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Elena L [17]
3 years ago
12

How many neutrons are found in a stable isotope of uranium-235?

Chemistry
1 answer:
bezimeni [28]3 years ago
8 0

thx but it's actually 143

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Ethanol, CH.0, is common beverage alcohol. At its boiling point of 78.5 °C, the enthalpy of vaporization of ethanol is 38.6 kJ/m
garik1379 [7]

Answer:

209.8 kilo Joules heat is required to vaporize 250 g of ethanol at 78.5 °C.

Explanation:

Mass of an ethanol = 250 g

Molar mass of ethanol = 46 g/mol

Moles of an ethanol = n=\frac{250 g}{46 g/mol}=5.435 mol

Enthalpy of vaporization of ethanol = \Delta H_{vap} =38.6 kJ/mol

Heat required to vaporize 250 g of ethanol at 78.5 °C : Q

Q=n\times \Delta H_{vap}

=5.435 mol\times 38.6 kJ/mol=209.7826 kJ\approx 209.8kJ

Q = 209.8 kilo Joules

6 0
3 years ago
2 Points
Rudiy27
The answer of this question is D
3 0
3 years ago
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What is the net cell reaction for the cobalt-silver voltaic cell? express your answer as a chemical equation?
slava [35]

Answer: The net ionic equation for the cobalt-silver voltaic cell is Co(s)+3Ag^+(aq.)\rightarrow Co^{3+}(aq.)+3Ag(s)

Explanation: In cobalt-silver voltaic cell, one half of the cell consists of cobalt electrode immersed in Co(NO_3)_3 solution ( which means that Co^{3+} are present in the solution) and other half of the cell consists of the Ag electrode immersed in AgNO_3 solution ( which means that Ag^+ is present in the solution)

The two electrodes are joined by the copper wire. The cobalt electrode acts as an anode and the silver electrode acts a  cathode.

At anode, oxidation reaction takes place and at cathode, reduction reaction takes place.

At Anode :                    Co(s)\rightarrow Co^{3+}(aq.)+3e^-

At Cathode:                   [Ag^+(aq.)+e^-\rightarrow Ag(s)]\times 3

Net ionic equation:   Co(s)+3Ag^+(aq.)\rightarrow Co^{3+}(aq.)+3Ag(s)

6 0
3 years ago
What is the mass of oxygen in 25.0 grams of potassium permanganate , KMnO4?
Katen [24]

Answer:

10.76 grams

Explanation:

Given that the amount of KMnO_4 is 25.0 grams.

Mass of 1 mole of KMnO_4 = 158 grams

The number of atoms in 1 mole of KMnO_4 is 4.

Mass of oxygen in 1 mole of KMnO_4 = 16\times 4 = 68 grams.

Here, 158 grams of KMnO_4 has 68 grams of oxygen

So, the amount of oxygen in 1 gram of KMnO_4 = 68/158 grams

Therefore,  the amount of oxygen in 1 gram of KMnO_4 = \frac {68}{158}\times 25 grams

=10.76 grams

Hence, 25 grams of KMnO_4 has 10.76 grams of oxygen.

3 0
2 years ago
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What is the process of using one or more of your five senses to measure or collect data?
sveta [45]

Answer:

B no observation pls mark me branilest

7 0
3 years ago
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