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3 years ago

Which characteristic should describe all evidence collected by a scientist?

Chemistry

1 answer:

scZoUnD [109]3 years ago

8 0

Answer:

Option C

Objective

Explanation:

When dealing with the study of natural phenomena, a purely objective approach must be made by scientists. This is because natural phenomena exist, devoid of human feelings or emotions. There is no room for subjectivity as the natural occurrences do not vary based on human thoughts and perceptions; rather they exist on their own.

Therefore, every evidence collected by scientists should be objective and analysis should be made based on characteristics displayed and not from their previous experiences or personal point of view.

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a gas cylinder contains 2.0 mol of gas X and 6.0 mol of gas Y at a total pressute of 2.1 atm. what is he partial pressure of gas

Igoryamba

The answer is 1.6 atm. Let's first calculate the mole fraction of gas Y.

The mole fraction (x) is: x = n1/n, where n1 is a number of moles of an

individual gas in a mixture and n is total moles of the gas mixture. We

know that n1 of gas Y is 6.0 mol (n1 = 6.0 mol) and that there are in

total 8. mol of the gas mixture (n = 2.0 + 6.0 = 8.0 mol). Now calculate

the mole fraction of gas Y. x = 6.0/8.0 = 0.75. Now, let's use the mole

fraction of gas Y (x) and the total pressure (P) to calculate the

partial pressure of gas Y (P1): x = P1/P. P1 = x * P. If x = 0.75 and P =

2.1 atm, then the partial pressure of gas Y is: P1 = 0.75 * 2.1 atm =

1.6 atm.

6 0

4 years ago

Read 2 more answers

C5H6 reacts with itself to form C10H12 according to the following process

swat32

When carbon compounds react with themselves to form a larger molecule the process is called polymerization. Specifically addition polymerization. sometimes molecules containing containing carbon to carbon double bonds can join together to form longer chains. The double bond is broken and the electrons in it join to neighboring molecules.

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3 years ago

Carbonic acid is a mixture of

kirill115 [55]

Answer:

a compound of the elements hydrogen, carbon, and oxygen. (also carbonic acid is a compound not a mixture ^-^)

Explanation:

hope this helps dude

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3 years ago

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When 3.4 moles of hydrogen gas react with excess oxygen how many moles of water will be produced

s344n2d4d5 [400]

1. start with balanced equation.
2 H2(g) + O2(g) 2 H2O(g)

2. Use stoichiometry
(3.4moles of H)(2moles of H2O/2moles of H)
The moles of H will cancel, leaving you with moles of H2O.
The answer is 3.4 moles of H2O

6 0

4 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of $B$ at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of $C$ at equilibrium = x = 0.168 M

3 0

4 years ago