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Gelneren [198K]
3 years ago
5

Help

Chemistry
1 answer:
Lemur [1.5K]3 years ago
8 0

Answer:

A molecule of oxygen can have atoms of carbon in it.

Explanation: For CO2 there is one atom of carbon and two atoms of oxygen.

hope this help

plz mark brainliest

You might be interested in
What titrant will be used to titrate the 0. 02 m hcl phenol red solution?
RideAnS [48]

The NaOH will be used What titrant  to titrate the 0. 02 m hcl phenol red solution.

Acid-base titrations may be the most typical titrations, although there are numerous more forms as well. Take a look at this illustration where sodium hydroxide is used to titrate a sample of hydrochloric acid (HCl) (NaOH). The titrant (NaOH), which is added gradually throughout the duration of the titration, has been added to the unknown solution.

Titrants are solutions with known concentrations that are added to solutions whose concentrations must be determined. The solution for whom the concentration needs to be determined is known as a titrant as well as analyte.

Therefore, the NaOH will be used as a titrant to titrate the 0. 02 m hcl phenol red solution.

To know more about titrant

brainly.com/question/21504465

#SPJ4

7 0
2 years ago
What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?​
trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution  = 2.5L

Mass of Calcium phosphate  = 600g

Unknown:

Concentration  = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

        Formula of calcium phosphate  = Ca₃PO₄

         molar mass = 3(40) + 31 + 4(16) = 215g/mol  

Number of moles of  Ca₃PO₄  = \frac{600}{215}   = 2.79moles

   Now;

  Concentration  = \frac{Number of moles }{volume }  

 Concentration  = \frac{2.79}{2.5}   = 1.12M

7 0
3 years ago
Elements join together to form_____.
sineoko [7]

Answer:

compounds

Explanation:

6 0
3 years ago
22. Most plates have a convergent boundary on one side and a
Alex73 [517]

Answer:

the answer for this question is true

8 0
3 years ago
Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g
Charra [1.4K]

Answer:

4.71 g of sodium and 7.25 g of chlorine

Explanation:

<em>Note: The question is incomplete. the complete question is given below:</em>

<em>Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 2.55 g of sodium and 3.93 g of chlorine. Being consistent with the law of constant composition, also called the law of definite proportions or law of definite composition, which set of masses could be the result of the decomposition of the other sample?</em>

<em>4.71 g of sodium and 3.30 g of chlorine</em>

<em>4.71 g of sodium and 7.25 g of chlorine</em>

<em>4.71 g of sodium and 1.31 g of chlorine</em>

<em>4.71 g of sodium and 13.7 g of chlorine</em>

The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass.

This means that irrespective of the source of any sample of a pure chemical compound, the constituents elements are always combined in the same mass ratio.

In the first sample, the mass ratio of Sodium to chlorine is given below:

mass of sodium = 2.55 g

mass of chlorine = 3.93 g

mass ratio; sodium to chlorine = 2.55 / 3.93 = 0.65

From the set of masses give above, we can determine the result of the decomposition of the second sample.

4.71 g of sodium and 3.30 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 3.30 = 1.43

4.71 g of sodium and 7.25 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 7.25 = 0.65

4.71 g of sodium and 1.31 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 1.31 = 3.59

4.71 g of sodium and 13.7 g of chlorine

mass ratio; sodium to chlorine = 4.71 / 13.7 = 0.34

From the results above, the correct set of masses for the second sample is 4.71 g of sodium and 7.25 g of chlorine

3 0
3 years ago
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