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2 years ago

5

Organisms have different ways of obtaining water. Bacteria obtain his water by allowing individual water molecules to enter thro

ugh the

1 answer:

Troyanec [42]2 years ago

6 0

Cell membrane is the answer.

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9. During an experiment the students prepared three mixtures A)Starch in water B) Sodium chloride solution C) Tincture of Iodine

il63 [147K]

Answer:

See explanation

Explanation:

Tyndall effect refers to the scattering of light in a solution. Tyndall effect occurs when the size of particles in the solution exceeds 1 nm in diameter. Such solutions are actually called false solutions.

In tincture of iodine, the size of particles in solution is less than 1 nm in diameter hence the solution does not exhibit Tyndall effect. Hence, tincture of iodine is a true solution.

Therefore, if the size of particles in solution exceeded 1nm in diameter, Tyndall effect is observed.

7 0

2 years ago

Which of the following are NOT enclosed by a cell wall.

Varvara68 [4.7K]

D. Cells from a dogs paw

6 0

2 years ago

The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil

KIM [24]

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5730 years

Putting values in above equation, we get:

$k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 25) = 75 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs$

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0

3 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

Dovator [93]

Answer: 406 hours

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 39.5 A

t= time in seconds = ?

The deposition of copper at cathode is represented by:

$Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000$ Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = $\frac{193000}{63.5}\times 19000=57748032$ Coloumb

$57748032=39.5\times t$

$t=1461975sec=406hours$ (1hour=3600s)

Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A

3 0

3 years ago

A certain object has a volume of 25.0 ml and a mass of 100 g. what is the density of the object

Tpy6a [65]

Density= mass/volume

= 100/25
density = 4g/ml

8 0

3 years ago