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Ivahew [28]
3 years ago
10

A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing th

e hydroxide to aluminum oxide by heating. How many grams of aluminum oxide should the student obtain if her solution contains 40.0 mL of 0.541 M aluminum nitrate?
Chemistry
1 answer:
Vladimir [108]3 years ago
4 0

Answer: The student should obtain <u>1.103 g of aluminum oxide </u>

Explanation:

  • First we write down the equations that represent the aluminum hydroxide precipitation  from the reaction between the aluminum nitrate and the sodium hydroxide:

Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3

Now,  the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.

2Al(OH)3 → Al2O3 + 3H2O

  • Second, we gather the information what we are going to use in our calculations.

Volumen of  Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 =  0.541M

Molecular Weight Al2O3 = 101.96 g/mol

  • Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

\frac{0.541moles Al(NO3)3}{1L} x\frac{1L}{1000mL} x 40mL Al(NO3)3 =  0.022moles Al(NO3)3

then we use the molar coefficients from the  equations to discover the amount of Al2O3  moles produced

0.022moles Al(NO3)3 x \frac{1mol Al(OH)3}{1mol Al(NO3)3} X\frac{1mol Al2O3}{2molAl(OH)3} = 0.011 moles Al2O3

finally, we use the molecular weight of the Al2O3  to calculate the final mass produced.

0.011moles Al2O3 x \frac{101.96g Al2O3}{1mol Al2O3} = 1.103g Al2O3

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PLEASE SOMEONE HELP ME WITH THIS!
Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877  and 4) 1.04x10^{3\\}

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x  \frac{1 L}{1000 mL} = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x \frac{10 dg}{ 1 g} = 1.291 dg

For the next exercises, you need to follow some rules:

1.  All numbers  that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062  - 638.1848  = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.  

1.04x10^{3}

4 0
3 years ago
Which formula is an empirical formula? (1) CH4 (2) C2 H6 (3) C3 H6 (4) C4 H10
rosijanka [135]
CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).
3 0
3 years ago
Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.
Vsevolod [243]

The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of acid, H₃PO₄ (Ma) = 0.390 M
  • Volume of acid, H₃PO₄ (Va) = 24.5 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

Learn more about titration:

brainly.com/question/14356286

5 0
2 years ago
How did Mendeleev set up his Periodic Table of Elements?
Ksju [112]
He set up his periodic table by the atomic mass
8 0
2 years ago
Choose all the answers that apply.
max2010maxim [7]

Answer: has properties similar to other elements in group 18, does not react readily with other elements, is part of the noble gas group

Explanation: I’ve done on edg before

6 0
2 years ago
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