All categories

3 years ago

How many moles of silver chloride, produced from 100 g of silver nitrate reacting with barium chloride BaCl2?

Chemistry

1 answer:

Dmitrij [34]3 years ago

7 0

Answer:

Equation of Reaction

2AgNO3 + BaCl2 === 2AgCl + Ba(NO3)2

Molar Mass of AgNO3 = 170g/mol

Moles of reacting AgNO3 = 100g/170gmol-¹

=0.588moles of AgNO3

From the equation of reaction...2moles of AgNO3 reacts to Produce 2Moles of Silver Chloride

So Their ratio is 2:2.

This means that 0.588Moles of AgCl Will be produced too.

ANSWER...0.588MOLES OF AgCl WILL BE PRODUCED.

You might be interested in

The best determinant of how an individual measures their quality of life is _______.

Ronch [10]

C is the correct answer

4 0

3 years ago

Read 2 more answers

83ef0c8

kumpel [21]

Answer:

0.17325 moles per liter per second

Explanation:

For a first order reaction;

in[A] = in[A]o - kt

Where;

[A]= concentration at time t

[A]o = initial concentration

k= rate constant

t= time taken

ln0.5 =ln1 - 2k

2k = ln1 - ln0.5

k= ln1 - ln0.5/2

k= 0 -(0.693)/2

k= 0.693/2

k= 0.3465 s-1

Rate of reaction = k[A]

Rate = 0.3465 s-1 × 0.50 mol/L

Rate = 0.17325 moles per liter per second

5 0

3 years ago

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(d) 5.363

(e) 5.570

(f) 12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886

(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH = √(kb x (A⁻) where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0

3 years ago

What two motions combine to produce orbit?

Alex73 [517]

<span>Forward & falling. Hope this helps!</span>

6 0

3 years ago

View the diagram below

Sauron [17]

Answer : The correct answer is the Bonds were broken on the reactants and new bonds were formed on the products.

Explanation :

In the chemical reaction, some substances react together are called reactant and the substance are formed are called product.

During the chemical reaction, the atoms of reactants rearranged to make products. There are on atoms are added or taken away in the reaction. This is known as the conservation of atoms.

For example : carbon atom react with the oxygen to form carbon dioxide.

$C+O_2\rightarrow CO_2$

From the given diagram, we conclude that the arrangement of molecules are different on both side of the mixture of reaction.

On the reactant side, the red molecules bonded with red molecule and the black molecule with white molecules. On the other hand i.e product side, the red molecule bonded with black molecule and white molecule bonded with red molecules. The molecular arrangement are different on both side of the reaction mixture.

Therefore, the correct answer is the Bonds were broken on the reactants and new bonds were formed on the products.

6 0

3 years ago