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aksik [14]
1 year ago
14

How many grams of solute would you use to prepare the following solutions?241.0 mL of 1.11 M NaOHExpress your answer with the ap

propriate units.
Chemistry
1 answer:
Ludmilka [50]1 year ago
8 0
<h2>Answer:</h2>

10.7 grams

<h2>Explanations:</h2>

The formula for calculating the molarity of a solution is given as:

\begin{gathered} molarity=\frac{moles}{volume} \\ moles=molarity\times volume \end{gathered}

Given the following parameters

molarity of NaOH = 1.11M

volume of solution = 241.0mL = 0.241L

Find the moles of NaOH

\begin{gathered} moles\text{ of Na}OH=\frac{1.11mol}{L}\times0.241L \\ moles\text{ of Na}OH=0.2675moles \end{gathered}

Determine the mass of solute (NaOH)

\begin{gathered} Mass\text{ of NaOH}=mole\times molar\text{ mass} \\ Mass\text{ of NaOH}=0.2675\times40 \\ Mass\text{ of NaOH}=10.7grams \end{gathered}

Hence the grams of solute that would be used to prepare the solution is 10.7grams

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Many tests to distinguish aldehydes and ketones involve the addition of an oxidant. Only __________ can be easily oxidized becau
tatiyna

Answer:

  1. Aldehydes
  2. A hydrogen atom
  3. Oxygen

Explanation:

Many tests to distinguish aldehydes and ketones involve the addition of an oxidant. Only <u>aldehydes</u> can be easily oxidized because there is<u> a hydrogen atom</u> next to the carbonyl and oxidation does not require<u> oxygen </u>

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3 years ago
How many moles in 110g of nahco3?
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Explain why ethics and skepticism are integral to science.
sdas [7]

Ethics and Skepticism are integral to science because they are the basis for the need to lay explanations on various fields of science through experiment. This will enable science bring the right data and arguments through scientific methods in an accurate and a straight forward manner.

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8 0
3 years ago
Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc
max2010maxim [7]

<u>Answer:</u> The percent yield of the water is 31.98 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol

The chemical equation for the combustion of methane is:

CH_4+2O_2\rightarrow CO_2+2H_2O

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = \frac{1}{2}\times 0.45=0.225mol of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = \frac{2}{2}\times 0.45=0.45 moles of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%

Hence, the percent yield of the water is 31.98 %

4 0
3 years ago
Electrons are:
Svetllana [295]

Answer:

B. Are fixed within electron orbitals

Explanation:

Electrons are found in the energy levels

4 0
3 years ago
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