What are the products of a reaction between ammonium iodide and magnesium sulfate?

The work function of palladium (Pd) is 5.22 eV. What is the minimum frequency of light to observe the photoelectric effect from

Ivan

Answer:

1)4×10^13Hz

2) 9.95×10-9J

Explanation:

From the image attached, it is clear that the work function of the palladium metal must first be obtained in joules. Then, the frequency is obtained from E=hf.

The kinetic energy if the photoelectrons is obtained as the difference between the energy of the photon and the work function of the metal.

4 0

3 years ago

Need help with 14 and 16 pls asap!! this is my friends test and im taking it tomorrow!!

marin [14]

Answer:

Q14: 17,140 g = 17.14 kg.

Q16: 504 J.

Explanation:

Q14:

To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).

m is the mass of the ice (m = ??? g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).

∵ Q = m.c.ΔT

∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)

∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.

Q16:

To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 12.0 g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).

∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.

6 0

3 years ago

How many moles of Kl are needed to make 4.07 L of a 1.138 M solution?

Nata [24]

Answer:

A

Explanation:

cause

5 0

3 years ago

When scientists use one of their five senses to gather information, they are A. making an observation. B. making an inference. C

Goshia [24]

A. making an observation.

Explanation:

When scientists use one of their five sense to gather information, they are simply making an observation.

Most physical properties of matter are studied using sense of taste, of feel, sight, hearing and smell.

Observation is the act of making a measurement or studying.
It helps to investigate particular phenomenon in nature.
Most observations made with our senses are usually qualitative in nature.
They cannot be assigned values like quantitative ones.

learn more:

Observations brainly.com/question/5096428

#learnwithBrainly

7 0

3 years ago

Read 2 more answers

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago