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zalisa [80]
3 years ago
14

As the number of electrons added to the same principal energy level increases, atomic size generally

Chemistry
2 answers:
sweet-ann [11.9K]3 years ago
8 0

decreases i took the test

iVinArrow [24]3 years ago
4 0

As the number of electrons added to the same principal energy level increases, atomic size generally

C. Decreases.

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At 25◦C a 4 L sample of H2 exerts a pressure of 5 atm. What pressure would the same sample exert in a 2 L container at 25◦C?
kykrilka [37]

Answer:

10 atm.

Explanation:

Using the combined gas law equation as follows;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 5 atm

P2 = ?

V1 = 4L

V2 = 2L

T1 = 25°C = 25 + 273 = 298K

T2 = 25°C = 298K

Using P1V1/T1 = P2V2/T2

5 × 4/298 = P2 × 2/298

20/298 = 2P2/298

Cross multiply

298 × 20 = 298 × 2P2

5960 = 596P2

P2 = 5960 ÷ 596

P2 = 10 atm.

4 0
3 years ago
How many chlorine atoms are in each set?<br> six calcium chloride formula units.
Aleonysh [2.5K]

Answer:

tae ko malutong

Explanation:

dahil utong ko maitim

3 0
3 years ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

3 0
3 years ago
Read 2 more answers
How would you expect a positive particle approaching another positive particle to behave?
JulijaS [17]

Answer:

They would produce a repulsive force to another

Explanation:

A positive particle approaching another positive particle will repulse it.

According to coulomb's law "like charges repel one another and unlike charges attract".

A charge is an intrinsic property of any matter.

When like charges e.g positive and positive or negative and negative charges are in the vicinity of one another, they repel each other.

When unlike charges; positive and negative are brought together, they simply attract one another.

Therefore, we expect that a positive particle approaching another positive particle will repel one another.  

8 0
3 years ago
Convert 850 mm Hg to kPa. *
Feliz [49]

Answer:

113.324 KPA

Explanation:

4 0
3 years ago
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