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REY [17]
3 years ago
5

Iron (III) oxide - what does the III mean?

Chemistry
1 answer:
Aleonysh [2.5K]3 years ago
7 0

Answer:

The iron is in the +3 oxidation state, which is what the III means. 

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What is the total number of electrons in a<br> Mg2+ ion?(1) 10 (3) 14<br> (2) 12 (4) 24
Citrus2011 [14]
The answer is (1) 10. The protons of Mg atom is 12. So the Mg atom has 12 electrons. The Mg2+ ion has lost two electrons so it has two positive charge. Then the answer is 10 electrons.
5 0
3 years ago
A student mixed a small amount of iron filings and sulphur powder in a dish. He could not
Temka [501]

Answer:

\huge \boxed{\mathrm{Carbon \ disulphide}}

Explanation:

Carbon disulphide is the liquid that can be used to separate iron fillings and sulphur powder.

When carbon disulphide is poured into the dish, the sulphur powder gets easily dissolved in the carbon disulfide. The iron fillings are left to settle on the bottom of the dish.

The iron fillings can get seperated through filtration. When the mixture of sulphur powder and carbon disulphide gets completely evaporated, the sulphur powder is left over.

7 0
3 years ago
Which statement is true about the products of two reactants that combine chemically?
Art [367]
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7 0
3 years ago
How much energy (joules) is needed to heat 250 grams of copper from 22 °C to 99 °C? The specific heat capacity (C) of copper is
Naddik [55]

Answer:

7430.5 Joules (7.4*10^4 Joules)

Explanation:

Q=mc∆T

where Q is energy in Joules.

Now m=250 g

c= 0.386 J/g°C

∆T = 99 - 22 = 77 °C

plugging the values in gives

Q=250*0.386*77=7430.5 Joules

(7.4*10^4 Joules, if 2 significant figures)

5 0
3 years ago
What is the pH of 0.30 M ethanolamine, HOCH2CH2NH2, (Kb = 3.2 x 10−5)?
Dmitry [639]

Answer:

pH= 11.49

Explanation:

Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.

From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5

Using the formula below;

[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)

[OH^-] =√(3.2×10^-5 × 0.30M)

[OH^-]= √(9.6×10^-6)

[OH^-]=3.0984×10^-3

pOH= -log[OH^-]

pOH= -log 3.1×10^-3

pOH= 3-log 3.1

pH= 14-pOH

pH= 14-(3-log3.1)

pH= 11+log 3.1

pH= 11+ 0.4914

pH= 11.49

8 0
3 years ago
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