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3 years ago

A gas has v= 4.0L, T =27 degree C and P=2.0 atm. What is the V if T=327 degree C and P=8.0 atm?

1 answer:

3 0

As I am reading the problem,i can tell the question gives you two temperatures, two pressures, one volume and asking for the other. this should be an indication that you need to use the following gas formula

P1V1/T1= P2V2/T2

P1= 2.0 atm
V2= 4.0 L
T1= 27= 300 K
P2= 8.0 atm
V2= ?
T2= 327= 600 k

let's plug in the values into the formula

(2.0 x 4.0)/ 300= (8.0 x V2)/ 600 K

V2= 2.0 Liters

answer is D

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Minerals are very useful to us. Which of the following is a false statement? Question options: Pyrite is an ore of gold and is u

lilavasa [31]

Answer:

The false statement is :

Pyrite is an ore of gold and is used in jewelry making.
Cobalt is not used in medicine.

Explanation:

Pyrite is an ore of iron sulfide found on the Earth and in coal, limestone and in many deposits of metallic ores. Due to its luster and pale yellowish color it appears as gold due to which it is also termed as Fool's gold.

It is used in firearms , in production of sulfur dioxide gas, etc.

Cobalt is used as medicine to treat cancer patient. Co-60 is used in treatment of cancer in which gamma rays produced by Co-60 are used to kill tumor cells in a cancer patient.

Where as talc is used in industries like : paper making, food, paints, plastics etc. Quartz is primary material used to prepare glass and cobalt is used as

4 0

3 years ago

Drag each tile to the correct box.

77julia77 [94]

this is my attachment answer hope it's helpful to you

3 0

3 years ago

Question 4 point

Ronch [10]

Answer:

$5393.4gH_2O$

Explanation:

Hello there!

In this case, for the described chemical reaction, we can write:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Thus, by considering the 1:2 mole ratio of methane to water, and the molar mass of the latter (18.02 g/mol), the following is useful to calculate the mass of water that is produced:

$149.65molCH_4*\frac{2molH_2O}{1molCH_4} *\frac{18.02gH_2O}{1molH_2O}$

Which is equal to:

$5393.4gH_2O$

Which is not among the choices.

Regards!

5 0

3 years ago

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

Studentka2010 [4]

Answer : The initial rate for a reaction will be $3.63\times 10^{-4}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$1.2\times 10^{-4}=k(0.40)^a(0.40)^b(0.40)^c$ ....(1)

Expression for rate law for second observation:

$3.6\times 10^{-4}=k(0.40)^a(0.40)^b(1.20)^c$ ....(2)

Expression for rate law for third observation:

$4.8\times 10^{-4}=k(0.80)^a(0.40)^b(0.40)^c$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-4}=k(0.80)^a(0.80)^b(0.40)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{3.6\times 10^{-4}}{1.2\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(1.20)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{4.8\times 10^{-4}}{1.2\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{4.8\times 10^{-4}}{4.8\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.80)^a(0.80)^b(0.40)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$1.2\times 10^{-4}=k(0.40)^2(0.40)^0(0.40)^1$

$k=1.875\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.55 M of reagent A and 0.80 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(1.875\times 10^{-3})\times (0.55)^2(0.80)^0(0.80)^1$

$\text{Rate}=3.63\times 10^{-4}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.63\times 10^{-4}Ms^{-1}$

4 0

3 years ago

Consider the following generic reaction for which Kp = 5.51 × 105 at 25°C:

lora16 [44]

Answer:

$K_c=1.35\times 10^7$

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2R_{(g)}+A_{(g)}\rightleftharpoons2Z_{(g)}$

Given: Kp = $5.51\times 10^5$

Temperature = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

T = (25 + 273.15) K = 298.15 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1

Thus, Kc is:

$5.51\times 10^5= K_c\times (0.082057\times 299)^{-1}$

$K_c\frac{1}{24.535043}=551000$

$K_c=13518808.69=1.35\times 10^7$

3 0

3 years ago