All categories

3 years ago

What is the density of 35 ml of a liquid that has a mass of 28.7 g?

Chemistry

2 answers:

omeli [17]3 years ago

4 0

Hey there!:

Mass = 28.7 g

Volume = 35 mL

Therefore:

D = m / V

D = 28.7 / 35

D = 0.82 g/mL

Nadya [2.5K]3 years ago

3 0

Answer: The density of the liquid will be 0.82 g/ml.

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of liquid = 28.7 grams

Volume of liquid = 35 ml

Putting in the values we get:

$Density=\frac{28.7g}{35ml}=0.82g/ml$

Thus density of the liquid will be $0.82g/ml$

You might be interested in

8. When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. I

Tamiku [17]

Answer:

The boiling point of the water by adding 2.5 moles of aluminium nitrate will be changed by 4° C.

Explanation:

Step 1: define the formula for an elevation of the boiling point

Δ T b = i *K b *bB

⇒with Δ T b = the elevation of the boiling point ( in °C or Kelvin)

⇒with i = van't Hoff i -factor for the solute. The factor i shows the number of individual particles (typically ions) that are formed by a compound in the solution.

⇒ with Kb = ebullioscopic constant, which depends on the properties of the solvent. (Eventually can be calculated).

⇒ with bB= the molal concentration (molality) of the solute.

Since we have 2 different solutions ( component 1 = sugar and component 2 = aluminium nitrate) of the same molality in the same solvent ( water). We can express this as followed:

Δ T b,2 / Δ T b,1 = (i2 * Kb * bB ) / (i1 * Kb *bB)

⇒after he was simplifying this becomes:

Δ T b,2 / Δ T b,1 = i2 / i1

⇒ Now we can isolate either Δ T b,2 or Δ T b,1:

Δ T b,2 =(i2/i1) x Δ T b,1

Sugar

⇒ is a covalent compound, so it doesn't dissociate in water ⇔ i1=1

Aluminium nitrate

⇒ it's a soluble ionic compound and in solution it will dissociate as the following equation:

Al(NO3)3 (aq) → Al3+(aq) + 3NO3- (aq)

⇒thus here are 4 particels formed : 1 Al3+ + 3NO3-

i2=4

As given, we also know that the temperature was raised by 1°C after adding sugar ⇒ Δ T b,1 = 1°C

Step 2: insert all the numbers in the formula for boiling point elevation

(i2/i1) x Δ T b,1

(4/1 ) x 1°C = 4°C

⇒ The boiling point of the water by adding 2.5 moles of aluminium nitrate will be changed by 4°C

6 0

2 years ago

Is gold amalgamation a chemical reaction with gold and mercury?

iren [92.7K]

Answer:

gold amalgamation is the mixing of gold and Mercury.

4 0

3 years ago

an aqueous solution has a mass of 490 grams containing 8.5×5103 gram of calcium ions. the concentration of calcium ions is in th

Alex777 [14]

There are several various of expressing concentration. For instance, mass percent, volume percent, Molarity, Normality, Molality, etc.

In present case, weight of solute and solvent are given, so it will be convenient to express concentration in terms of mass percent.

Given: weight of solute (Ca2+) = 8500 g
weight of solvent (water) = 490 g.
Therefore, mass of solution = 8500 + 490 = 8990 g

Now, mass percent = $\frac{\tect{Weight of solute}}{\text{Weight of solution}} X 100$
= $\frac{\tect{8500}}{\text{8990}} X 100$
= 94.55 %

Answer: Concentration of calcium ions is in this solution is 94.55 % (w/W)

8 0

3 years ago

Sodium metal and water react to form hydrogen and sodium hydroxide. If 1.99 g of sodium react with water to form 0.087 g of hydr

zavuch27 [327]

Answer:

1.56 g of water was involved in the reaction

Explanation:

From the stoichiometric equation

2Na + 2H2O = 2NaOH + H2

NB : Mm Na= 23, Mm H2O = ( 2+16)= 18

2(23) of Na requires 2(18) of water

Hence 1.99g of Na will require 1.99×2×18/2(23) of water = 1.56 g of water

7 0

3 years ago

Read 2 more answers

What is the value of for this aqueous reaction at 298 K? A+B↽⇀C+D ΔG°=12.86 kJ/mol K=

kogti [31]

Answer:

Kc = 0.5951 (4 sig. figs.)

Explanation:

For A + B ⇄ C + D at standard thermodynamic conditions (298K, 1atm)

ΔG = ΔG° + R·T·lnQ => 0 = ΔG° + R·T·lnKc => ΔG° = - R·T·lnKc

=> lnKc = - ΔG°/R·T

ΔG° = +12.86 Kj/mol

R = 8.314 Kj/mol·K

T = 298K

lnKc = - (+12.86Kj) / (8.314Kj/mol·K)(298K) = - 0.519 mol⁻¹

Kc = e⁻⁰°⁵¹⁹ mol⁻¹ = 0.5957 mol⁻¹ (4 sig. figs.)

5 0

3 years ago