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3 years ago

Can anyone help me part 1

Chemistry

1 answer:

valina [46]3 years ago

8 0

I can help you with part 1

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Please help !!

ipn [44]

Answer:

For 2: The % yield of the product is 92.34 %

For 3: 12.208 L of carbon dioxide will be formed.

Explanation:

For 2:

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(1)

Given values:

Actual value of the product = 78.4 g

Theoretical value of the product = 84.9 g

Plugging values in equation 1:

$\% \text{yield}=\frac{78.4 g}{84.9g}\times 100\\\\\% \text{yield}=92.34\%$

Hence, the % yield of the product is 92.34 %

For 3:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Given mass of carbon dioxide = 24 g

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

$\text{Moles of carbon dioxide}=\frac{24g}{44g/mol}=0.545 mol$

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.545 moles of carbon dioxide will occupy = $\frac{22.4L}{1mol}\times 0.545mol=12.208L$ of volume

Hence, 12.208 L of carbon dioxide will be formed.

5 0

2 years ago

What is the mass of 3.00 moles of magnesium chloride, MgCl2? Express your answer with the appropriate units.

Tresset [83]

To answer this question you need to know how to calculate the molecular weight of a molecule. The compound is having 1 Mg atom and 2 Cl atom, thus the molecular weight should be:
24.305+ 2*35.453= 95.211 gram/mol.

Then, the mass of 3 mol of MgCl would be:
3 mol * 95.211 gram/mol= 285.633 gram

7 0

3 years ago

Need help please Balancing the equation in the pic

IRINA_888 [86]

2KClO3=2KCl+3O2
The equation

7 0

3 years ago

What is the proper name for the following compound CaCl2?

RoseWind [281]

Calcium chloride.

It might seem reasonable to call it calcium dichloride, but the IUPAC name for it is calcium chloride.

7 0

3 years ago

Read 2 more answers

In basic solution, se2− and so32− ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this

Annette [7]

(a) Write balanced half-reactions for the process:

Oxidation: Se^2- (aq) → Se (s) + 2e-

Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- → S2O3^2- + 6OH- (aq)

(b) If E sulfite is 0.57 V, calculate E selenium:

E anode = E cathode – E cell

= -0.57 – 0.35

= -.092

3 0

3 years ago