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ryzh [129]
3 years ago
14

Select the correct answer Giving brainliest

Chemistry
1 answer:
ANEK [815]3 years ago
6 0

Answer:

it is the bowling ball is the correct one

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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
Determine the change in volume that takes place when a 2.00 L sample of N2 gas is heated from 250 C to 500 C.
Effectus [21]

Answer:

V₂ = 2.96 L

Explanation:

Given data:

Initial volume = 2.00 L

Initial temperature =  250°C

Final volume = ?

Final temperature = 500°C

Solution:

First of all we will convert the temperature into kelvin.

250+273 = 523 k

500+273= 773 k

According to Charles's law,

V∝ T

V = KT

V₁/T₁ = V₂/T₂

V₂ = T₂V₁/T₁

V₂ = 2 L × 773 K / 523 k

V₂ =  1546 L.K / 523 k

V₂ = 2.96 L

4 0
3 years ago
Can Someone help me please…
erma4kov [3.2K]

Answer:

1.Ecologists often estimate the size and density of. populations using quadrats and the mark-recapture method.

2. The two key components of ecosystem stability are resilience and resistance.

3.The availability of abiotic factors (such as water, oxygen, and space) and biotic factors (such as food) affect population

4. When the climate is changing in the colder direction it effects the planet in the opposite way than it does when it is changing in the warmer direction. Ad nauseam the contemporary, it’ll keep getting hotter and at a greater rate so the ecosystem is going to die.

7 0
2 years ago
One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w
expeople1 [14]

Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

similarly,

weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

hence, molecular weight of the gas = 32.256g

vapor density = molecular weight/2

= 32.256/2 = 16.128g

hence, vapor density of the gas is 16.128g.

Explanation:

4 0
2 years ago
How many moles of Carbon are in 3.06 g of Carbon
natta225 [31]

Answer:

\boxed {\boxed {\sf 0.255 \ mol \ C }}

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

  • Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Since we are converting 3.06 grams to moles, we multiply by that value.

3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}                      

3.06 * \frac{1 \ mol \ C}{12.011 }    

\frac {3.06}{12.011 } \ mol \ C                                

0.25476646 \ mol \ C

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

  • 0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

0.255 \ mol \ C

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>

3 0
2 years ago
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