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3 years ago

. Using the pH scale below what would be the pH value of an Acid? *

Chemistry

1 answer:

earnstyle [38]3 years ago

6 0

Answer : The pH value of an acid is below 7.

Explanation :

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

Mathematically,

$pH=-\log [H^+]$

When the pH less than 7 then the solution is acidic and the concentration of hydrogen ion is greater than hydroxide ion.

When the pH more than 7 then the solution is basic and the concentration of hydrogen ion is less than hydroxide ion.

When the pH is equal to 7 then the solution is neutral and the concentration of hydrogen ion is equal to the hydroxide ion.

Hence, the pH value of an acid is below 7.

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The number at the end of an isotope’s name is the ____ number. A) Atomic B)neutron C)mass D)electron

storchak [24]

each isotope of a given element has a different mass number. For example, carbon-12, carbon-13 and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13 and 14

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3 years ago

Read 2 more answers

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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3 years ago

13. Fill in the following table

Mandarinka [93]

Answer:

Potassium Bromide = KBr

Nitrogen dioxide = NO₂

Lithium oxide = Li₂O

Explanation:

Potassium Bromide (KBr)

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3 years ago

Which sample contains a total of 3.0 x 10^23 molecules?

belka [17]

Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:

A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.

Therefore the only choice that fits is A. 80 g of Br2.

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3 years ago

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2 years ago