The hypothetical explosion that created the universe is called the big bash. true or false

How many moles of LiOH are required to make 4.2 liters of a 0.98 M

ASHA 777 [7]

Answer:

4.12 mol

Explanation:

Given data:

Moles of LiOH required = ?

Volume of solution = 4.2 L

Molarity of solution = 0.98 M

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

we will calculate the moles from above given formula.

0.98 M = number of moles / 4.2 L

0.98 M × 4.2 L = number of moles

Number of moles = 0.98 M × 4.2 L

Number of moles = 4.12 mol (M = mol/L)

7 0

4 years ago

Radioactive isotope has a half-life of 1.2 billion years. as measured by the presence of the isotope and its stable decay produc

Virty [35]

The radioactive isotope is 15 years old

6 0

4 years ago

What is the density of an object if its mass is 10 g and a volume of 2 mL?

Lilit [14]

Hi there! Let's solve this problem shall we!

⠀Volume = 10g

Mass = 2 mL

In this specific problem, they are asking us to find the density of the object. So, using the information given to us (volume and mass), let's solve the problem!

Now, if you remember, D = M ÷ V

So, let's fill in the blanks!

D = Our unknown value

M = 2mL

V = 10g

Here is the filled out formula:

D = M ÷ V

D = 2mL ÷ 10g

D = 5 g/mL

*Make sure you put the units for your final solution!*

4 0

2 years ago

A voltaic cell is constructed from an Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s) half-cell. The initial concentration of Ni2+

SCORPION-xisa [38]

Explanation:

For what I can see, is missing the concentration of [Ag+] in the half-cell. To calculate it:

Niquel half-cell

Oxidation reaction: $Ni \longrightarrow Ni^{2+}+2 e^-$

$E=E^0 - \frac{R*T}{n*F}*ln(1/[Ni^{2+}])$

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

$E=-0.23V - \frac{8.314*298}{2*96500}*ln(1/0.014M)$

$E=-0.285V$

Silver half-cell

Reduction reaction: $Ag^+ + e^- \longrightarrow Ag$

$E=E^0 - \frac{R*T}{n*F}*ln(1/[Ag+])$

$E_{cell}=E_{red} - E_{ox}$

$E_{red}=1.12 V + (-0.855V)=0.835V$

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

$0.835V=0.8V - \frac{8.314*298}{1*96500}*ln(1/[Ag+])$

$[Ag+]=0.26 M$

3 0

4 years ago

A common car battery consists of six identical cells each of which carries out the reaction: Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 +

docker41 [41]

Answer:

The correct answer is 4.58 grams.

Explanation:

Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).

For the reaction, n * 96500 C = molar mass

1C = molar mass/n*96500 = Equivalent wt / 96500

w = Equivalent wt / 96500 * I * t

In the given reaction,

Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.

Now putting the values in the equation we get,

w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)

w = 4.58 gm.

6 0

4 years ago