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poizon [28]
3 years ago
10

Chapter 3 reactivity of metals mcq question and question and answer and important notes? ​

Chemistry
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

resend again that question dear

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Heating 235 g of water from 22.6°C to 94.4°C in a microwave oven requires 7.06 × 104 J of energy. If the microwave frequency is
Darya [45]

Answer: 3.69 × 10^27

Explanation:

Amount of energy required = 7.06 × 10^4 J

Frequency of microwave (f) = 2.88 × 10^10 s−1

Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum

Recall ;

Energy of photon = hf

Therefore, energy of photon :

(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1

= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J

Hence, number of quanta required :

(7.06 × 10^4)J / (19.0944 × 10^-24)J

= 0.369 × 10^(4 + 24) = 0.369×10^28

= 3.69 × 10^27

6 0
3 years ago
What is the top of a transverse wave
Andru [333]
It would be called the crest.

Happy to help! Have a great evening.

~Brooke❤️
8 0
4 years ago
The ""size"" of an atom is sometimes defined by the radius of a sphere that contains 90% of the charge density of the electrons.
vfiekz [6]

Answer:

3/2a

Explanation:

The complete step by step answer is found in the attachment

8 0
4 years ago
In a position time graph, a steep slope indicates _____.
sweet-ann [11.9K]
Im pretty positive its A , hope this helps!
5 0
4 years ago
Read 2 more answers
Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen
Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen=n_1

Moles of acetylene =n_2

\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}

Volume of large tank with oxygen gas, V_1 = 5.00 L

Pressure of the oxygen gas inside the tank = P_1=135 atm

RT=\frac{P_1V_1}{n_1} ..[1]

Volume of small tank with acetylene gas ,V_2= 3.50 L

Pressure of the acetylene gas inside the tank = P_2=?

RT=\frac{P_2V_1}{n_2} ..[2]

Considering both the gases having same temperature T, [1]=[2]

\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}

P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}

=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm

77.14 atm of pressure should be of an acetylene in the tank.

8 0
3 years ago
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