Answer the question below. Use the rubric in the materials for help if needed.

Need help with this question plsssss

GaryK [48]

Answer:

234

Explanation:

8 0

2 years ago

Why dilute H2SO4 is added to oxali acid solution in redox reaction

Sphinxa [80]

To prevent the hydrolysis and to catalyse the reaction.

Explanation:

Sulphuric acid is the catalyst and also a dehydrating agent in this reaction.

Sulphuric acid is using in redox reaction because sulphuric acid is providing H+ ions which is necessary for this reaction to occur more quickly, but the sulphate ions from the sulphuric acid barely react during this process. So H2SO4 is adding in this reaction to make it more acidic.

H2SO4 is preventing hydrolysis by providing excess H+ ions into the reaction. H2SO4 is stable towards the direction of oxidation.

7 0

3 years ago

Using aluminium as an example, describe the key properties of p-block metals.

olga nikolaevna [1]

Answer:

P-block metals have classic metal characteristics like they are shiny, they are good conductors of heat and electricity, and they lose electrons easily. These metals have high melting points and readily react with nonmetals to form ionic compounds.

Explanation:

6 0

2 years ago

Read 2 more answers

Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

STEP I

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

STEP II

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

STEP III

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

STEP IV

Canceling spectator ions, the ions that appear the same on either side of the equation

(note: in the above step the ions in bold have gotten canceled.)

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

3 years ago

A nuclear power plant operates at 40.0% efficiency with a continuous production of 1042 MW of usable power in 1.00 year and cons

Sergio039 [100]

Answer:

3.00 x 10^-11 joules / atom of U-235

Explanation:

We know that the formula for Power = Work done (w)/Time (t)

We need to get the joules from power , since Joules is the SI unit of work.

From the formula P = W/t

W = Power (P) * Time (t)

The SI unit for Time is seconds, hence we change 1 year in seconds

1yr * 365 days/yr * 24hrs/day * 60mins/hr * 60 secs/min = 31536000 secs

It was stated in the question that the plant operates at an efficiency of 40%,

Thus to get the true power we divide the power provided in the question by 0.4 or 40%

= X(0.4) = 1042MW

True Power X = 1042/0.4 = 2605MW

Thus true power = 2605 * 10^6 Watts

Now we have the time in seconds and true power in Watts, we then find the work done.

From our above formula P = W/t

W = P*t = (2605* 10^6) (31536000) =

Finally, we can solve for our energy (work):

P = W / T PT = W = (2880x10^6) (31536000) = 8.22 x 10^16 joules

We then calculate the amount of energy released by only 1 single uranium-235 atom.

= 8.22 x 10^16 joules / 1.07x10^6 g U-235 (235 g / 1 mol)(1 mol/6.0210^23 atoms)

= 3.00 x 10^-11 joules / atom of U-235

5 0

3 years ago