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Taya2010 [7]
3 years ago
14

How many moles are contained in the stp 22.4l of n2?

Chemistry
1 answer:
statuscvo [17]3 years ago
8 0
One mole of any gas under STP has volume 22.4 L,
so 22.4 L N2 under STP contains 1 mole of molecules N2.
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According to Pascal's principle, how does pressure from a hand squeezing a ketchup bottle move throughout the ketchup?
Pepsi [2]

- According to Pascal’s principle, for a certain fluid in a totally enclosed system, a change in pressure at a given point in the fluid is transferred to all points in the fluid, as well as to the enclosing walls.

- This is illustrated by the fact that the pressure inside an enclosed system is the same according to the relation [ pressure = force/area ]. Therefore, the change in pressure resulting from squeezing a ketchup bottle will be transferred equally to all parts of that bottle as well as its internal content.

- That’s how hydraulic machines, such as garbage trucks and hydraulic lifts function..


3 0
3 years ago
Read 2 more answers
If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th
qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

K_c=4.74\times 10^4

Concentration at equilibrium:

[COBr_2]=1.58\times 10^{-6}M

[Co]=2.78\times 10^{-3}M

[Br_2]=2.51\times 10^{-5}M

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0
3 years ago
Predict the products and then balance the equation
Aneli [31]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the balanced equation for given Double displacement reaction ~

\sf Pb(NO_3)_2 +2 \:  KI = PbI_2  +2 \:  KNO_3

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )

4 0
2 years ago
Read 2 more answers
The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con
bonufazy [111]

<u>Answer:</u> The concentration of radon after the given time is 3.83\times 10^{-30}mol/L

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

We are given:

t_{1/2}=3.82days

Putting values in above equation, we get:

k=\frac{0.693}{3.82}=0.181days^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant = 0.181days^{-1}

t = time taken for decay process = 3.00 days

[A_o] = initial amount of the reactant = 1.45\times 10^{-6}mol/L

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}

[A]=3.83\times 10^{-30}mol/L

Hence, the concentration of radon after the given time is 3.83\times 10^{-30}mol/L

7 0
3 years ago
Why do we crush solids before putting them into a solvent
ahrayia [7]
So that they will mix in better, or more easily. However you want to say it.
7 0
2 years ago
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