All categories

2 years ago

The question is in the photo

Chemistry

1 answer:

nadya68 [22]2 years ago

4 0

The required moles of AgBr precipitate produced by given moles of silver nitrate is 0.0123.

<h3>How do we calculate moles from molarity?</h3>

Molarity of any solution is define as the moles of solute present in per liter of the solution and it will be represented as:
M = n/V

Given that, molarity of AgNO₃ = 0.250M

Volume of AgNO₃ = 49.5mL = 0.0495L

Moles of AgNO₃ = (0.25)(0.0495) = 0.0123mol

Given chemical reaction is:

2AgNO₃(aq) + CaBr(aq) → 2AgBr(s) + Ca(NO₃)₂(aq)

As it is mention that CaBr is present in excess quantity and AgNO₃ is the limiting reagent so the formation of precipitate will depend on the AgNO₃.

From the stoichiometry of the reaction, it is clear that:

2 moles of AgNO₃ = produces 2 moles of AgBr

0.0123 moles of AgNO₃ = produces 2/2×0.0123=0.0123 moles of AgBr

Hence 0.0123 is the required moles of precipitate.

To know more about moles, visit the below link:

brainly.com/question/19099163

#SPJ1

You might be interested in

what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require

vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$ . So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940 0 0

C: -x +x +x

E: 0.1940-x x x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or, $x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$

5 0

3 years ago

Calculate the wavelength of the photon that would be absorbed or emitted. Round your answer to 3 significant digits.

Flauer [41]

This is an incomplete question, the image for the given question is attached below.

Answer : The wavelength of photon would be absorbed, $3.06\times 10^{-7}m$

Explanation :

From the given diagram of energy we conclude that,

Energy at ground state, A = 400 zJ

Energy of 2nd excited state, C = 1050 zJ

Now we have to calculate the energy of the photon.

$E=E_c-E_A$

$E=(1050-400)zJ= 650zJ=650\times 10^{-21}J$

Now we have to calculate the wavelength of the photon.

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

E = energy of photon = $650\times 10^{-21}J$

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon = ?

c = speed of light = $3\times 10^8m/s$

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

$650\times 10^{-21}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=3.06\times 10^{-7}m$

Therefore, the wavelength of photon would be absorbed, $3.06\times 10^{-7}m$

5 0

3 years ago

A balloon is filled with 3.8 l of helium gas at stp. approximately how many moles of helium are contained in the balloon?

love history [14]

Hope this helps you.

7 0

3 years ago

Read 2 more answers

An electrochemical cell is constructed using two half-cells: Al(s) in Al(NO2)3(aq) and Cu(s) in Cu(NO3)2(aq). The two half cells

wlad13 [49]

Answer:

Cu(s) in Cu(NO₃)₂(aq)

Explanation:

The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.

The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.

4 0

3 years ago

A solution of pH 5 is diluted 100 times. Find the pH of the resulting solution.

liq [111]

First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:

pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M

M1V1 = M2V2
<span>1 x 10^-5 M (V1) = M2(100V1)
</span>M2 = 1 x 10^-7

pH = -log[<span>1 x 10^-7</span>]
pH = 7

3 0

3 years ago

Read 2 more answers