Help me please I really need help on this?

The reaction nacl(s) → nacl(aq) is performed in a coffee cup calorimeter, using 100 ml of h2o(l) and 5.00g of nacl. if the tempe

MrRissso [65]

Since the density of water is 1 g /mL, hence there is 100 g of H2O. So total mass is:

m = 100 g + 5 g = 105 g

=> The heat of reaction can be calculated using the formula:

δhrxn = m C ΔT

where m is mass, C is heap capacity and ΔT is change in temperature = negative since there is a decrease

δhrxn = 105 g * 4.18 J/g°C * (-2.30°C)

δhrxn = -1,009.47 J

=> However this is still in units of J, so calculate the number of moles of NaCl.

moles NaCl = 5 g / (58.44 g / mol)

moles NaCl = 0.0856 mol

=> So the heat of reaction per mole is:

δhrxn = -1,009.47 J / 0.0856 mol

δhrxn = -11,798.69 J/mol = -11.8 kJ/mol

5 0

3 years ago

After being thoroughly stirred at 10.°C, which mixture is heterogenous?(1) 25.0 g of KCl and 100. g of H2O

zheka24 [161]

The answer is (2) KNO3. This depends on the solubility of these four compounds at 10℃. For NaCl, it is 35.8 g, For NaNO3, 80.8 g. KCl, 31.2 g. KNO3, 21.9g. So only KNO3 is less than 25.0 g.

5 0

3 years ago

Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th

aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = ?

Depression in freezing point = $\Delta T_f$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample = 0.275 mol/kg

i = van't Hoff factor

We have: $K_f$ = 40°C kg/mol

i = 1 ( non electrolyte)

$\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg$

$\Delta T_f=11^oC$

$\Delta T_f=T- T_f$

$T_f=T- \Delta T_f=179^oC-11^oC=168^oC$

168°C is the melting point of your impure sample.

4 0

3 years ago

How many kilograms of solvent would contain 0.43 mol of CaO in a 2.5 molality solution

dalvyx [7]

<h3>Answer:</h3>

0.024 kg CaO

<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>Aqueous Solutions</u>

Molarity = moles of solute / liters of solution

<u>Atomic Structure</u>

Reading a Periodic Tables
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.41 mol CaO

2.5 M Solution

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

1000 g = 1 kg

Molar Mass of Ca - 40.08 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.43 \ mol \ CaO(\frac{56.08 \ g \ Cao}{1 \ mol \ CaO})(\frac{1 \ kg \ CaO}{1000 \ g \ CaO})$
Multiply: $\displaystyle 0.024114 \ kg \ CaO$