Complete and balance the following reactions ?<br> uo^2+ +cr2o^2-7 ــــــــــــــــ uo2+ +cr3+

What is the average speed vavg of the molecules in the gas? express your answer numerically to three significant digits?

ira [324]

In order to determine the velocity of gas molecules, it would not be unreasonable to measure the average velocity, vavg, of the gas molecules. However, the molecules of gas are constantly moving in completely random directions and this results in the average velocity being equal to 0 m/s.

8 0

2 years ago

Suppose hydrochloric acid reacts with potassium sulfite yielding water, sulfur dioxide, and potassium chloride. Suppose 4 moles

anzhelika [568]

Answer:

128 grams of sulfur dioxide are produced.

Explanation:

$2HCl+K_2SO_3\rightarrow SO_2+2KCl+H_2O$

Moles of HCl = 4 moles

According to reaction, 2 moles of HCl gives 1 mole of sulfur dioxide gas.

Then 4 moles of HCl will give:

$\frac{1}{2}\times 4 moles=2 moles$ of sulfur dioxide gas.

Mass of sulfur dioxide gas = 2 mol × 64 g/mol = 128 g

128 grams of sulfur dioxide are produced.

4 0

2 years ago

I NEED THIS NOW AND NO LINKS OR ILL REPORT

kolbaska11 [484]

Aluminum or glass I think

3 0

2 years ago

Read 2 more answers

Consider the following reaction where Kc = 154 at 298 K.2NO(g) + Br2(g) 2NOBr(g)A reaction mixture was found to contain 4.64×10-

IgorLugansk [536]

Answer:

The reaction is not at equilibrium and reaction must run in forward direction.

Explanation:

At the given interval, concentration of NO = $\frac{4.64\times 10^{-2}}{1}M=4.64\times 10^{-2}M$

Concentration of $Br_{2}$ = $\frac{4.56\times 10^{-2}}{1}M=4.56\times 10^{-2}M$

Concentration of NOBr = $\frac{0.102}{1}M=0.102M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(0.102)^{2}}{(4.64\times 10^{-2})^{2}\times (4.56\times 10^{-2})}=106$

So, the reaction is not at equilibrium.

As $Q_{c}< K_{c}$ therefore reaction must run in forward direction to increase $Q_{c}$ and make it equal to $K_{c}$ .

4 0

2 years ago

Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1: a

Talja [164]

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

$\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)$ .

$\rm NH_3$ gains one hydrogen ion to produce the ammonium ion $\rm {NH_4}^{+}$ . In other words, $\rm {NH_4}^{+}$ is the conjugate acid of the weak base $\rm NH_3$ .

Both buffer 1 and 2 include

the weak base ammonia $\rm NH_3$ , and
the conjugate acid of the weak base $\rm {NH_4}^{+}$ .

The ammonia $\rm NH_3$ in the solution will react with hydrogen ions as they are added to the solution:

$\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)$ .

There are more $\rm NH_3$ in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of $\rm NH_3$ in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.

5 0

3 years ago

Complete and balance the following reactions ? uo^2+ +cr2o^2-7 ــــــــــــــــ uo2+ +cr3+