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3 years ago

Elements located in the vertical column of the periodic table are of the same _ and have _ chemical properties.

Chemistry

2 answers:

Oksanka [162]3 years ago

7 0

Answer:

Elements located in the vertical column of the periodic table are of the same

group

and have

similar

chemical properties.

Firdavs [7]3 years ago

6 0

They are of the same period and are similar in some form

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

What is the mass percent of nitrogen in ammonium carbonate, (NH4)2CO3?

Margaret [11]

In order for you to calculate for the mass of ammonium carbonate, you need to know the molar mass of it and the nitrogen atoms in the compound. Ammonium carbonate has a molar mass of 96.08 grams per mole. There are two nitrogen atoms in ammonium carbonate which is equal to 28.02 grams per mole. Divide the molar mass of nitrogen to the ammonium carbonate, 28.02/96.08 x 100, we get 29.16wt% nitrogen.

6 0

3 years ago

At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat

wariber [46]

Answer:

0.808 M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$ ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$ ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$ $= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity = $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

$= 0.001616*1000$

$= 1.616 M$

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have $\frac{1.616}{2} =0.808 M$

3 0

3 years ago

Read 2 more answers

List 3 particles of an atom (w/their symbol & charge)

Ulleksa [173]

Answer:

there are three fundamental particles of atom

which are :

proton ( positively charged (+) )
electron ( negatively charged ( - ) )
neutron ( having no charge on it ( neutral ))

Particle Symbol

electron =》e-

proton =》p+

neutron =》n°

6 0

2 years ago

Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction. Ag+(aq)+e−→Ag(s) Part A Silver can be e

dmitriy555 [2]

Answer: Mass of silver deposited at the cathode is 37.1g

Explanation: According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode (cathode or anode) is directly proportional to quantity of electricity passed through the electrolyte

Faraday has found that to liberate one gm eq. of substance from an electrolyte, 96500C of electricity is required.

$Ag^+$ +e− ==> Ag(s)

Given that

Current (I) = 8.5A

Time (t) = 65 *60 = 3900s

Quantity of electricity passed = 8.5*3900 =33150C

Molar mass of Ag= 108g

96500C will liberate 108g

33150C will liberate Xg

Xg= (108*33150)/96500

=37.1g

Therefore the mass of Ag deposited at the cathode is 37.1g.

3 0

3 years ago

Elements located in the vertical column of the periodic table are of the same ___ and have ___ chemical properties.

Elements located in the vertical column of the periodic table are of the same _ and have _ chemical properties.