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3 years ago

The volume of a sample of nitrogen is 6.00 liters at 35oC and 0.75 atm. What volume will it occupy at STP?

1 answer:

3 0

The volume that will occupy at STP is calculated as follows
by use of ideal gas equation
that is PV=nRT where n is number of moles calculate number of moles

n= PV/RT
p=0.75 atm
V=6.0 L
R = 0.0821 L.atm/k.mol
T= 35 +273= 308k
n=?

n= (o.75 atm x 6.0 L)/( 0.0821 L.atm/k.mol x 308 k)= 0.178 moles

Agt STP 1 mole= 22.4 L what obout 0.178 moles

= 22.4 x0.178moles/ 1moles =3.98 L( answer C)

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What is the relationship between Force and Velocity?

PIT_PIT [208]

Answer:

Cause/ effect relationship.

Explanation:

because the force is the cause of energy that is present and velocity is the effect (speed) that is present.

7 0

2 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

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

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by ma

Montano1993 [528]

Answer:

The answer to your question is C₂HO₃

Explanation:

Data

Hydrogen = 3.25%

Carbon = 19.36%

Oxygen = 77.39%

Process

1.- Write the percent as grams

Hydrogen = 3.25 g

Carbon = 19.36 g

Oxygen = 77.39 g

2.- Convert the grams to moles

1 g of H ----------------- 1 mol

3,25 g of H ------------- x

x = (3.25 x 1) / 1

x = 3.25 moles

12 g of C ---------------- 1 mol

19.36 g of C ---------- x

x = (19.36 x 1) / 12

x = 1.61 moles

16g of O --------------- 1 mol

77.39 g of O --------- x

x = (77.39 x 1)/16

x = 4.83

3.- Divide by the lowest number of moles

Carbon = 3.25/1.61 = 2

Hydrogen = 1.61/1.61 = 1

Oxygen = 4.83/1.61 = 3

4.- Write the empirical formula

C₂HO₃

4 0

3 years ago

A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life

mixas84 [53]

T is amount after time t
Ao is initial amount 
t is time 
HL is half life 

log (At) = log [ Ao x (1/2)^(t/HL) ] 
log (At) = log Ao + log (1/2)^(t/HL) 
log (At) = log Ao + (t/HL) x log (1/2) 

( log At - log Ao) / log (1/2) = t / HL 
log (At/Ao) / log (1/2) = t / HL 

HL = t / [( log (At / Ao)) / log (1/2) ] 

HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] 

HL = 14.4 s / 2 = 7.2 seconds

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3 years ago

When NH NO3(aq) reacts to form N2O(g) and H2O(1), 150 kJ of energy are evolved for each mole of NH NO3(aq) that reacts.

amid [387]

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3 years ago