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pochemuha
3 years ago
7

The electronegativity is 2.1 for H and 3.0 for N. Based on these electronegativities, NH4 would be expected to The electronegati

vity is 2.1 for H and 3.0 for N. Based on these electronegativities, NH4 would be expected to have polar covalent bonds with a partial positive charges on the H atoms. be ionic and contain H- ions. have polar covalent bonds with a partial negative charges on the H atoms. be ionic and contain H ions.
Chemistry
2 answers:
Morgarella [4.7K]3 years ago
8 0

Answer:

Electronegativity difference of N and H is: 0.9

Its covalent

Electronegativity difference of F and CI is: 1.0

Its covalent

Explanation:

amid [387]3 years ago
3 0

Answer:

First statement is true:

  • <u><em>NH₄ would be expected to have polar covalent bonds with a partial positive charges on the H atoms.</em></u>

Explanation:

Nitrogen (N) and hydrogen (H) are two nonmetals; thus, it should be a hint to assert that they form a covalent bond.

The difference in <em>electronegativities</em> between<em> N</em> and <em>H </em>is <em>3.0</em> - <em>2.1</em> = 0.9.

That makes that the bonds be <em>polar</em>. Ions are not formed because to transfer the electrons from one atom to the other (to form ionic bonds) requires an electronegativity difference greater than about 1.7.

Since N is more electronegative than H, in the molecule <em>NH₄</em>, each N atom will pull the electron density toward itself acquiring a partial negative charge and leaving H atoms with a partial positive charge.

Thus, each NH bond is<em> polar covalent with a partial positive charge on the H atoms</em> (first statement from the choices).

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Explanation:

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3 years ago
You are lying in bed list the internal forces acting on the mattress
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Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa
kumpel [21]

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

P_{50} = 22 mm Hg

h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

Hence, the fractional saturation for hemoglobin is 0.86.

4 0
3 years ago
What is the solution of pOH if the pH is 9.50
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3 years ago
The mass of 0.125 mol of nitrogen is <br> A 224g<br> B 1.75g<br> C 3.50g<br> D 112g
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Answer:C - 3.50g

Explanation:

To solve this problem, nitrogen is a gas and it's always two atoms of it in every reaction.

Using the formula

n = m/ Mm

n - number of mole

m - number of mass

Mm - number of molar mass

n = 0.125 mole

The molar mass of N2 = 14.0067 * 2 = 28.0134g/mol

(N - 14.0067 )

n = mass / molar mass

0.125 mol = mass / 28.0134g/mol

Mass = 0.125 * 28.0134

= 3.501675

= 3.50g

5 0
3 years ago
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