Which explanation of the solar system best fits the observations of the planets and how they orbit the sun?

Suppose a current of is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for s

WARRIOR [948]

The given question is incomplete. The complete question is:

Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.

Answer: 0.0484 g

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 0.920 A

t= time in seconds = 47.0 sec

$Q=0.920A\times 47.0s=43.24C$

$AgNO_3\rightarrow Ag^++NO_3^-$

$Ag^++e^-\rightarrow Ag$

96500 Coloumb of electricity electrolyzes 1 mole of Ag

43.24 C of electricity deposits = $\frac{1}{96500}\times 43.24=0.00045moles$ of Ag

$\text{ mass of Ag}={\text{no of moles}\times {\text{Molar mass}}=0.00045mol\times 108g=0.0484g$

Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g

6 0

2 years ago

Can you Guys please help me on 14,and 15

jok3333 [9.3K]

14. is a and i think 15. is b

7 0

2 years ago

Read 2 more answers

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

2 years ago

I’m struggling with dimensional analysis in chemistry! Will someone please help me with .74 Kcal/min to cal/sec ? Explain and gi

valina [46]

Answer:

12.33 cal/sec

Explanation:

As we know,

1 Kcal = 1000 cal

So,

0.74 Kcal = X cal

Solving for X,

X = (0.74 Kcal × 1000 cal) ÷ 1 Kcal

X = 740 cal

Also we know that,

1 Minute = 60 Seconds

Therefore, in order to derive cal/sec unit replace 0.74 Kcal by 740 cal and 1 min by 60 sec in given unit as,

= 740 cal / 60 sec

= 12.33 cal/sec

3 0

3 years ago

26.6 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL. A 16.0 mL sample of the resulting solution is then diluted to

frozen [14]

Answer:

In the final solution, the concentration of sucrose is 0.126 M

Explanation:

Hi there!

The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:

Ci · Vi = Cf · Vf

Where:

Ci = concentration of the original solution

Vi = volume of the solution taken to prepare the more diluted solution.

Cf = concentration of the more diluted solution.

Vf = volume of the more diluted solution.

For the first dillution:

26.6 ml · 2.50 M = 50.0 ml · Cf

Cf = 26.6 ml · 2.50 M / 50.0 ml

Cf = 1.33 M

For the second dilution:

16.0 ml · 1.33 M = 45.0 ml · Cf

Cf = 16.0 ml · 1.33 M / 45.0 ml

Cf = 0.473 M

For the third dilution:

20.0 ml · 0.473 M = 75.0 ml · Cf

Cf = 20.0 ml · 0.473 M / 75.0 ml

Cf = 0.126 M

In the final solution, the concentration of sucrose is 0.126 M

7 0

2 years ago