Ask question

Ask question

All categories

2 years ago

9

What are the strengths and limitations of a cell?

1 answer:

Dmitry [639]2 years ago

4 0

Cells have limitations, if the cell is stronger it targets other cells , when a cell is weak it possibly mean it’s disease.

You might be interested in

Number 7 plz tysm guys

morpeh [17]

It is letter C: Continental crust is thicker than oceanic crust.

6 0

2 years ago

Read 2 more answers

0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an

Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic , its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30 x 10⁻³ moles .

8 0

3 years ago

100 points! Please help! Will give brainliest!!

patriot [66]

1- change of state (boiling)
2-increase in acidity
3-methylxanthines (caffeine) theobromine and theophylline
4- lemon juice

3 0

2 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

2 years ago

When the glucose transport maximum is reached, _______.?

n200080 [17]

Glucose is extracted in the urine and all of the glucose is reabsorbed.

8 0

2 years ago