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3 years ago

What allows nutrients to enter Plants and animals are able to be transported

Chemistry

1 answer:

elena55 [62]3 years ago

3 0

I think it’s the cell membrane if you’re talking about animal cells and plant cells.

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A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc

Paul [167]

Answer : The volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of $O_2$ gas = (740-22.4) torr = 717.6 torr

$P_2$ = final pressure of $O_2$ gas at STP= 760 torr

$V_1$ = initial volume of $O_2$ gas = 280 mL

$V_2$ = final volume of $O_2$ gas at STP = ?

$T_1$ = initial temperature of $O_2$ gas = $25^oC=273+25=298K$

$T_2$ = final temperature of $O_2$ gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}$

$V_2=242.2mL=0.2422L$

Therefore, the volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

5 0

3 years ago

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

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

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0

3 years ago

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.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,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

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

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

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

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago

Which of the following is the simplest ketone used as an organic solvent?

Lyrx [107]

Answer: Option (d) is the correct answer.

Explanation:

A ketone is an organic functional group that contains a carbon and oxygen atom bonded together through a double bond, that is, C=O.

For example, acetone $CH_{3}-CO-CH_{3}$ is a ketone.

Whereas a hydrocodone is a drug which is used to relieve from moderate to sever pain.It is mostly combined with other drugs and resulting in a chemical formula $C_{18}H_{21}NO_{3}.C_{4}H_{6}O_{6}$ .