Which indicator will be the same color in hydrochloric acid as it is in pure water?1.methyl orange
2.phenolphthalein
3.bromcresol green

Answers

Answer 1

Answer: Phenolphthalein
In acid it is colourless and in water also colourless

Related Questions

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What is the hydrogen-ion concentration of the ph is 3.7
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A pharmacist wishes to strengthen a mixture from 10%alcohol to 30% alcohol. How much pure alcohol should be added to 7 liters of the 10% mixture?

Answers

Answer:

2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.

Explanation:

Suppose x is the number of litres added to the 10% mixture than the quantity of new mixture is given as below

$n_(old)=7$ litres
$n_(new)=7+x$ litres

Also the quantity of alcohol is given as

$q_(old)=10 \% \, of \, 7 \, litres =0.7$

$q_(added)=x$

$q_(new)= 30 \% \,of \,new\, quantity = 0.3(7+x)$

Now the equation is as

$q_(old)+q_(added)=q_(new)\n0.7+x=0.3(7+x)\n0.7+x=2.1+0.3x\nx-0.3x=2.1-0.7\n0.7x=1.4\nx=2 \, litres$

So 2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.

A chemist prepares hydrogen fluoride by means of the following reaction:CaF2 + H2SO4 --> CaSO4 + 2HF

The chemist uses 11 g of CaF2 and an excess of H2SO4, and the reaction produces 2.2 g of HF.
(a) Calculate the theoretical yield of HF.

(b) Calculate the percent yield of HF.

Answers

Answer:

39.3%

Explanation:

CaF2 + H2SO4 --> CaSO4 + 2HF

We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:

For CaF2;

Number of moles reacted= mass/molar mass

Molar mass of CaF2= 78.07 g/mol

Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass

Molar mass of hydrogen fluoride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

Which of the following examples illustrates a number that is correctly rounded to three significant figures? a. 109 526 g to 109 500 g
b. 0.03954 g to 0.040 g
c. 20.0332 g to 20,0 g
d. 04.05438 g to 4,054 g
e. 103.692 g to 103.7g

Answers

Answer:

c. 20.0332 g to 20,0 g

Explanation:

A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.

Which of the following examples illustrates a number that is correctly rounded to three significant figures?

a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.

b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.

c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.

d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.

e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.

What is the concentration of hydronium ion ( [H3O+]) in a solution with a PH of _1,3?

Answers

Answer: [H3O+]= 0.05 M

Explanation:

Message me for extra explanation.

snap- parkguy786

Give an example of an element and an example of a compound.

Answers

Answer:

an element is a atom like titanium and a compound is like a water, glucose, alcohol and salt

Explanation:

the titanium is a element and water, glucose, alcohol, and salt those are a compound

Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answers

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions: 1 M , 1 atm).

Nernst equation at 25ºC is:

$E = E^(0) - [((0.0592)/(n) · log Q)]$

where

E: Cell potential (non standard conditions)

$E^(0)$ = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

$Q = (C^(c) * D^(d) )/(A^(a)*B^(b) )$

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: $E^(0)$ , n and Q

Let´s calcule potential in nomal conditions ( $E^(0)$ ):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions: Eo (v):

$(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ -0.23

$Ni --> Ni^(2+)$ + 2 e- +0.99

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions: Eo (v):

2 * [ $(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ ] -0.23

1 * [ $Ni --> Ni^(2+)$ + 2 e- ] +0.99

------------------------------------------------------------------- -------------

2 $(VO_(2))^(2+)$ + 2e- + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$ + 2e- 0.76 v

Then global balanced chemical reaction is:

2 $(VO_(2))^(2+)$ + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$

and the potential in nomal conditions is:

$E^(0)$ = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

$Q = ([(VO_(2)+]^(2)* [Ni^(2+) )/([VO_(2+)]^(2) * Ni )]$

From the exercise we know:

$[VO_(2) ^(2+)] = 2.5 M$

$[VO_(2)+] = 0.083 M$

$[Ni^(2+)] = 2.5 M$

$[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.$

$Q = ((2.5)^(2)* 2.5 )/((0.083)^(2) * 1) = 2,268.11$

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

$E = E^(0) - ((0.0592)/(n) - log Q)\n \nE = 0.76 - ((0.0592)/(2)-log (2,268.11) \nE = 0.76 - (0.0296 - 3.36)\nE = 4.09 v$

Cell potential under non standard concentration is 4.09 v

Final answer:

To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.

Explanation:

For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.

Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.

Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).

After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.