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3 years ago

Which of the below elements are part of a group? Feel free to use the periodic table.

Chemistry

2 answers:

LUCKY_DIMON [66]3 years ago

5 0

Answer:

B. Be, Mg, Ca, Sr

Explanation:

Be, Mg, Ca, Sr are parts of the alkaline Earth metal family/group. So they are the second most reactive elements following behind alkali metals. Furthermore, Be, Mg, Ca, Sr all have 2 valence electrons that lose them to form cations. They have low melting points, low boiling points, can conduct electricity, have high malleability and ductility.

Hope it helped!

gogolik [260]3 years ago

4 0

B. Be, Mg, Ca, Sr (hope this helps)

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What is the percent yield of the purification system

Jlenok [28]

Answer: 90.3

Explanation:

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3 years ago

In which situation is no work considered to be done by a force?

Masja [62]

If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.

If the angle is 45, then there is still some work involved.

The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90.

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3 years ago

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What Element is found in column 16, period 2?

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3 years ago

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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

I hope it helps!

5 0

3 years ago

You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl

bulgar [2K]

Answer: Mass of $CO_2$ produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)$

Mass or reactants = Mass of $CaCO_3$ + mass of $HCl$ = 16.00 + 64.80 = 80.80 g

Mass of products = mass of aqueous solution + mass of $CO_2$ + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of $CO_2$ produced in this reaction was 6.56 grams

7 0

3 years ago