How would the energy needs of a cell be compared to the energy needs of a city ?

1 answer:

3 0

The energy needs of a cell is very important because they need energy to breakdown,synthesis, and transport nutrients and molecules so an living organism can survive just like how the city needs energy to power its cities so it’s people can do what they need to do to survive. HOPE IT HELPS

You might be interested in

How many moles of H2 would be required to produce 5.0 moles of water

earnstyle [38]

5 moles of H2
..............

3 0

3 years ago

Define<br>Equivalent weight of a<br>Chemical substance

SSSSS [86.1K]

Equivalent weight, in chemistry, the quantity of a substance that exactly reacts with, or is equal to the combining value of, an arbitrarily fixed quantity of another substance in a particular reaction. ... The equivalent weight of an element is its gram atomic weight divided by its valence (combining power).

Hope it helps!! Have a great day❤️✨

4 0

2 years ago

Please answer asap!

hram777 [196]

1.) A compound is a thing that is composed of two or more (separate) elements.

2.) A compound cannot be separated easily, while its properties differ from where it originates from. It also is formed from chemical reactions.

3.) A pure substance, also known as chemical substance, is a material with constant composition and consistant properties.

4.) If a substance is not a pure substance, then it is an impure substance.

Hope this helped. :)

4 0

3 years ago

NEEEEEEEEEEEEEEEEED HELP

lakkis [162]

<h3> <u>Answer</u></h3>

4. loses 2 electrons

<h3><u>EXPLANATION</u></h3>

Mg²+ has 10 electrons while Mg has 12 electrons.

7 0

2 years ago

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

3 years ago