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Dafna1 [17]
3 years ago
7

The density of Mg is 1.74g/cm3. The density of strontium is 2.60g/cm3. What would you expect the density of Ca to be?

Chemistry
1 answer:
Tomtit [17]3 years ago
6 0

The density of Ca will be between that of Mg and Sr

Explanation:

Ca, Mg and Sr are group II elements. They are called alkali earth metals. The correct order of the elements in this group are: Be, Mg, Ca, Sr, Ba and Ra.

Density is an intensive property of matter which describes the amount of matter(mass) per volume of a substance.

  • Density varies proportionally with mass. The higher the mass, the higher its density.
  • On the periodic table, atomic mass which the number of protons and neutrons in the nucleus of an atom increases down the group.
  • This implies a progradation in the value of density down the group. Therefore one expects that the value of density of Ca will fall between that of Mg and Sr. It cannot be more than 2.6g/cm³ nor less than 1.74g/cm³.

Learn more:

density brainly.com/question/2658982

mass number brainly.com/question/2597088

#learnwithBrainly

                   

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Explain why CaCl2 is likely to have properties similar to those of CaBr2
qaws [65]

Answer:

Because both CaCl2 and CaBr2 both contain elements (Chlorine and Bromine) from the same group (group 7)

Explanation:

Elements are placed into different groups in the periodic table. Elements in the same group are those that have the same number of valence electrons in their outermost shell and as a result will behave similar chemically i.e. will react with other elements in the same manner.

Chlorine and Bromine are two elements belonging to group 7 of the periodic table. They are called HALOGENS and they have seven valence electrons in their outermost shell. Hence, when they form a compound with Calcium, a group two element, these compounds (CaCl2 and CaBr2) will possess similar properties because they have elements that are from the same group (halogen group).

4 0
3 years ago
Chemistry question please help!!
LiRa [457]
Answer: It’s the first one
8 0
2 years ago
A sample of an ionic compound NaA, where A- is the anion of a
vitfil [10]

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

6 0
3 years ago
A gas that was cooled to 200 Kelvin has a volume of 65.8 L. If its initial volume was 132.4 L, what was its initial temperature?
Mandarinka [93]

Answer:

Initial temperature, T1 = 99.4 Kelvin

Explanation:

<u>Given the following data;</u>

  • Initial volume, V1 = 65.8 Litres
  • Final temperature, T2 = 200 Kelvin
  • Final volume, V2 = 132.4 Litres

To find the initial temperature (T1), we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

\frac {V}{T} = K

\frac {V_{1}}{T_{1}} = \frac {V_{2}}{T_{2}}

Making T1 as the subject formula, we have;

T_{1} = \frac {V_{1}T_{2}}{V_{2}}

Substituting the values into the formula, we have;

T_{1} = \frac {65.8 * 200}{132.4}

T_{1} = \frac {13160}{132.4}

<em>Initial temperature, T1 = 99.4 Kelvin</em>

6 0
3 years ago
How many moles are 4.20 * 10 ^ 25 atoms of Ca?
aleksandr82 [10.1K]

Answer:

~69.744 moles of Ca

Explanation:

Using Avogadro's constant , we know that:

1 mole = 6.022 x 10^23 atoms

S0, the number of moles in 4.20 x 10^25 atoms of Ca:

=(4.20 x 10^25 x 1 )/(6.022 x 10^23)

~69.744 moles of Ca

Q2:How many atoms are in 0.35 moles of oxygen?

1 mole = 6.022 x 10^23 atoms

S0, the number of atoms in 0.35 moles of  oxygen:

=[0.35 x (6.022 x 10^23)]

=2.1077 x 10^23 atoms of Oxygen

Hope it helps:)

4 0
2 years ago
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