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2 years ago

9

I'm trying to do question number 2.56

1 answer:

Lunna [17]2 years ago

8 0

The answer is in the picture which is given below:

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Balance this equation Hgo —> Hg + O2

Nesterboy [21]

Answer:

2Hgo —>2Hg + O2

Explanation:

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2 years ago

How would granite be classified as

vredina [299]

Answer:

Granite is an intrusive igneous rock which means it is cooled slowly deep upper the Earth's crust. It is composed of 25% to 35% quartz and over 50% potassium- and sodium rich feldspars.

Explanation:

4 0

2 years ago

Read 2 more answers

A balloon is filled to a volume of 1.50 L with 3.00 mol of gas at 25 C. with pressure and temperature held constant, what will b

bixtya [17]

Answer:

3.2L

Explanation:

PV=nRT

since pressure and temperature are held constant we have V=nR

R is a constant also,

Thus; $\frac{v1}{n1}=\frac{v2}{n2}$

v1=1.5L , n1=3mol, n2=1.4mol

$\frac{1.5}{3}=\frac{v2}{1.4}$

v2= $\frac{1.5}{3}*1.4$

v2=3.2L

7 0

2 years ago

If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai

Rama09 [41]

We are given –

Final velocity of car is, v= 0
Initial velocity of car is, u= 100 km/hr
Time taken, t is = 3 minutes or 180 sec

Here–

$\qquad$ $\pink{\bf \longrightarrow Initial\: velocity = 100 \:km/hr}$

$\qquad$ $\sf \longrightarrow Initial\: velocity = \dfrac{ 100 \times 1000}{3600} \:m/s$

$\qquad$ $\pink{ \bf \longrightarrow Initial\: velocity = 27.78\: m/s}$

Now –

$\qquad$ ____________________________

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{Final\: Velocity -Initial \:Velocity }{Time}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{v -u}{t}}$

$\qquad$ $\sf \longrightarrow Acceleration = \dfrac{(0- 27.78)}{1800}$

$\qquad$ $\sf \longrightarrow Acceleration =\cancel{ \dfrac{- 27.78}{1800}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = -0.15 \: m/s^2}$

$\qquad$ _______________________________

6 0

2 years ago

Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio

umka21 [38]

Answer : The value of $K_p$ at this temperature is 66.7

Explanation : Given,

Pressure of $PCl_3$ at equilibrium = 0.348 atm

Pressure of $Cl_2$ at equilibrium = 0.441 atm

Pressure of $PCl_5$ at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}$

Now put all the values in this expression, we get :

$K_p=\frac{(10.24)}{(0.348)(0.441)}$

$K_p=66.7$

Therefore, the value of $K_p$ at this temperature is 66.7

4 0

2 years ago