Which two particles would be attracted to each other?

Consider the half reactions below for a chemical reaction.

ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0

3 years ago

A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.11 g/mL, 1.8

ehidna [41]

Answer:

D. her results are neither precise, nor accurate

Explanation:

Reason being that: First, she obtained four different result for an experiment ( obviously it's supposed to be one value for the different procedure or atleast very close range) sugar solution and all four values were far apart.

6 0

3 years ago

1.3-3 Determine whether each of the following statements is true or false. If the statement is false, demonstrate this by proof

Inga [223]

Answer:

(a)False (b)False (c)False (d)False (e)False

Explanation:

a) Continuous time signal can also be digital. e.g continuous time square waveform (see first picture)

b) A discrete time signal can also be analog. e.g discrete time exponential function can take any value (see second picture)

c) When both energy and power of signal are infinite, it is neither power signal nor energy signal e.g ramp signals, $e^{t}u(t)$ signal

d)A signal of infinite length can also have finite energy. e. g $e^{t}u(t)$

e)Step waveform u(t) is a power signal but also causal

3 0

3 years ago

How many moles of N₂O₅ are produced when 3.00g of O₂ react completely in the following equation? 2N₂ + 5O₂ → 2N₂O₅ A. 0.0375 B.

a_sh-v [17]

Answer: A. 0.0375

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} O_2=\frac{3.00g}{32g/mol}=0.094moles$

$2N_2+5O_2(g)\rightarrow 2N_2O_5$

According to stoichiometry :

As 5 moles of $O_2$ give = 2 moles of $N_2O_5$

Thus 0.094 moles of $O_2$ give = $\frac{2}{5}\times 0.094=0.0375moles$ of $N_2O_5$

Thus 0.0375 moles of $N_2O_5$ will be produced when 3.00g of O₂ react completely

4 0

4 years ago

Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a 0.4 M HA originally a

Allushta [10]

This question is not complete, the complete question is;

Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a 0.4 M HA originally at pH = 5.0 ( pKa = 5.0)

Neglect the volume change.

Options:

a) 6.10

b) 5.09

c) 7.00

d) 5.02

Answer:

the final pH of a solution is 5.02

Option d) 5.02 is the Correct Answer

Explanation:

Given the data in the question,

Initially pKa = pH; so ratio is 1:1

thus, 0.4 M acid and base

Now, moles of NaOH = molarity × volume = 0.5 × 10 = 5 mmol = 5 × 10⁻³ mol.

Going into 500 mL ( 0.5 L ) of solution

new molarity will be;

⇒ moles / volume = 5 × 10⁻³ / 0.5 = 0.01 M

ACID reacting with BASE

original concentration of acid = 0.4 - 0.01 = 0.39 M

original concentration of base = 0.4 + 0.01 = 0.41 M

so

pH = 5 + log( base/acid)

= 5 + log ( 0.41/0.39)

= 5 + log ( 1.0512)

= 5 + 0.021

pH = 5.02

Therefore the final pH of a solution is 5.02

Option d) 5.02 is the Correct Answer

5 0

3 years ago