Why do we ignore noble gases usually when we discuss electronegativity values?

Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of –7.2°C. Bromine can be classified as a

andreev551 [17]

Answer:

Pure substance.

Explanation:

7 0

2 years ago

If element x forms the oxides xo and x2o3 the oxidation numbers of element x are

Deffense [45]

Answer:

b) +2 and +3.

Explanation:

Hello,

In this case, given the molecular formulas:

$XO$

And:

$X_2O_3$

We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:

$X^{+2}O^{-2}$

Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:

$X_2^{+3}O_3^{-2}$

Thus, answer is b)+2 and +3

Best regards.

3 0

2 years ago

What is the percent cr, by mass, in the steel sample? express your answer numerically as a percentage?

djverab [1.8K]

Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−

Assuming all the Cr is contained in the BaCrO4 at the end.
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr

8 0

2 years ago

Read 2 more answers

A train travels at a speed of 30 miles per hour. If 1 mile = 1.6 kilometers, how fast is the train traveling in kilometers per m

Alina [70]

The answer
the speed of the train traveling in kilometers per minute can be found by using the following method:
v = 30 miles per hour
1 mile = 1.6 kilometers

just do the calculus by changing miles to kilometers
1 mile = 1.6 kilometers, so 30 miles= 1.6 km x 30 = 48 km
so the speed of the train is
v=30 miles / hour = 48km / h = 48km / 60 mn = 0.8 km/ mn
the answer is
C- 0.8 km/min

7 0

2 years ago

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

likoan [24]

Balanced Half reactions are:

At anode 2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$

At Cathode: 2 $H^{+}$ + $2e^{-}$ ==> H₂

Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, $H^{+}$ , $Mn^{2+}$ and $Cl^{-}$

Now at Anode reaction will occur as given:

2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$ (will occur)

At Cathode:

2 $H^{+}$ + $2e^{-}$ ==> H₂ (will occur)

At Cathode:

$Mn^{2+}$ + $2e^{-}$ ==> Mn (This reaction will not occur)

The deposition of solid Mn will not occur because in aqueous solution, $H^{+}$ will be reduced before $Mn^{2+}$ .

The reduction potentials for $H^{+}$ is zero whereas reduction potential for $Mn^{2+}$ is - 1.18V.

The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.

To learn more about the half reaction please click on the link brainly.com/question/13186640

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8 0

1 year ago