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3 years ago

Which of these equations is balanced? a. 3 H2SO4 + 2Al → Al2(SO4)3 + 3H2 b. 2KCl + Pb(NO3)2 → KNO3 + PbCl2 a b a and b neither

2 answers:

8 0

A.) 3 H2SO4 + 2 Al > Al2(SO4)3 + 3 H2
6 H+ +3 S +12 O +2 Al> 2 Al + 3S + 12O + 6H+
This one is balanced, with 6 H, 3 S, 12 O, and 2 Al on both sides.
B.) 2KCl + Pb(NO3)2 > KNO3 + PbCl2
2K + 2Cl + Pb + 2N +6 O > 1 K + 1 N+ 3 O + Pb + 2 Cl
This one is not balanced as there is 2 K, 6 O and 2 N on the reactants side but only 1 K, 3 O, and 1 N on the products side.
Therefore, only A.) is balanced

Nataly_w [17]3 years ago

3 0

Only equation A is balanced.
To balance equation B, simply add another molecule of KNO3, so the final equation would be:
2KCl + Pb(NO3)2 ---> 2KNO3 + PbCl2

Please mark my answer as brainliest

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A and B are two imaginary substances. The molecules of substance A have weak electrical forces between them. The

GrogVix [38]

I believe the answer is A, if they have a stronger connection lower the vapor pressure so higher the boiling point

7 0

1 year ago

How many moles of molecules are in 40g of H2O?

soldier1979 [14.2K]

Molecular mass of H2O is: 16+2*1=18
so moles of 40g of H2O is: 40/18=2.22

6 0

3 years ago

Read 2 more answers

What is the molar mass of (NH4)3 PO4? 113g, 121g, 149g, 339g

allochka39001 [22]

(NH4)3PO4 :

N = 14 * 3 = 42
H = 1 * 12 = 12
P = 31 * 1 = 31
O = 16 * 4 = 64
-------------------------
42+12+31+64 = 149 g / mol

Hope this helps!.

7 0

3 years ago

The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r

Vinil7 [7]

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

5 0

3 years ago

Lithium reacts with bromine (Br2) in a synthesis reaction to produce lithium bromide. Determine the limiting reactant if 25.0 gr

Lunna [17]

Answer: Bromine is the limiting reactant

Explanation:

First of all let's generate a balanced equation for the reaction

2Li + Br2 —> 2LiBr

Molar Mass of Li = 7g/mol

Molar Mass of Br2 = 2x80 = 160g/mol

From the question given, were told that 25g of Li and 25g Br2 were present at the take-off of the reaction. Converting these Masses to mole, we have:

Number of mole of Li = 25/7 = 3.6moles

Number of mole of Br2 = 25/160 = 0.156mol.

To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2

For the experimental yield:

Li : Br2 = 3.6/ 0.156 = 23 : 1

For the theoretical yield:

Li : Br = 2 : 1

From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.

4 0

3 years ago