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tiny-mole [99]
4 years ago
6

Cheryl has a mug that she says is made up of matter. Heather says that the hot chocolate inside the cup is made up of matter, to

o. Keaton says that the steam coming from the cup is also made up of matter. Mikayla says that none of it is matter because matter is too small to see. Which best describes who is correct?
Chemistry
2 answers:
svet-max [94.6K]4 years ago
5 0
Keaton is right because steam is water is gas form.

koban [17]4 years ago
4 0

Cheryl has a mug that she says is made up of matter. Heather says that the hot chocolate inside the cup is made up of matter, too. Keaton says that the steam coming from the cup is also made up of matter. Mikayla says that none of it is matter because matter is too small to see. Which best describes who is correct?

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Determine the bonding type for boron trihydride given the electronegativity on the Pauling scale for boron is 2.0 and for hydrog
Anika [276]

Answer: The bond between boron and hydrogen in boron trihydride is covalent bond.

Explanation:

The type of bonding between the atoms forming a compound is determined by using the electronegativity difference between the atoms. According to the pauling's electronegativity rule:

  • If \Delta \chi=0, then the bond is non-polar.
  • If \Delta \chi\leq 1.7, then the bond will be covalent.
  • If \Delta \chi>1.7, then the bond will be ionic.

We are given:

Electronegativity for boron = 2.0

Electronegativity for hydrogen = 2.1

\Delta \chi=\chi_{H}-\chi_{B}\\\\\Delta \chi=2.1-2.0=0.1

As, \Delta \chi is less than 1.7 and not equal to 0. Hence, the bond between boron and hydrogen is covalent bond.

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3 years ago
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3 years ago
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5 0
3 years ago
Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B
Ivenika [448]

Answer:

-2546 kJ

Explanation:

It is possible to obtain the enthalpy of a reaction from the sum of different intermediate reactions.

For the reaction:

4B(s) + 3O₂(g) → 2B₂O₃(s)

The intermediate reactions are:

A- B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g), ΔH°A= +2035 kJ

B- 2B(s) + 3H₂(g) → B₂H₆(g), ΔH°B= +36 kJ

C- H₂(g) + 1/2O₂(g) → H₂O(l), ΔH°C= -285 kJ

D- H₂O(l) → H₂O(g), ΔH°D= +44 kJ

2B = 4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH°2B= +78 kJ

-2A = 6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH°-2A= -4070 kJ

-6C = 6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH°-6C= +1710 kJ

-6D = 6H₂O(g) → 6H₂O(l) ΔH°-6D = -264 kJ

The sum of 2B - 2A - 6C - 6D produce:

4B(s) + 3O₂(g) → 2B₂O₃(s)

And the enthalpy is: ΔH°2B + ΔH°-2A + ΔH°-6C + ΔH°-6D = <em>-2546 kJ</em>

I hope it helps!

3 0
3 years ago
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