

Volume = 250 x 10⁻³L

Input the parameters and solve for number of moles;

Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol

From the equation;

1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O

So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

= 0.5 x 249.6

= 124.8g

The mass of CuSO₄.5H₂O is 124.8g

5 0

3 years ago

Define position and reference point in your own words. Dont copy and paste!

irga5000 [103]

Answer: The beginning stage you decide to depict the area, or position, of an object.

Explanation: starting point or position

4 0

2 years ago

Can someone help please

Serhud [2]

Let the mass be X g

percentage = X/ 6.50 * 100 =2.2%

X= 0.143 g

The mass is 0.143 g

8 0

2 years ago

Read 2 more answers

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0

2 years ago