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Dvinal [7]
3 years ago
10

2Al + 3Br2 --> 2AlBr3

Chemistry
1 answer:
vlada-n [284]3 years ago
4 0
A) The limiting reactant is Al
b) Br2 is the excess reactant
c) The amount moles of AlBr3 that get produced will be equal to the number of moles of Al to begin with.
d) 0
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What causes some aqueous solutions to have a low PH?
noname [10]
<span> Low pH means the solution is acidic; if the solution is acidic, that means that there are hydronium ions in solution (H3O+). For example, hydrochloric acid dissolves into H+ ions and Cl- ions, and the H+ ions rreact with water like this:
H+ + H2O --> H3O+

If you want to get mathematical, pH is defined as the negative log of the concentration of hydronium ions.
Thus, if there are a lot of hydronium ions, the solution will have a low, or acidic, pH.  Hope this helps^-^</span>
5 0
3 years ago
Explain the difference between cocl2 6h2o and anhydrous cobalt chloride
lana66690 [7]
<span>Colbat (ii) which is a compound birth out of the combination of chlorine and colbat to form Cocl2.6h2o has water in it as we can see from it's chemical it's hexahydrate Anhydrous cobalt chloride as the word anhydrous clearly states , does not have water in</span>
8 0
4 years ago
Is there a qualitative difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process
Vesna [10]

Answer:

No

Explanation:

given that, enthalpy is a state function, that means it depends only on the initial and final states,  there is no difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process, when the cooling or heating process finish in a change of phase.

It does not  matter which way we take to cool or heat the substances the Enthalpy of this process will be the same.

3 0
3 years ago
? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2
Novay_Z [31]
Data: 
M_{concentrated} = 2.5\:mol
V_{concentrated} = ?
M_{dilute} = 0.50\:mol
V_{dilute} = 100\:mL\to0.100\:L
<span>
Formula: Dilution Calculations

</span>M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
<span>
Solving:

</span>
M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}
2.5 * V_{concentrated} = 0.50 * 0.100
2.5V_{concentrated} = 0.05
V_{concentrated} =  \frac{0.05}{2.5}
\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark<span>







</span>
3 0
4 years ago
2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.
AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

7 0
3 years ago
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