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3 years ago

If 11.0 g of ccl3f is enclosed in a 1.1 −l container, will any liquid be present?if so, what mass of liquid?

1 answer:

8 0

Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. 

Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol 

Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. 

Then : 0.05x 137.5 = 6.88gm of vapor 

If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d

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Which aqueous solution has the lowest freezing point c6h12o6, c2h5oh, ch3cooh, or nacl?

Eddi Din [679]

Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates only partially.

So, most particles of the solute is going to be in the solution of NaCl,
so the lowest freezing point has the aqueous solution of NaCl.

3 0

3 years ago

The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the

IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

Explanation :

The expression for first order reaction is:

$[C_t]=[C_o]e^{-kt}$

where,

$[C_t]$ = concentration at time 't' (final) = ?

$[C_o]$ = concentration at time '0' (initial) = 0.100 M

k = rate constant = $5.40\times 10^{-3}s^{-1}$

t = time = 17.0 min = 1020 s (1 min = 60 s)

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

$[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}$

$[C_t]=4.05\times 10^{-4}M$

Thus, the concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

4 0

3 years ago

How many grams of MgCl2 (molar mass = 95.20 g/mol) will be formed from 25.6 mL of a 0.100 M HCL solution reacting with excess ma

lorasvet [3.4K]

The balanced equation for the above reaction is as follows;
Mg + 2HCl ---> MgCl₂ + H₂
stoichiometry of HCl to MgCl₂ is 2:1
we have been told that Mg is in excess therefore HCl is the limiting reactant
number of HCl moles reacted - 0.100 mol/L x 0.0256 L = 0.00256 mol
according to molar ratio, number of MgCl₂ moles formed - 0.00256/2
Therefore number of MgCl₂ moles formed - 0.00128 mol
mass of MgCl formed - 0.00128 mol x 95.20 g/mol = 0.122 g

3 0

3 years ago

Read 2 more answers

A tank of gas is found to exert 8.6 atm at 38°C. What would be the required

Vesna [10]

Answer:

36.2 K

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 8.6 atm

Initial temperature of the gas (T₁): 38°C

Final pressure of the gas (P₂): 1.0 atm (standard pressure)

Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

6 0

3 years ago

What is the mRNA for TTGACAAGTTAC

sertanlavr [38]

Answer:

What is the question? I don't understand what you are trying to say.

3 0

2 years ago