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Nuetrik [128]
3 years ago
7

platinum has a mass number of 195 and contain 74 electron. how many proton and neutrons does it contain?​

Chemistry
1 answer:
yuradex [85]3 years ago
4 0

Answer:

...

Explanation:

p is equal to atomic number so p=74

electron also equal to atomic number if it doesn't have charge so electron =74

neuralon=atomic mass -atomic number so 195-74=121is neutron

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The inner part of the ovary that contains an egg is the ovule<br> true or false
tigry1 [53]

Answer:

The answer is true

Explanation:

the ovary is a tubular structure and contains ovules

3 0
2 years ago
A photon has an energy of 2.78 x 1032 J. What is its wavelength?
8090 [49]

Answer:

7.14 × 10^(-58) m

Explanation:

We are given;

Energy; E = 2.78 x 10^(32) J

Formula to find this energy is;

E = hc/λ

Where;

E is energy

h is Planck's constant with a constant value of 6.62 × 10^(-34) J/s

c is speed of light with a constant value of 3 × 10^(8) m/s

λ is wavelength

Making λ the subject, we have;

λ = hc/E

λ = (6.62 × 10^(-34) × 3 × 10^(8))/(2.78 x 10^(32))

λ = 7.14 × 10^(-58) m

3 0
3 years ago
A mining company wants to build a modern processing plant that is safe for workers.
777dan777 [17]
To determine the level of radionuclides present 
3 0
4 years ago
A geochemist measures the concentration of salt dissolved in Lake Parsons and finds a concentration of 21.0 gL⁻¹. The geochemist
dedylja [7]

Answer:

The percentage of Lake Parsons which has evaporated since it became isolated is 68.24%.

Explanation:

The concentration of salt dissolved in Lake Parson = 21.0 g/L

= In 1 L of lake Parson  water = 21.0 g of salt

The concentration of salt in several nearby non-isolated lakes = 6.67  g/L

The salt concentration in Lake Parsons before it became isolated = 6.67 g/L

In 1 L of lake water = 6.67 g of salt

If 1 L of lake Parson  water has 6.67 g of salt. Then 21.0 grams of salt will be in ;

\frac{1}{6.67}\times 21.0 L = 3.15 L

Water evaporated = 3.15 L - 1 L = 2.15 L

The percentage of Lake Parsons which has evaporated since it became isolated:

\frac{2.15 L}{3.15 L}\times 100=68.25\%

5 0
4 years ago
Water heater contains 56 part a how many kilowatt-hours of energy are necessary to heat the water in the water heater by 25 ?c?
DochEvi [55]

Answer is: 550,021 kWh of energy is needed to heat the water 

V(water) = 51 gal = 51·3,78 = 189,3 L.

ΔT(water) = 25°C.

d(water) = 1000 g/L.

m(water) = V(water) · d(water)

m(water) = 189,3 L · 1000 g/L

m(water) = 189300 g.

Q = m(water) · ΔT(water) · C(water)

Q = 189300 g · 25°C · 4,184 J/°C·g

Q = 19800780 J = 19800,78 kJ ÷ 3600 = 550,021 kWh.



6 0
3 years ago
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