When BOTH organisms benefit from a relationship it is called?

A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il

Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

P= 2 atm
V= ?
n=0.223 moles
R= 0.0821 $\frac{L*atm}{mol*K}$
T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 $\frac{L*atm}{mol*K}$ * 306 K

Solving:

$V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\$

V= 2.80 L

<u><em>The volume of the gas is 2.80 L.</em></u>

7 0

3 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago

How can we tell the distance sediments traveled due to erosion ?

meriva

Answer:

more rounded the grains are the more they have been moved around

Explanation:

Generally – the more rounded the grains are the more they have been moved around (i.e. the longer the length of time or distance they have moved). Angular grains cannot have travelled far

geolsoc.org.uk

4 0

2 years ago

The energy needed to remove an electron from an atom is called

Arlecino [84]

Ionization energy, also called ionization potential, in chemistry, the amount of energy required to remove an electron from an isolated atom or molecule

3 0

3 years ago

A mixture of MgCl2 and NaCl has an initial mass of 0.4015 g. Excess AgNO3(aq) was added and 1.032 g of precipitate, AgCl(s), was

zlopas [31]

Answer:

8.909*10^-4 moles

Explanation:

The mixture contains MgCl $_{2}$ and NaCl with a total mass of 0.4015 g. The mass of precipitate AgCl(aq) is 1.032 g and it has a molar mass of 143.32 g/mol. Therefore, the moles of the precipitate is:

n = 1.032/143.32 = 7.2007*10^-3 moles

Molar mass of NaCl = 58.44 g/mol and the molar mass of MgCl $_{2}$ is 95.21 g/mol. Let the mass (g) of MgCl $_{2}$ in the original mixture be 'x'. Thus:

7.2007*10^-3 = (0.4015-x)/58.44 + 2x/95.21

(7.2007*10^-3)*58.44*95.21 = 95.21(0.4015-x) + 2x(58.44)

40.06505 = 38.227 -95.21x + 116.88x

40.06505 - 38.227 = -95.21x + 116.88x

1.83805 = 21.67x

x = 1.83805/21.67 = 0.0848 g

moles of MgCl $_{2}$ = 0.0848g/95.21 g/mol = 8.909*10^-4 moles

4 0

3 years ago