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sweet [91]
3 years ago
7

A solution that has a pOH of 5.36 is-

Chemistry
1 answer:
sdas [7]3 years ago
7 0

Answer:

Option D

Explanation:

A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.

A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale.

The hydroxide ion molarity may be expressed as a p-function, or pOH.

pOH = −log[OH−]

Basic solutions are those with hydronium ion molarities less than 1.0 × 10−7 M and hydroxide ion molarities greater than 1.0 × 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 7.00).

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Flauer [41]
 <span>1. 1 molecule of C6H12O6(dextrose sugar), 2 molecles of c2h6o (ethyl alcohol), 2 molecules of Co2 

2. 48 hydrogen atoms </span>
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Define Condensation .​
ozzi

Answer:

Condensation is the change of the physical state of matter from the gas phase into the liquid phase.

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Name the following compound:<br> CH3-CH2-CH2-CH2-CH3<br> CH3 CH3
stepan [7]

Answer:

<u><em>Pentane </em></u>

Explanation:

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6 0
3 years ago
One of the waste products produced by kidneys is creatinine. A healthy range ofcreatinine in human blood is between 0.50 mg/dL a
Anna35 [415]

The normal range of creatinine in human blood is between 0.50 mg/dL and 1.1 mg/dL. The patient's blood has a concentration of 0.0082 g/L. Let's convert that value into mg/dL.

We kwnot that there are 1000 mg in 1 g. And there are 10 dL in 1 L. We have to use those conversions.

1000 mg = 1 g 10 dL = 1 L

0.0082 g/L = 0.0082 g/L * 1000 mg/g = 8.2 mg/L * 1 L/ (10 dL) = 0.82 mg/dL

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8 0
1 year ago
Which is the limiting reactant if we start with 30.0 g Al and 30.0 h O2​
lorasvet [3.4K]

Al

Explanation:

The limiting reactant will be Al:

        4Al + 3O₂ → 2Al₂O₃

The limiting reactant is the reactant in short supply in a chemical reaction.

     Given parameters:

  Mass of Al = 30g  Molar mass = 27g/mol

         Number of moles = \frac{mass}{molar mass} =  \frac{30}{27}

         Number of moles of Al = 1.111 mole

  Mass of O₂ = 30g, molar mass = 32g/mol

   Number of moles =  \frac{30}{32} = 0.94mol

In the reaction:

          4 moles of Al reacted with 3 moles of O₂

         1.11moles of Al will require \frac{1.11 x3}{4} = 0.83mole to react

But we have been given 0.94mole of O₂. This is more than required.

Therefore O₂ is in excess and Al is the limiting reactant.

Learn more:

Limiting reagents brainly.com/question/6078553

#learnwithBrainly

4 0
3 years ago
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