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3 years ago

How many moles of CaCI are present in 60.0mL of a 0.10 M CaCI2 solution?

Chemistry

1 answer:

balandron [24]3 years ago

4 0

Answer:

0.006 mol of CaCl2 are present

Explanation:

Use dimensional analysis for this:

60.0 mL CaCl2/1 * 1 L/1000 mL * 0.10 mol/1 L = 0.006 mol

If you were to write this as fractions out on paper, you'll see that all of the units will cancel out and you'll be left with moles of CaCl2.

**REMEMBER** that molarity (M) is talking about how many moles of that substance is present in one Liter of solvent.

Hope this helps :)

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A manganese electrode was oxidized electrically. If the mass of the electrode decreased by 225 mg during the passage of 1580 cou

qwelly [4]

Answer:

<u>Oxidation state of Mn = +4</u>

Explanation:

Atomic mass of Mn = 55g/mol

From Faraday's law of electrolysis,

Electrochemical equivalent = $\frac{mass}{charge}$

i.e Z = $\frac{m}{Q}$ = $\frac{0.225g}{1580C}$ = 0.0001424 g/C

But Equivalent weight, E = atomic mass ÷ valency = Z × 96,485

⇒ $\frac{55}{valency}$ = 0.0001424 × 96,485

<u>∴ Valency of Mn = +4</u>

8 0

2 years ago

A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.

MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0

2 years ago

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

4 0

3 years ago

Odor extensive or intensive

slava [35]

Answer:

Odour is intensive

Explanation:

intensive properties are Independent of the amount of matter

6 0

3 years ago

Please Help me on this question

kirza4 [7]

Acids: taste sour, has a pH less than 7
bases: tastes bitter, has a pH greater than 7

6 0

3 years ago