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bonufazy [111]
3 years ago
12

What should you do during a titration when you notice the indicator start to indicate the approach of the equilibrium point? Add

the second reactant faster. Add the second reactant slower. Add more indicator.
Chemistry
2 answers:
Ksju [112]3 years ago
7 0

Answer:

Add the second reactant faster.

Explanation:

In titration we take, we take a reactant (titrand) in conical flask. The second reactant (titrant) is filled in burette.

Now we add indicator in the first reactant (taken in flask) and start adding the second reactant from burette.

The reaction starts and when it appears that we are approaching the end point, we start adding the second reactant slower, in order to avoid missing the end point. If we add the second reactant fast, it may lead to addition of excess of reagent and thus may affect our calculations and readings.

Leni [432]3 years ago
6 0

B. Add the second reactant slower.

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3 years ago
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands
Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

E=\frac{R_y}{n^2}

R_y=2.18\times 10^{-18} J =  Rydberg energy

n =  principal quantum number of the orbital

Energy of 11th orbit = E_{11}

E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J

Energy of 10th orbit = E_{10}

E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J

=E'=0.38\times 10^{-20} J

\lambda =\frac{hc}{E'} (Planck's' equation)

\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}

\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

3 0
3 years ago
3 examples of potential energy
spin [16.1K]
The potential energy .. 
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3 years ago
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Basile [38]

Answer:

B-2

Explanation:

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4 0
2 years ago
Read 2 more answers
A 6.40 g sample of a compound is burned to produce 8.37 g CO_2, 2.75 g H_2O, 1.06 g N_2, and 1.23 g SO_2. What is the empirical
LuckyWell [14K]

The empirical formula :

C₁₀H₁₆N₄SO₇

<h3>Further explanation</h3>

Given

6.4 g sample

Required

The empirical formula

Solution

mass C :

= 12/44 x 8.37 g

= 2.28

mass H :

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mass N = 1.06

mass S :

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= 0.615

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Mol ratio :

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= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019

= 10 : 16 : 4 : 1 : 7

The empirical formula :

C₁₀H₁₆N₄SO₇

3 0
3 years ago
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