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spayn [35]
3 years ago
8

Molecular shape of XeOF4​​

Chemistry
1 answer:
Scrat [10]3 years ago
6 0

42 valence electrons 8 valence electrons are coming from the xenon atom 6 valence electrons are coming from the oxygen atom 7 valence electrons from each of the four fluorine atoms. Answer: <em>Square pyramidal</em>.

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Culinary arts HELP FAST 1. Most people don't often think about science and restaurants or the food industry as
Andrews [41]

Answer:

i dont know

Explanation:

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For the reaction H2PO4- HAsO4 2-HPO4 2-+H2AsO4- <br> what species are a conjugate acid-base pair?
aleksklad [387]

Ionic equation:

\text{H}_2\text{PO}_4^{-} + \text{HAsO}_4^{2-} \to \text{HPO}_4^{2-} + \text{H}_2\text{AsO}_4^{-}

The acid and base in a conjugate pair differ by only one proton \text{H}^{+}. The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.

\text{H}_2\text{PO}_4^{-} loses one proton to produce \text{HPO}_4^{2-} in this reaction.

\text{H}_2\text{PO}_4^{-} \to \text{H}^{+} + \text{HPO}_4^{2-}

Meanwhile, \text{HAsO}_4^{2-} gains one proton to form \text{H}_2\text{AsO}_4^{-}.

\text{HAsO}_4^{2-} + \text{H}^{+} \to \text{H}_2\text{AsO}_4^{-}

Therefore

  • \text{H}_2\text{PO}_4^{-} is the conjugate acid  \text{HPO}_4^{2-}, its conjugate base.
  • \text{HAsO}_4^{2-} is the conjugate base of \text{H}_2\text{AsO}_4^{-}, its conjugate acid.
8 0
3 years ago
Read 2 more answers
When an erythrocyte is removed from circulation the iron ion of hemoglobin?
omeli [17]
Transporting metals, ions, water-insoluble molecules, and hormones. .... When erythrocytes are removed from circulation,
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3 years ago
What does Avogadro's law state and what is the formula​
sertanlavr [38]

The law is approximately valid for real gases at sufficiently low pressures and high temperatures. The specific number of molecules in one gram-mole of a substance, defined as the molecular weight in grams, is 6.02214076 × 1023, a quantity called Avogadro's number, or the Avogadro constant.

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3 years ago
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A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200
Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

7 0
3 years ago
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