Answer to AP chem question

As the number of protons increases but the distance between the protons and electron REMAINS THE SAME...

Aloiza [94]

Answer:

The correct option is;

The electronegativity increases

Explanation:

The electronegativity is the measure of an atom's ability to attract a shared electron pair. The electronegativity of an atom is dependent on the atom's atomic number and the separation distance between the electrons in the valence shell and the positively charged nucleus such that an increase in the atomic number results in an increase in electronegativity and an increase in the distance between the valence electrons and the nucleus, leads to a decrease in electronegativity.

8 0

3 years ago

Help! I am in a hurry! Will get brainliest if correct!

sashaice [31]

Answer:

nonmetel

Explanation:

3 0

3 years ago

Read 2 more answers

If gravity ceased to exist on Earth, what would happen to our life supporting atmosphere ?

melamori03 [73]

C is correct. right now gravity holds the atmosphere so if there is no gravity then no atmospher

4 0

3 years ago

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen,

grin007 [14]

Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}$

$\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}$

$\text{Rate of formation of }CO=+\frac{d[CO]}{dt}$

$\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}$

The rate of reaction expression is:

$\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}$

As we are given that:

$+\frac{d[CO]}{dt}=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}$

$+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s$

$\frac{d[H_2]}{dt}=3\times 0.35M/s$

$\frac{d[H_2]}{dt}=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s

6 0

3 years ago

The rate of disappearance of HCl was measured for the following reaction:

AURORKA [14]

Answer: (a) 0.000083M/s, 0.000069M/s, 0.000052M/s, 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min =0.000049M/s

(c) The average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

Explanation:

Average reaction rate = change in concentration / time taken

(a) after 54mins, t = 54*60s = 3240s

average reaction rate = (1.58 - 1.85)M / (3240 * 0.0)s

= -0.27M/3240

= 0.000083M/s

after 107mins, t = 107*60s = 6420s

average reaction rate = (1.36 - 1.58)M/ (6420 - 3240)s

= -0.22M/3180s

= 0.000069M/s

after 215mins, t = 215*60s = 12900s

average reaction rate = (1.02 - 1.36)M/ (12900 - 6420)s

= -0.34M/6480s

= 0.000052M/s

after 430mins,t = 430*60 = 25800s

average reaction rate = (0.580 - 1.02)M / (25800 - 12900)s

= -0.44M/12900s

= 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min

= (0.580 - 1.85)M/ (25800 - 0.0)s

= -1.27M/25800s

=0.000049M/s

(c) average reaction rate between t = 54.0min and t = 215.0min

= (1.02 - 1.58)M / (12900 - 3240)s

= -0.56M/9660s

= 0.000058M/s

average reaction rate between t=107.0 and t=430.0min

= (0.580 - 1.36)M / (25800 - 6420)s

= 0.78M /19380s

= 0.000040M/s

Therefore the average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

6 0

3 years ago