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Svetradugi [14.3K]
3 years ago
11

What is the molar mass of AlCl3?

Chemistry
1 answer:
miss Akunina [59]3 years ago
8 0

Answer:

133.33 g/mole

Explanation:

To get the molar mass, you will need to first determine the atoms in the compound.

AlCl₃

We have aluminum (Al) and Chlorine (Cl)

Next step is to determine how many atoms of each element you have in the compound. The subscript will tell you how many. If you see no subscript, that means 1 atom.

Al = 1 atom

Cl = 3 atoms

Then you multiply the number of atoms by the atomic weight to get the molar mass of each element and add them up to get the total molar mass of the compound:

Element          # of atoms        atomic weight           molar mass

Al                         1               x     26.98 g/mole =    26.98 g/mole

Cl                         3              x     35.45 g/mole =   <u>106.35 g/mole</u>

                                                             TOTAL =   133.33 g/mole

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Hemoglobin in your blood does not use elemental iron. It uses iron in the form of Fe2+(aq).
valina [46]

Explanation:

Balanced chemical reaction equation will be as follows.

     2Fe^{2+}(aq) + 2H^{+}(aq) \rightleftharpoons 2Fe^{3+}(aq) + H_{2}(g)

In human body, the neutral iron changes into Fe^{2+}(aq) cation. There will be an oxidation-half reaction and a reduction-half reaction. Equations for this reaction are as follows.

Oxidation: 2Fe^{2+}(aq) \rightleftharpoons 2Fe^{3+}(aq) + 2e^{-}[/tex] .... (1)

Reduction: 2H^{+}(aq) + 2e^{-} \rightleftharpoons H_{2}(g) ...... (2)

On adding both equation (1) and (2), the overall reaction equation will be as follows.

     2Fe^{2+}(aq) + 2H^{+}(aq) \rightleftharpoons 2Fe^{3+}(aq) + H_{2}(g)

Therefore, neutral iron is a part of Heme - b group of Hemoglobin and in an aqueous solution it dissolutes as a part of Heme group. Hence, then it becomes an Fe^{2+} cation.

3 0
3 years ago
How do you do part B?
galina1969 [7]
I cannot see the whole equation.Therefore I shall not answer

7 0
2 years ago
You are carefully watching the temperature of your melting point apparatus as it is heating up. At 132 C it is still a white sol
Lisa [10]

Answer:

See the answer below

Explanation:

<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>

1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.

2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.

6 0
3 years ago
A mixture of hydrogen (2.02 g) and chlorine (35.90 g) in a container at 300 K has a total gas pressure of 748 mm Hg. What is the
Llana [10]

The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.

<h3>How do we calculate the partial pressure of gas?</h3>

Partial pressure of particular gas will be calculated as:

p = nP, where

  • P = total pressure = 748 mmHg
  • n is the mole fraction which can be calculated as:
  • n = moles of gas / total moles of gas

Moles will be calculated as:

  • n = W/M, where
  • W = given mass
  • M = molar mass

Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole

Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole

Mole fraction of hydrogen = 1 / (1+0.5) = 0.6

Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm

Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.

To know more about partial pressure, visit the below link:
brainly.com/question/15302032

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3 0
1 year ago
The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
3 years ago
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