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allochka39001 [22]
3 years ago
6

What’s a conjugative Bas of OH-

Chemistry
1 answer:
LiRa [457]3 years ago
8 0

Answer:

the answer is O²- hopefully

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What type of bond is formed when atoms share electrons?
navik [9.2K]
Covalent bond is formed when atoms share electrons.
Hope this helps you.
5 0
3 years ago
Match the following terms to their correct examples
VARVARA [1.3K]

conduction : a spoon resting in a cup of tea gets hot

convection: water warming on the stove

Radiation: The sun warming the earth

3 0
3 years ago
If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?
myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30.17}\ year^{-1}

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Initial concentration [A_0] = ?

Final concentration [A_t] = 12.5 grams

Applying in the above equation, we get that:-

12.5\ g=[A_0]e^{-0.02297\times 90.6}

[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g

<u>Mass of original sample = 100 g</u>

8 0
3 years ago
What is the energy of a microwave photon that has a frequency of 9.86 x 1012 Hz?
BlackZzzverrR [31]

Answer:

6.533 × 10^-21J

Explanation:

The energy of the microwave photon can be calculated using:

E = hf

Where;

E = energy of photon (J)

h = Planck's constant (6.626 × 10^-34 J/s)

f = frequency (9.86 x 10^12 Hz)

Hence, E = hf

E = 6.626 × 10^-34 × 9.86 x 10^12

E = 65.33 × 10^(-34 + 12)

E = 65.33 × 10^(-22)

E = 6.533 × 10^-21J

The energy of the microwave photon is

6.533 × 10^-21J

5 0
3 years ago
What do these images model ?
jek_recluse [69]
The answer is A

Mark Brainliest ☺☻
4 0
3 years ago
Read 2 more answers
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