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KATRIN_1
2 months ago
14

After heating the solution, you decant the liquid and wash the remaining solid with distilled water. What you are removing by th

is washing?
Chemistry
1 answer:
lorasvet [2.7K]2 months ago
7 0
What is being removed during the wash is the solvent.
You might be interested in
Next we need to determine the mass of oleic acid in the monolayer. The concentration of the oleic acid/benzene solution is 0.02g
Alekssandra [3086]

Response:

m=1x10^{-6}g

Clarification:

Hello,

In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:

m=0.02\frac{g}{L}*0.05mL*\frac{1L}{1000mL}\\ \\m=1x10^{-6}g

Best wishes.

3 0
3 months ago
A container is filled with 4.0 g and 5.0 g 02. The mixture is ignited produced water how much water is produced
VMariaS [2998]
The chemical equation can be expressed as:

2H2 + O2 = 2H2O

Given the amounts of the reactants, we need to identify the limiting reactant before calculating the amount of product generated.

4.0 g H2 ( 1 mol / 2.02 g ) = 1.98 mol H2
5.0 g O2 ( 1 mol / 32 g ) = 0.1563 mol O2

The limiting reactant is O2, as it will be fully consumed in the reaction.

0.1563 mol O2 ( 2 mol H2O / 1 mol O2 ) ( 18.02 g / mol ) = 5.6 g H2O will be produced
6 0
2 months ago
13.3 g of benzene (C6H6) is dissolved in 282 g of carbon tetrachloride. What is the molal concentration of benzene in this solut
Tems11 [2777]

Answer:

0.605 molal

Explanation:

Molality indicates the solute amount in a specific solvent mass.

Let’s find the amount of benzene solute.

Mass of benzene = 13.3g

Molar mass of C6H6 = 12*6 +1*6 =72+7=78g/mol

Amount of benzene = mass/molar mass

                           =13.3/78

                          =0.1705mol

Molality = amount of solute/mass of solvent in kg

Mass of solvent = 282g = 0.282kg

Molality = 0.1705/0.282

    =0.605 molal

6 0
2 months ago
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
KiRa [2933]

Response:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Clarification:

Weight of the alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Weight of the water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

m_{a} C_{a} [[T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This gives us the specific heat of the alloy.

4 0
2 months ago
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