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Ainat [17]
2 years ago
12

6.3x10-5M. What is the pH?

Chemistry
1 answer:
BabaBlast [244]2 years ago
5 0

Answer:

https://sensorex.com/ph-calculator/

Explanation:

go to this link and then scroll down a little bit and then you'll find the answer

You might be interested in
A 27.3 g marble sliding to the right at 21.0 cm/s overtakes and collides with a 11.7 g marble moving in the same direction at 12
NISA [10]

Answer:

The answer to this is

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

Explanation:

To solve the question, let us list out the given variables and their values

Mass of first marble m1 = 27.3g

Velocity of the first marble v1 = 21.0 cm/s

Mass of second marble m2 = 11.7g

Velocity of the second marble v2 = 12.6 cm/s

After collision va1 = unknown and va2 = 23.7 cm/s

From Newton's second law of motion, force = rate of change of momentum produced

Hence m1v1 + m2v2 = m1va1 + m2va2 or

va1 = (m1v1 + m2v2 - m2va2)÷m2 or (720. 72-277.29)÷m1 → va1 = 16.24 cm/s

The velocity of the 27.3Kg marble after collision is = 16.24 cm/s

4 0
3 years ago
Can you identify some techniques in separating mixtures? In what
olga55 [171]

Answer:

Boiling

Filtration

Distillation

Evaporation

5 0
3 years ago
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
DONT SKIPP What type of bond will occur between lithium and bromine?
den301095 [7]
An ionic bond is formed between lithium and bromine.
4 0
3 years ago
A star is estimated to have a mass of 2.0 x 10 ^36kg. Assuming it to be a sphere of average radius of 7.0 x 10 ^5 km. Calculate
Montano1993 [528]

Answer:

<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>

<em>b) 8.69 x 10^7 lb/ft^3</em>

<em></em>

Explanation:

mass of the star m =  2.0 x 10^36 kg

radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m

The density of substance ρ = mass/volume

The volume of the star = volume of a sphere = \frac{4}{3}\pi  r^{3}

==> V = \frac{4}{3}*3.142*(7.0*10^8)^{3} = 1.437 x 10^27 m^3

density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3

in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>

in lb/ft^3 =  (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>

6 0
3 years ago
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