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Mice21 [21]
3 years ago
12

Question attached pls help

Chemistry
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:

D sorry it's wrong

Explanation:

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In the titration of a solution of Sodium hydroxide, an acid titrant was prepared by diluting 125.0 mL of 10.00 mol/L Nitric acid
harina [27]

Answer:

1.395M NaOH

Explanation:

Sodium hydroxide, NaOH, reacts with nitric acid, HNO3, as follows:

NaOH + HNO3 → NaNO3 + H2O

<em>Where 1mol of NaOH reacts with 1mol of HNO3</em>

To solve this question we must find the concentration of the titrant. With the concentration and the needed acid we can find the moles of HNO3 added = moles NaOH in the solution. With the moles of NaOH and its volume we can find its concentration as follows:

<em>HNO3 concentration:</em>

10.00mol/L HNO3 * (125.0mL/500.0mL) = 2.500M HNO3

<em>Moles HNO3 = Moles NaOH:</em>

16.74mL = 0.01674L*(2.500mol/L) = 0.04185 moles HNO3 = Moles NaOH

<em>Concentration NaOH:</em>

0.04185 moles / 0.0300L =

1.395M NaOH

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3 years ago
Give the value of the quantum number ℓ, if one exists, for a hydrogen atom whose orbital angular momentum has a magnitude of 6√(
MakcuM [25]

Answer:

For this angular momentum, no quantum number exist

Explanation:

From the question we are told that

   The magnitude of the angular momentum is  L  = 6\sqrt{\frac{h}{2 \pi} }

The generally formula for Orbital angular momentum is mathematically represented as

           L  = \sqrt{(l * (l + 1)) } *  \frac{h}{2 \pi}

Where l is the quantum number

now  

We can look at the given angular momentum in this form as

      L  = 6\sqrt{\frac{h}{2 \pi} }    =  \sqrt{36}  * \sqrt{\frac{h}{2 \pi}} }

comparing this equation to the generally equation for Orbital angular momentum

     We see that there is no quantum number that would satisfy this equation

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4 years ago
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Explanation: Canada has 3,500 tons while the U.S. has 6,000

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3 years ago
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Multiple them divided by 2.50
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What are the sex cells of the plants that Mendel studied?
guajiro [1.7K]

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alleles

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