3 years ago

An aqueous HCL solution has a proton concentration equal to 6.00 mol/L. The HCL concentration in this solution is ————M.

Chemistry

1 answer:

Harman [31]3 years ago

4 0

Answer:

HCl conc.= 6.0mol/L

Explanation:

From the dissociation of HCl= 1 mole H+ and 1mol Cl-, which is equivalent stoichiometrically in concentration to that of 1 mol HCl,

You might be interested in

Nascar fans love race day when they get a chance to cheer on there favorite racer if a driver was able to travel 600 miles in 3

hjlf

Answer:

200 mph

Explanation:

600/3=200

5 0

3 years ago

Part A In a particular experiment at 300 ∘C, [NO2] drops from 0.0138 to 0.00886 M in 374 s . The rate of disappearance of NO2 fo

sweet-ann [11.9K]

Answer:

The rate of disappearance of $NO_2$ for this period is $-1.32\times 10^{-5} M/s$

Explanation:

Initial concentration of $NO_2$ = x = 0.0138 M

Final concentration of $NO_2$ = y = 0.00886 M

Time elapsed during change in concentration = Δt = 374 s

Change in concentration , $\Delta [NO_2]$ = y - x = 0.00886 - 0.0138 M = -0.00494 M

The rate of disappearance of $NO_2$ for this period is:

$\frac{\Delta [NO_2]}{\Delta t}=\frac{-0.00494 M}{374 s}=-1.32\times 10^{-5} M/s$

4 0

3 years ago

Convert 4.500000000 to scientific notation

aniked [119]

I think it’s 4.5 x 10^9

7 0

3 years ago

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?

Rom4ik [11]

M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

M(HCl) = 0.35
V(HCl) = 45mL
M(NaOH)= 0.35

now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

0.35 * 45 = 0.35 * V(NaOH)

V(NaOH) = 45 mL

8 0

3 years ago

Read 2 more answers

if you could take 1 step every second and one step is equal to 3 feel how many miles could you walk in a year of 365 days if you

skad [1K]

If you never stopped you could walk 17918.2 miles

6 0

3 years ago