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Serjik [45]
3 years ago
13

What kind of energy transformations occur when you hit a coconut with a hammer

Chemistry
1 answer:
hram777 [196]3 years ago
3 0

Answer: I dont really know

Explanation:

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Li2SO4 _____ an electrolyte in solution.<br> A. Is <br> B. Is not
Rasek [7]

Answer:

a

Explanation:

it is an electrolyte because of its strong polar chemical bond

8 0
2 years ago
Read 2 more answers
How many electrons will a fluorine (f) atom gain or lose in forming an ion?
777dan777 [17]
It will gain one electrons to form the fluorine ion
6 0
3 years ago
In which orbital does an electron in a bromine atom experience the greatest effective nuclear charge?
asambeis [7]

First let us determine the electronic configuration of Bromine (Br). This is written as:

Br = [Ar] 3d10 4s2 4p5

 

Then we must recall that the greatest effective nuclear charge (also referred to as shielding) greatly increases as distance of the orbital to the nucleus also increases. So therefore the electron in the farthest shell will experience the greatest nuclear charge hence the answer is:

<span>4p orbital</span>

4 0
3 years ago
Read 2 more answers
3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?
Dafna1 [17]

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J

Thus, the total energy turns out to be:

Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J

Best regards!

5 0
3 years ago
(Thermodynamics)
frutty [35]

Answer:

3853 g

Step-by-step explanation:

M_r: 107.87

         16Ag + S₈ ⟶ 8Ag₂S; ΔH°f =  -31.8 kJ·mol⁻¹

1. Calculate the moles of Ag₂S

Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S

2. Calculate the moles of Ag

Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag

3. Calculate the mass of Ag

Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag

You must react 3853 g of Ag to produce 567.9 kJ of heat

3 0
3 years ago
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