All categories

3 years ago

(Please what is the answer to this )How do atoms relate to cells ?

Chemistry

2 answers:

ziro4ka [17]3 years ago

8 0

A cells made of molecules and a molecule is made Adams.that is simplest Way of putting it. made up of macromolecules, such as proteins, lipids, etc. A molecule is a particular configuration of atoms.

Elden [556K]3 years ago

5 0

Molecules are made of atoms. Cells are made by a multitude of molecules. ... So cells are made of molecules and, consequently of atoms .

You might be interested in

What is the molarity of a solution in which 7.1 g of sodium sulfate is dissolved in enough water to make 100. mL of solution?

Sidana [21]

Answer:

0.50 M

Explanation:

Given data

Mass of sodium sulfate (solute): 7.1 g

Volume of solution: 100 mL

Step 1: Calculate the moles of the solute

The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

$7.1g \times \frac{1mol}{142.02g} = 0.050mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

$100mL \times \frac{1L}{1000mL} =0.100L$

Step 3: Calculate the molarity of the solution

$M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M$

3 0

3 years ago

Read 2 more answers

What is the density of 0.50 grams of gaseous carbon stored under 1.5 atm of pressure at a temperature of -20.0 C?

Colt1911 [192]

Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.

Explanation:

d = m/V, where d is the density, m is the mass and V is the volume.

We have the mass m = 0.50 g, so we must get the volume V.

To get the volume of a gas, we apply the general gas law PV = nRT

P is the pressure in atm (P = 1.5 atm)

V is the volume in L (V = ??? L)

n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).

Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.

Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.

6 0

3 years ago

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Romashka-Z-Leto [24]

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we want to find $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?$ when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

$\displaystyle V =\frac{1}{3} \pi (5h)^2 h$
$\displaystyle V =\frac{1}{3} \pi \ 25h^3$

Differentiate this equation with respect to time t.

$\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}$
$\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}$

Plug known values into the equation and solve for dh/dt.

$\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}$
$\displaystyle 10 = 100 \pi \ \frac{dh}{dt}$

Divide both sides by 100π to solve for dh/dt.

$\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0

3 years ago

A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liber

In-s [12.5K]

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass )/(molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34= 0.00532 (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3 = (mass of BaCO3 )/(mass of sample )*100

= 0.242/(0.846 )*100

= 28.605%

5 0

3 years ago

21,980 mg of phosphorus reacts in a gas jar containing oxygen. The resulting oxide weighed 50.000 g. The percent composition of

stealth61 [152]

Answer: False

Explanation:

8 0

3 years ago