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____ [38]
3 years ago
8

What is the mass percentage for copper ?

Chemistry
1 answer:
RideAnS [48]3 years ago
3 0

Answer:

Explanation:

To find the mass percent composition of an element, divide the mass contribution of the element by the total molecular mass. This number must then be multiplied by 100% to be expressed as a percent.

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6.
ki77a [65]

Answer:

4Na  +   O₂  →   2Na₂O  

6.82g

Explanation:

The combustion of sodium involves combining the element with oxygen gas:

              4Na  +   O₂  →   2Na₂O  

We have been given 5g of Na.

The balanced reaction equation is given as shown above. This reaction is also a synthesis or combination reaction.

b. Maximum amount of product that can be obtained.

Sodium is the limiting reactant and it will determine the amount of product that can be formed.

  To solve this, we find the number of moles of the given sodium.

  Mass of given sodium  = 5g

    Molar mass of Na  = 23g/mol

  Number of moles of Na  = \frac{mass}{molar mass}  

Insert the parameters and solve;

 Number of moles of Na  = \frac{5}{23}    = 0.217moles

  From the balanced reaction equation:

         4 moles of Na will produce 2 moles of Na₂O  

      0.217 moles of Na will produce \frac{0.217 x 2}{4}   = 0.11mole of Na₂O  

So;

 Mass of Na₂O   = number of moles x molar mass

    Molar mass of Na₂O   = 2(23) + 16  = 62g/mol

 Mass of Na₂O   = 0.11 x 62  = 6.82g

6 0
3 years ago
How does sleeping help us maintain homeostasis?
hjlf

Answer:

Sleep homeostasis denotes a basic principle of sleep regulation. A sleep deficit elicits a compensatory increase in the intensity and duration of sleep, while excessive sleep reduces sleep propensity

Explanation:

:]

7 0
3 years ago
A mutation occurs in a gene causing a pea plant produce 75% fewer leaves. this would be considered what kind of mutation?
denpristay [2]

Answer:

negative

Explanation:

There is decrease of a character in the new progeny so it is negative mutation

5 0
3 years ago
The phophorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by nurning phosphorus in oxygen.
Radda [10]

Answer : The percent yield of P_4O_{10} is, 87.7 %

Solution : Given,

Moles of P_4 = 0.200 mole

Moles of O_2 = 0.200 mole

Molar mass of P_4O_{10} = 283.9 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

P_4+5O_2\rightarrow P_4O_{10}

From the balanced reaction we conclude that

As, 5 mole of O_2 react with 1 mole of P_4

So, 0.200 moles of O_2 react with \frac{0.200}{5}=0.04 moles of P_4

From this we conclude that, P_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of P_4O_{10}

From the reaction, we conclude that

As, 5 mole of O_2 react to give 1 mole of P_4O_{10}

So, 0.200 moles of O_2 react to give \frac{0.200}{5}=0.04 moles of P_4O_{10}

Now we have to calculate the mass of P_4O_{10}

\text{ Mass of }P_4O_{10}=\text{ Moles of }P_4O_{10}\times \text{ Molar mass of }P_4O_{10}

\text{ Mass of }P_4O_{10}=(0.04moles)\times (283.9g/mole)=11.4g

Theoretical yield of P_4O_{10} = 11.4 g

Experimental yield of P_4O_{10} = 10.0 g

Now we have to calculate the percent yield of P_4O_{10}

\% \text{ yield of }P_4O_{10}=\frac{\text{ Experimental yield of }P_4O_{10}}{\text{ Theretical yield of }P_4O_{10}}\times 100

\% \text{ yield of }P_4O_{10}=\frac{10.0g}{11.4g}\times 100=87.7\%

Therefore, the percent yield of P_4O_{10} is, 87.7 %

5 0
3 years ago
A light twig is barely floating on the surface of water. What will most likely happen to the twig if a drop of dishwashing deter
sveta [45]
"<span>It will sink because soap disrupts the intermolecular forces of water" is the statement among the choices given in the question that describes what will  likely happen to the twig  </span><span>if a drop of dishwashing detergent is added near it. The correct option among all the options that are given in the question is the third option. </span>
6 0
3 years ago
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