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adelina 88 [10]
3 years ago
14

Anions are those elements on the upper right side of the periodic table, excluding the Noble gases, that gain electrons when the

y ionize. T or F?
Chemistry
1 answer:
Gala2k [10]3 years ago
3 0

Answer:

True

Explanation:

The given statement is true. The elements on the upper right side of periodic table accept the electrons because of higher electronegativity values and form anions.

For example:

Consider the halogens. All halogens accept the one electron per atom to complete the octet and form anion. when they combine with metals metals loses their electrons to halogens atoms and form cations while halogens accept the electrons and form anion.

Consider the sodium chloride salt. The sodium metal loses its one valance electron which is accepted by chlorine.

Na⁺Cl⁻

Magnesium bromide:

To complete the octet bromine required one electron. By gaining the one electron it showed -1 oxidation state.

When bromine combine with magnesium both formed MgBr₂.

The one atom of Mg have +2 oxidation state while bromine have -1 that's why to make the overall compound neutral two bromine atoms combine with one atom of magnesium.

You might be interested in
Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U
Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of nitrogen gas = 79mL/s

R_2 = rate of effusion of sulfur dioxide gas = ?

M_1 = molar mass of nitrogen gas  = 28 g/mole

M_2 = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}

R_2=52mL/s

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0
3 years ago
A mole of oxygen and a mole of hydrogen (at STP) have all of the following in common EXCEPT
Drupady [299]

Answer:

Root mean squared velocity is different.

Explanation:

Hello!

In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:

v_{rms}=\sqrt{\frac{3RT}{MM} }

Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.

Best regards!

8 0
2 years ago
A sample of a vapor occupies a volume of 500 mL at 65°C. If pressure remains constant, what is the volume of the gas at standard
DedPeter [7]

Answer:

403 mL

Explanation:

First, I will assume that the mole is 1, because you are not specifing this.

Now, with the innitial data, we need to get the pressure:

T = 65+273 = 338 K

V = 500 / 1000 = 0.5 L

Now if:

PV = nRT

Then:

P = nRT/V   and V = nRT/P

Let's calculate the P:

P = 1 * 0.082 * 338 / 0.5 = 55.432 atm

The standard temperature is 0° C or 273 K so, the volume is:

V = 1 * 0.082 * 273 / 55.432

V = 0.40384 L or simply 403.84 mL

8 0
3 years ago
A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene
muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

\Delta T_f = K_f \times m

Where,

\Delta T_f = Depression in freezing point

K_f = Molal depression constant

m = Molality

\Delta T_f = K_f \times m

1.33 = 5.12 \times m

m = 0.26

Molality = \frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}

Mass of solvent (toluene) = 15.0 g = 0.015 kg

0.26 = \frac{Mole\ of\ compound}{0.015}

Moles of compound = 0.015 × 0.26 = 0.00389 mol

Mol = \frac{Mass\ in\ g}{Molecular\ weight}

Mass of the compound = 1.450 g

Molecular\ weight = \frac{Mass\ in\ g}{Moles}

Molecular weight = \frac{1.450}{0.00389} = 372.13\ g/mol

4 0
2 years ago
What is the term for a procedure that allows scientists to obtain good clear analysis and draw
rusak2 [61]
Answer is D(An experiment
6 0
3 years ago
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