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3 years ago

7

How many grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl? Show all of the work needed to solve this

problem.

1 answer:

Natasha_Volkova [10]3 years ago

3 0

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A particle has 5 protons, 6 neutrons, and 5 electrons. What is its net charge.

Trava [24]

It would have a zero charge because it is a neutral atom. The number of electrons which is a negative charge is equal to the number of protons which is positive, they will cancel each other out hence meaning it will become neutral

4 0

3 years ago

An irregularly shaped stone was lowered into a gradulated cylinder holding a volume of water equal to 12mL.the height of the wat

denis23 [38]

The density of the stone is 5 because the formula is mass/volume and the volume is 5 and the mass is 25

4 0

3 years ago

Find percent yield:

saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

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

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

<u>For $B_5H_9$ :</u>

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

<u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

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

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago

A chemist carefully measures the amount of heat needed to raise the temperature of a 894.0g sample of a pure substance from to −

astraxan [27]

Answer:

C = 0.2349 J/ (g °C)

Explanation:

Mass, m = 894.0g

Initial Temperature = −5.8°C

Final Temperature = 17.5°C

Temperature change = 17.5°C - (−5.8°C) = 23.3

Heat, H = 4.90kJ = 4900 J

Specific heat capacit, C = ?

The relationship between these quantities is given by the equation;

H = mCΔT

C = H / mΔT

C = 4900 / (894)(23.3)

C = 0.2349 J/ (g °C)

4 0

3 years ago

In step 2, of the experiment, the procedure uses 3.0M NaOH. However, the student notices that the only solution of NaOH is conce

Luda [366]

Answer:

We need 78.9 mL of the 19.0 M NaOH solution

Explanation:

Step 1: Data given

Molarity of the original NaOH solution = 19.0 M

Molarity of the NaOH solution we want to prepare = 3.0 M

Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

Step 2: Calculate volume of the 19.0 M NaOH solution needed

C1*V1 = C2*V2

⇒with C1 = the concentration of the original NaOH solution = 19.0 M

⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED

⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M

⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

19.0 M * V2 = 3.0 M * 0.500 L

V2 = (3.0 M * 0.500L) / 19.0 M

V2 = 0.0789 L

We need 0.0789 L

This is 0.0789 * 10^3 mL = 78.9 mL

We need 78.9 mL of the 19.0 M NaOH solution

8 0

3 years ago