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azamat
3 years ago
15

Which group is more reactive than the alkaline earth metals?

Chemistry
2 answers:
SIZIF [17.4K]3 years ago
8 0

Answer: Alkaline metals(Group 1)

Explanation:

These elements react vigorously with cold water to give hydrogen gas and they have one electron in their outer-most shell which allows them to give electron freely and are good reducing agents.

examples of these elements are:

Hydrogen

Lithium

Sodium

Potassium

Rubidium

Hydrogen is the only gas in group 1.

Vlad [161]3 years ago
5 0

Answer: <em>The </em><em>alkali metals</em><em> reactivity is generally higher than the alkaline earth metals. </em>

Explanation: <em>This makes the alkaline earth metals with their two valence electrons less reactive than alkali metals with their one valence electron.</em>

<u>Alkali metals is ns1 and alkaline earth metals is ns2</u>

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6 0
3 years ago
Solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the
aksik [14]

Answer:

83,40 (w/w) %

Explanation:

The osmotic pressure (π) is defined as:

π = iMRT

Where i is Van't Hoff factor (3 for MgCl₂ and 2 for NaCl), M is molarity, R is gas contant (0,082atmL/molK) and T is temperature (298,15K)

iM = π / RT

iM = 0,01607mol/L

It is possible to write:

<em>3x+2y = 0,01607mol/L </em><em>(1)</em>

Where x are moles of MgCl₂ and y moles of NaCl.

<em>-M = moles of each compund because M is molarity (moles/L) and there is 1,000L-</em>

Knowing molar mass of MgCl₂ is 95,2 g/mol and for NaCl is 58,44g/mol:

<em>x×95,211 + y×58,44 = 0,5000g </em><em>(2)</em>

Replacing (2) in (1):

3x+2(0,0086 - 1,629x) = 0,01607mol/L

-0,258x = -0,00113

<em>x = 0,004380 moles of MgCl₂</em>

In grams:

0,004380 moles of MgCl₂×(95,211g/mol) = 0,417g of MgCl₂

Mass percent is:

(0,4170g of MgCl₂/0,5000g of solid) ×100 = <em>83,40 (w/w) %</em>

I hope it helps!

4 0
3 years ago
List at least 15 important events in the science of Astronomy. Include a description of the event including the year and what cu
lbvjy [14]
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7 0
3 years ago
Read 2 more answers
Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o
pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

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Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0
3 years ago
What kind of questions do scientists ask?
Usimov [2.4K]

ones that are answered through observing :)

7 0
3 years ago
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