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3 years ago

Calculate the heat required to convert 52.5 g of c2cl3f3 from a liquid at 11.05 ∘c to a gas at 76.65 ∘c.

1 answer:

8 0

<span>Answer: We heated up the liquid CFC from 11.05 degrees celsius to its boiling point (47.6 degrees) That means we heated it up 36.55 degrees The formula to solve this is C = Q / M x dT where C is the specific heat of the liquid CFC (.91 J/G/C) Q is the amount of heat required (what we're trying to find out) M is the mass (52.5 grams) and dT is the temperature difference (36.55 degrees) Now put in the values and find out what Q is .91 = Q / 52.5 x 36.55 .91 = Q / 1918.875 Q = 1746.17 joules to heat up the liquid CFC</span>

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6.) 50.0 mol H2O<br> ? molecules

8_murik_8 [283]

<h3>Answer:</h3>

3.01 × 10²⁵ molecules H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.0 mol H₂O

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u /> $\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} )$ = 3.011 × 10²⁵ molecules H₂O

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

5 0

2 years ago

Which radiation can only be stopped by thick layers of dense metal? (2 points)

AfilCa [17]

Gamma rays have the highest penetrating power so it can only be stopped by thick layers of dense metal. :)

5 0

3 years ago

Read 2 more answers

If you are calculating the grams to mL ratio, you are trying to find the object’s…

Alex

Answer:

Density

Explanation:

The ratio of mass to the volume of an object is called its density. Unit of mass is grams and that of volume is mL.

Density = mass/volume

$\text{Density}=\dfrac{\text{grams}}{\text{mL}}\\\\=\text{grams/mL}$

If you are calculating the grams to mL ratio, it means that we are trying to find the object's density.

4 0

2 years ago

a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill

baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRT$

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0

2 years ago

An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm

frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

xA= 0.34, yA= 0.55

ii) 0.50 = 0.55 nv + 0.34 nL

Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0

2 years ago