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pshichka [43]
3 years ago
11

Can you please help me

Chemistry
1 answer:
rosijanka [135]3 years ago
7 0

Shure what you need help with

You might be interested in
what is the molarity of an HCI solution if 25.0 ml of 0.185 M NaOH is required to neutralize 0.0200 L of HCI?​
butalik [34]

Answer:

Molarity of HCl solution = 0.25 M

Explanation:

Given data:

Volume of NaOH= V₁ = 25.0 mL (25/1000 = 0.025 L)

Molarity of NaOH solution=M₁ = 0.185 M

Volume of HCl solution = V₂ = 0.0200 L

Molarity of HCl solution =  M₂= ?

Solution:

M₁V₁   =  M₂V₂

0.185 M ×0.025 L =  M₂ × 0.0200 L

M₂  = 0.185 M ×0.025 L / 0.0200 L

M₂  = 0.005M.L /0.0200 L

M₂  =  0.25 M

7 0
3 years ago
why do you think the purple cabbage indicator used in this experiment is called "red cabbage"? where is it typically used in foo
tatiyna

Answer:

Red cabbage juice contains a natural pH indicator that changes colors depending on the acidity of the solution. The pigment in red cabbage that causes the red color change is called flavin (an anthocyanin). ... At a lower pH, more hydrogen ions are in solution, and therefore the solution is acidic

Actually I don't know but I tried so I hope it helped u!

4 0
3 years ago
If 200. mL of 0.60 M MgCl2(aq) is added to 400 mL of distilled water, what is the concentration of Mg and Cl in the resulting so
sladkih [1.3K]

Answer:

C. 0.20 M Mg ion & 0.40 M Cl ion

Explanation:

MgCl₂ is a ionic salt which is dissociated as this

MgCl₂  →  Mg²⁺  +  2Cl⁻

First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M

Molarity . volume = moles.

0.6 mol/l . 0.2l = 0.12 mol

  MgCl₂  →  Mg²⁺  +  2Cl⁻

0.12mol      0.12         0.24

This moles are also in 400mL of water, so the new concentration is

[Mg²⁺] = 0.12 m/0.6L = 0.2M

[Cl⁻] = 0.24 m/0.6L = 0.4M

Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)

8 0
3 years ago
I need help with this.
Phoenix [80]

Answer:

can you show a clearer picture of the chart?

Explanation:

4 0
3 years ago
Read 2 more answers
The combustion of ammonia in the presence of excess oxygen yields no2 and h2o: 4 nh3 (g) + 7 o2 (g) â 4 no2 (g) + 6 h2o (g) the
VLD [36.1K]
The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2 
so 1mole of NH3 will give 1 mole of NO2 
43.9 grams of NH3 contains 2.58 moles 
so 2.58 moles will be produced of NO2 
which is 118.7 grams this the amount of oxygen that is used.
8 0
3 years ago
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