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Margarita [4]
3 years ago
9

If methyl orange is added to a substance and it turns the substance red, what does this tell you about the pH of the substance?

Would this substance be an acid or a base?
Chemistry
2 answers:
prisoha [69]3 years ago
6 0

Explanation:

methyl orange turns pink in an acidic solution

yellow in basic and orange in neutral

MatroZZZ [7]3 years ago
5 0

Answer: The substance would be a strong acid.

Explanation:

Methyl orange turns orange when it is added to a medium acid and turns red when it is added to a strong acid. The pH of the acid would be above 4 because it is a strong acid. Methyl orange turns yellow when it is added to a medium base.

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Convert 4.6 atm to mmHg.
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3496 you can google this you know
3 0
3 years ago
A reaction has ∆H = −356 kJ and ∆S = −36 J/K. Calculate ∆G (kJ) at 25°C.
Cloud [144]

Answer: -345.2 KJ

Explanation: As we know that ,dG=dH-TdS

T=25+273=298 K

dG= -356 x1000-298(-36)= -356000+10728

=-345272 j

= -345.2 KJ

5 0
3 years ago
A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both s
marusya05 [52]

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9

4 0
3 years ago
Read 2 more answers
At Jim's auto shop, it takes him minutes to do an oil change and minutes to do a tire change. Let be the number of oil changes h
Juli2301 [7.4K]

Answer:

The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.

The time needed for x oil changes is 12 * x.

The time needed for y tire changes is 18 * y.

The total time is the sum of the above times and needs to be less than 180 that is

12 * x + 18 * y < 180 divide both sides of equation by 6

12/6 * x + 18/6*y < 180/6

2*x + 3*y < 30

2*x < 30 - 3*y divide both sides by 2 to get the inequality for x

x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15

2*x + 3*y < 30

3*y < 30 - 2*x divide both sides by 3 to get the inequality for y

y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10

Also we can write x + y < x+ 3/2 * y < 15.

Explanation:

Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.

5 0
3 years ago
A graduated cylinder contains 20.5 mL of water. What is the new water level after 35.2 g of silver metal with a density of 10.5
iragen [17]
 <span>V = 24.0 mL + (35.2 g)(mL/10.5g) = I think i'm not all that sure but I think its this.</span>
3 0
3 years ago
Read 2 more answers
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