K_C204(aq) + _<br> NaOH(aq) →

What must be known to determine the nature of the salt formed from a weak acid and weak base?

Umnica [9.8K]

D the strength of both acid and base

4 0

3 years ago

Buffers are substances that help resist shifts in pH by Question 10 options: 1) releasing H in acidic solutions. 2) donating H t

ehidna [41]

Answer:

5) donating H to a solution when they have been depleted and accepting H when they are in excess.

Explanation:

A buffer solution is a solution that undergoes a negligible change in pH in addition of moderate quantities of acid or alkali. In other words, a buffer solution is one that resists a change in pH on addition or dilution of small amounts of acids or alkalis.

So, from the given question:

Buffers are substances that help resist shifts in pH by donating H to a solution when they have been depleted and accepting H when they are in excess.

8 0

3 years ago

What is one of the ways a fission reaction can get out of control?

cricket20 [7]

Answer : When the nuclear reaction produces too many neutrons.

Explanation : In a nuclear fission reactor where the reaction is controlled by regulating the number of neutrons produced by the chain reaction. Production of too many neutrons may cause an accelerated uncontrolled chain reaction which may cause enormous heat to be generated in a very short period of time inside the reactor which may result into an explosion.

For, this purpose control rods, made of neutron absorbing species, like cadmium, are placed in the nuclear reactors to control the excess number of neutrons at any point of reaction when required.

5 0

4 years ago

Read 2 more answers

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viktelen [127]

Answer:

HmmmmmmmmmmI think the answer is 69

3 0

3 years ago

Read 2 more answers

0.0500 mol of gas occupies a cylinder which is sealed on top by a moveable piston. The piston is circular, with a mass of 30.0 k

Viefleur [7K]

Answer:

The workdone by both N₂ and neon gas is 49.3 J

The change in internal energy of N₂ and neon gas is 125.6 J and 73.54 J respectively

The heat for N₂ and neon gas is 171.9 J and 122.84 J respectively.

Explanation:

Given that:

number of moles = 0.05 mole

mass of the piston = 30 kg

diameter = 5.00 cm = 0.05 m

Area (A) = πr²

$Area (A) = \pi*(\frac{0.05}{2})^2 \\ \\ Area (A) = 0.0019635 \\ \\ Area (A) = 19.635*10^{-4} \ m^2$

The piston is said to move from 30 cm - 40 cm

So, the change in volume ΔV is calculated as:

$=(40-30)*10^{-2} *19.635*10^{-4}$

$= 1.9635*10^{-4} \ m^3$

Outside the cylinder; the pressure $P_{air}= 1 \ atm$ = 101325 Pa

Thus, workdone $w_1$ = PΔV

= $101325*1.9635*10^{-4}$

= 19.90 J

The gravitational work $w_2 = mgh$

Given that the height (h) = 10 cm = 0.1 m

Then; $w_2 = 30*9.8*0.1$

$w_2 = 29.4 \ J$

The total workdone $w_{total}$ for both cases is:

$w_{total } =w_1 + w_2$

$w = (19.90 + 29.4) \ J$

$w =49.3 \ J$

The pressure of gas inside the cylinder is determined as:

$P_{in}.A = P_{out}.A +mg$

$(P_{in}-P_{out}) = \frac{mg}{A} \\ \\ P_{in} -10^5 = \frac{30*9.8}{19.635*10^{-4}} \\ \\ P_{in} = 149732.6203+10^5 \\ \\ P_{in} = 2.497*10^5 \ Pa$

a). assuming that the gas is N₂.

$C_v =\frac{5}{2}R$

Thus, the change in internal energy ΔU is given as:

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{5}{2}R \delta T \\ \\ \delta U = \frac{5}{2}nR \delta T$

Since $P_{in} \delta V = nR \delta T ; \ Then$ ;

$\delta \ U = \frac{5}{2} P_{in} \delta V \\ \\ \delta \ U = \frac{5}{2}*2.497*10^5 *1.9635*10^{-4} \\ \\ \delta \ U = 122.57 \ J$

ΔU ≅ 125.6 J

The heat Q = ΔU + W

Q = (122.6 + 49.3) J

Q = 171.9 J

b) In Neon gas:

$C_v = \frac{3}{2}R$

∴

change in internal energy is;

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{3}{2}R \delta T \\ \\ \delta U = \frac{3}{2}P_{in}.V$

$\delta U = \frac{3}{2}*2.497*10^5*1.9635*10^{-4}$

ΔU = 73.54 J

The heat Q = ΔU + W

Q = (73.54 + 49.3) J

Q = 122.84 J

5 0

3 years ago

K_C204(aq) + _ NaOH(aq) →