Can anybody help please

We want to find how much charge is on the electrons in a nickel coin. follow this method. a nickel coin has a mass of about 5 g.

Lady bird [3.3K]

The mass of a nickel coin is 5 g.
1 mol of Ni weighs 58 g. 1 mol contains 6.022 x 10²³ atoms of Ni.
therefore in 58 g there are 6.022 x 10²³ atoms of Ni
then in 5 g the number of Ni atoms are - 6.022 x 10²³ /58 x 5 = 5.2 x 10²² Ni atoms
Therefore number of Ni atoms are 5.2 x 10²² atoms in a nickel coin

3 0

3 years ago

10 points. Please help.

ratelena [41]

Answer:

-191.7°C

Explanation:

P . V = n . R . T

That's the Ideal Gases Law. It can be useful to solve the question.

We replace data:

2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°

(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°

T° = 81.3 K

We convert T° from K to C°

81.3K - 273 = -191.7°C

6 0

3 years ago

Read 2 more answers

3.2 moles of H3PO4 to grams

Nimfa-mama [501]

Answer:

313, 6grams of H3PO4

Explanation:

We calculate the weight of 1 mol of H3PO4:

Weight 1 mol H3PO4= (Weight H)x3+ (Weight P)+(Weight 0)x4 =1gx3+31g+16gx4

Weight 1 mol H3PO4=98 g /mol

1 mol-----98 grams H3PO4

3,2mol----x= (3,2molx 98 grams H3PO4)/ 1mol=313,6 grams H3PO4

4 0

3 years ago

It takes to break an iodine-iodine single bond. Calculate the maximum wavelength of light for which an iodine-iodine single bond

Zolol [24]

The given question is incomplete. The complete question is :

It takes 151 kJ/mol to break an iodine-iodine single bond. Calculate the maximum wavelength of light for which an iodine-iodine single bond could be broken by absorbing a single photon. Be sure your answer has the correct number of significant digits.

Answer: 793 nm

Explanation:

The relation between energy and wavelength of light is given by Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

E = energy of the light = 151 kJ= 151000 J (1kJ=1000J)

N= moles = 1 = $6.023\times 10^{23}$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light = ?

Putting in the values:

$151000J=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda}$

${\lambda}=7.93\times 10^{-7}m=793nm$ $1m=10^{-9}nm$

Thus the maximum wavelength of light for which an iodine-iodine single bond could be broken by absorbing a single photon is 793 nm

3 0

3 years ago

If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti

ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0

3 years ago