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xenn [34]
3 years ago
13

Concentrated 8.2 of commercial chloride or chlorox diluted to 14% ?

Chemistry
1 answer:
ELEN [110]3 years ago
7 0
Is it 8.06?

Or 58.57?

Don't get mad if there wrong!!
But please let me know if it's right or wrong tho.
You might be interested in
Which formula represents 1,2-ethanediol?
Oliga [24]

Answer:

1) C2H4(OH)2

Explanation:

A 1,2-ethanediol has an ethane structure consisting of two Carbon atoms with a hydrogen from each carbon substituted by a hydroxyl group. This makes it a 1,2-diol.

8 0
2 years ago
Which substance can be decomposed by chemical means
jolli1 [7]

Answer:

the answer is Compounds

Explanation:

Compounds are pure substances formed by the combination of elements; they can be decomposed by ordinary chemical means.

5 0
3 years ago
3k+aici3 3kci+ai is what type of reaction
exis [7]
3K      +     AlCl3 -------->  3 KCl     +          Al
element    compound     compound       element

This is single replacement (displacement) reaction.
3 0
3 years ago
How many moles of ammonia gas occupy 50 mL at at 700 kPa and 30.0 0C?
nalin [4]

For the conversions

I will start with pressure
1atm=101.3kPa
x =700kPa
x=700kPa/101.3kPa
x=6.91atm

Temperature
273K+30.00C
303K

Volume
1L=1000ml
x =50ml
x=0.05L

PV=nRT
6.91*0.05=n*0.08206*303
0.3455=24.86418n
0.3455/24.86418=n
0.0138=n
number of moles = 0.0138moles

Note: 0.08206 is the gas constant in this case
6 0
3 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
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