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4 years ago

What family flouride in the table

Chemistry

1 answer:

andrew11 [14]4 years ago

6 0

I assume you're talking about fluorine. Fluorine is in the halogen family and it belongs to group number 17.

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3. Students measured the mass of the reactants and products for a combustion reaction they observed.

coldgirl [10]

Answer:

A. Students made a measurement error, because ending with more products is impossible.

Explanation:

The law of conversation of matter tells us that in a chemical reaction, matter is never created or destroyed, it's simply converted from one form to another. So the mass of reactants should always equal the mass of the products in a chemical reaction. If there is excess mass in the product, the students have made an error of some kind.

4 0

3 years ago

An isotope of yttrium has 39 protons and 59 neutrons. what is the atomic mass of that isotope?

Charra [1.4K]

We can define atomic mass the total of number of protons and number of neutrons in an atom or isotope.
So when an isotope of yttrium has 39 protons and 59 neutrons, its atomic mass is equal to;
number of protons + number of neutrons = 39 + 59 = 98

8 0

3 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: The mass of glycine that can be dissolved is $1.3\times 10^2g$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

5 0

3 years ago

Flerovium valence electron dots . Did I put the dots in the right place ? Did I do it right ? ANSWER FAST ! 10 points !

Lyrx [107]

Answer:

Flerovium has 4 valence electrons in its outer shell. So, you did it right. All the dots are in the right place. Just fill in the dots and you should be good.

Hope that helps.

7 0

3 years ago

Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.

Ronch [10]

Answer : The limiting reagent is $O_2$

Solution : Given,

Moles of methane = 2.8 moles

Moles of $O_2$ = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 5 moles of $O_2$ react with $\frac{5}{2}=2.5$ moles of $CH_4$

From this we conclude that, $CH_4$ 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.

Hence, the limiting reagent is $O_2$

8 0

3 years ago