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love history [14]
3 years ago
12

Enter the formula for the polyatomic ion in MnCO3.

Chemistry
1 answer:
joja [24]3 years ago
8 0
The formula is Mn2+CO2= MnCO3
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What is the energy of an electromagnetic wave with a electromagnetic wave with a frequency of 3 * 10 Hz ?
Olin [163]

Answer:

E=1.989*10^{-32}J

Explanation:

From the question we are told that:

Frequency f=3 * 10 Hz

Generally the equation for Energy of an electromagnetic wave is mathematically given by

 E=hf

Where

  h=Planck's constant

  h=6.62607015 * 10^{-34}

Therefore

 E=6.62607015 * 10^{-34}*3 * 10

 E=1.989*10^{-32}J

8 0
3 years ago
How many electrons can be held in the third orbital?.
Evgen [1.6K]
32 electrons. as the orbitals get father away from the nucleus, they hold more electrons.
6 0
3 years ago
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1. When 50.0 mL of water at 80.0°C was mixed with 50.0 mL of water in a calorimeter at
ddd [48]

Answer:

The heat capacity of the calorimeter is 5.11 J/g°C

Explanation:

Step 1: Given data

50.0 mL of water with temperature of 80.0 °C

Specific heat capacity of water = 4.184 J/g°C

Consider the density of water = 1g/mL

50.0 mL of water in a calorimeter at 20.0 °C

Final temperature = 47.0 °C

Step 2: Calculate specific heat capacity of the water in calorimeter

Q = Q(cal) + Q(water)

Q(cal) = mass * C(cal) * ΔT

Qwater = mass * Cwater * ΔT

Qcal = -Qwater

mass(cal) * C(cal) * ΔT(cal) =  mass(water) * C(water) * ΔT(water)

50 grams * C(cal) * (47.0 - 20.0) =- 50grams * 4.184 J/g°C * (47-80)

1350 * C(cal) = 6903.6

C(cal) = 5.11 J/g°C

The heat capacity of the calorimeter is 5.11 J/g°C

8 0
3 years ago
One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable.a. 39/19K an
marishachu [46]

Answer:

a) 39/19 K : stable nuclide, 40/19 K  : radioactive nuclide.

b) 209B: stable nuclide, 208Bi : radioactive nuclide

c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.

Explanation:

As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.

Mass number (A) = Atomic number (Z) + No. of neutrons (N)

Or, N = A - Z

a)

39/19 K and 40/19 K

Calculate no. of neutrons in 39/19 K as follows:

atomic no. = 19, mass no. 39

N = 39 - 19

   = 20 (even no.)

Calculate no. of neutrons in 40/19 K as follows:

atomic no. = 19, mass no. 40

N = 40 - 19

   = 21 (odd no.)

Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive

nuclide.

b)

209Bi and 208Bi

Calculate no. of neutrons in 209Bi as follows:

atomic no. = 83, mass no. 209

N = 209 - 83

   = 126 (even no.)

Calculate no. of neutrons in 208Bi as follows:

atomic no. = 83, mass no. 208

N = 208 - 83

   = 125 (odd no.)

Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.

c)

nickel-58 and nickel-65

Calculate no. of neutrons in nickel-58 as follows:

atomic no. = 28, mass no. 58

N = 58 - 28

   = 30 (even no.)

Calculate no. of neutrons in nickel as follows:

atomic no. = 28, mass no. 65

N = 65 - 28

   = 37 (odd no.)

Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.

5 0
3 years ago
Which of the statements correctly describes the reactivity of halogens, according to the octet rule?
zzz [600]
They have seven electrons in their valence shell, so halogens are very reactive.
Hope this helps! :)
8 0
3 years ago
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