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3 years ago

How many grams of nacl are required to make 500.0 ml of a 1.500 m solution?

Chemistry

2 answers:

Naddika [18.5K]3 years ago

7 0

Answer:

43.875 grams of NaCl are required to make 500.0 ml of a 1.500 M solution.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 500.0 mL = 0.500 L

Molarity of the NaCl solution = 1.500 M

$n=1.500 M\times 0.500 L=0.75 mol$

Mass of 0.75 moles of NaCl:

0.75 mol × 58.5 g/mol=43.875 g

43.875 grams of NaCl are required to make 500.0 ml of a 1.500 M solution.

mestny [16]3 years ago

4 0

43.83g of nacl are required to make 500.0 ml of a 1.500 m solution.

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Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

kkurt [141]

Answer: The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

Explanation:

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

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

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

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

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0

3 years ago

Can you show me the answer and explain?

Hitman42 [59]

Answer:

Explanation:

looking at a periodic table X is fluorine and Y is potassium

Fluorine is in group 7 and forms a 1- charge (which gains electrons) and potassium is in group 1 and forms a 1+ charge (which loses electrons)

Fluorine (X) has an electronic structure of 2,7 and needs to gain an electron from Potassium (Y) to have a full outer shell and potassium has an electronic structure of 2,8,8,1 so needs to lose an electron to have a full outer shell as well. This means that the electron that potassium (Y) has lost is given away to fluorine (X), so both elements become stable.

This is known as ionic bonding where metals (like potassium) lose electrons and non-metals (like fluorine) gain electrons to become more stable, forming ions

Any further clarification let me know

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2 years ago

How much water (H2O ) would form if 4.04 g of hydrogen (H2) reacted with 31.98 g of oxygen (O2 )?

Norma-Jean [14]

Answer:

Mass = 36 g

Explanation:

Given data:

Mass of water formed = ?

Mass of hydrogen = 4.04 g

Mass of oxygen = 31.98 g

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 4.04 g/ 2 g/mol

Number of moles = 2.02 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 31.98 g/ 32 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of water with hydrogen and oxygen.

O₂ : H₂O

1 : 2

H₂ : H₂O

2 : 2

2.02 : 2.02

Number of moles of water formed by oxygen are less thus oxygen will limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 2 mol × 18 g/mol

Mass = 36 g

8 0

2 years ago

The most likely van't Hoff factor for an 0.01 m calcium iodide solution is

monitta

This problem is providing us with the molality of a solution of calcium iodide as 0.01 m. So the most likely van't Hoff factor is required and theoretically found to be 3 due to the following:

<h3>Van't Hoff factor:</h3>

In chemistry, the correct characterization of solutions also imply the identification of the ions it will release in aqueous solution. For that reason, the van't Hoff factor gives us an idea of this number, according to the formula the solute has got.

In such a way, for calcium iodide, we write its ionization equation as shown below:

$CaI_2\rightarrow Ca^{2+}+2I^-$

Assuming it is able to ionize due to the low molality, because if it was higher, then it won't ionize. Hence, since we have three moles of ion products, one Ca²⁺ and two I⁻, we can conclude the van't Hoff factor would be 3, although calculations may lead to a different, yet close result.

Learn more about the van't Hoff factor: brainly.com/question/23764376

5 0

2 years ago

What kinds of substances besides water can be involved in hydrogen bonding?

MrMuchimi

Kind of substance besides water:
The best example of hydrogen bonding excluding water is DNA. The two strands of polymers are connected by hydrogen bonds between the nucleotide bases.

7 0

3 years ago