A. acidic b. basic
; neutral
d. will not be affected
Find the molarity of 10% NaOH.[MOE 2003
a. 1.5
b! 2.5
c. 4
d. 0.4
10 ml of 2.5 N NaOH is mixed with 20 ml of
N HCI. The mixture is diluted to 100 ml. WI
is the nature of mixture?
IMOF 20​

1. You didn't post the question to Number 1.

2.

10% means There's 10g in 100ml of this solution.

This is the weight/volume(w/v) expression of concentration.

So

We have Mass =10g

volume =100ml

Molarity = Moles of solute/volume of solution(in LITRES)

Moles=Mass/Molar mass

Molar mass of NaOH=40g/mol

Mole=10/40

=0.25mole

Volume =100ml =0.1Litres

MOLARITY=0.25/0.1 =

=2.5M

OPTION B.

3. NOTE: THE MOLARITY AND NORMALITY OF NAOH AND HCL ARE THE SAME(This doesn't happen for all compounds tho)

So We can take 2.5N(Normality) of NaOH to be 2.5M(Molarity) NaOH

I think you forgot to write the Normality of the second one. I'll take it to be 1N. Maybe you can then Input supposed value when you're solving on your own

So

1N HCl is same as 1M HCL

We were given their respective volumes

2.5M NaOH can also be written as 2.5mole/volume(in Liters)

The volume of NaOH =10ml or 0.01L

Moles = 2.5mole/L x 0.01L

You notice that Liters on top and bottom cancels out... leaving the moles

So

Mole=0.01x2.5 = 0.025moles of NaOH

we're gonna do the same for 1N HCl in 20ml(0.02L)

So

Mole = 1 x 0.02 =0.02moles

Total Mole = 0.02 + 0.025 =0.045moles

The Final Volume is 100ml as stated in the question. It was diluted to 100ml or 0.1L

So

Final Concentration In Molarity

= Total Moles/Volume in L

=0.045/0.1

=0.45M.

is it a purebred? it looks more like a red nose, but it doesn't have the red could it be a staffordshire bull terrior? unsure if that's the right name for it sorry

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g