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just olya [345]
3 years ago
14

When electrons are lost, a____ion is formed.

Chemistry
2 answers:
kati45 [8]3 years ago
3 0

Answer:

Positive; negative

Explanation:

When electrons are lost, a positive ion is formed. A positive ion is called a cation.

When electrons are gained, a negative ion is formed. A negative ion is called an anion.

Savatey [412]3 years ago
3 0
When electrons are lost a cations are formed

When electrons are gained a anions are formed
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If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? (2005 K)
jolli1 [7]

Answer:

205K

Explanation:

The following were obtained from the question:

n = 4moles

P = 5.6atm

V = 12L

R = 0.082atm.L/Kmol

T =?

PV = nRT

T = PV/nR

T = (5.6 x 12)/ (4 x 0.082)

T = 205K

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3 years ago
1. a) Name two examples of good conductors?
Pachacha [2.7K]
A. good conductors - copper, aluminium

b. fair conductors -  carbon,human body

c. insulator  -   paper, wood
7 0
4 years ago
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kj . f the change in e
TEA [102]

<u>Given:</u>

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

<u>To determine:</u>

The work done, W

<u>Explanation:</u>

Based on the first law of thermodynamics,

ΔH = ΔU + PΔV

the work done by a gas is given as:

W = -PΔV

Therefore:

ΔH = ΔU - W

W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

Ans: Work done is -9.8 kJ


6 0
3 years ago
In two or more complete sentences, describe the process of fractional distillation that used to refine crude oil.
Oksi-84 [34.3K]
I hope this helps you

3 0
3 years ago
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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes
frez [133]

Answer : The mass of MnCO_3 required are, 35 kg

Explanation :

First we have to calculate the mass of MnO_2.

The first step balanced chemical reaction is:

2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2

Molar mass of MnCO_3 = 115 g/mole

Molar mass of MnO_2 = 87 g/mole

Let the mass of MnCO_3 be, 'x' grams.

From the balanced reaction, we conclude that

As, (2\times 115)g of MnCO_3 react to give (2\times 87)g of MnO_2

So, xg of MnCO_3 react to give \frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg of MnO_2

And as we are given that the yield produced from the first step is, 65 % that means,

60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg

The mass of MnO_2 obtained = 0.4542x g

Now we have to calculate the mass of Mn.

The second step balanced chemical reaction is:

3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3

Molar mass of MnO_2 = 87 g/mole

Molar mass of Mn = 55 g/mole

From the balanced reaction, we conclude that

As, (3\times 87)g of MnO_2 react to give (3\times 55)g of Mn

So, 0.4542xg of MnO_2 react to give \frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg of Mn

And as we are given that the yield produced from the second step is, 80 % that means,

80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg

The mass of Mn obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of MnCO_3 required are, 35 kg

4 0
3 years ago
Read 2 more answers
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