3. If an item of trash is described as highly biodegradable, is that a physical property or

The following data is given to find the formula of a Hydrate:

umka21 [38]

The masses can be found by substractions:

Mass of CaSO₄.H2O (hydrate):

16.05 g - 13.56 g = 2.49 g

Mass of CaSO₄ anhydrate:

15.07 g - 13.56 g = 1.51 g

The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:

2.49 g - 1.51 g = 0.98 g

The percent of water is found by the formula:

massWater ÷ massHydrate * 100%

0.98 g ÷ 2.49 g * 100% = 39.36%

The mole of water is calculated using water's molecular weight (18g/mol):

0.98 g ÷ 18 g/mol = 0.054 mol water

A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)

1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄

The ratio of mole of water to mole of anhydrate is:

0.054 mol water / 0.011 mol CaSO₄ = 0.49

In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O

3 0

3 years ago

A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal

bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 25.0 L

V₂ = ?

P₁ = 2575 mm Hg

Also, P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}$

$\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}$

Solving for V₂ , we get:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

4 0

4 years ago

Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration

elixir [45]

Answer:

8.54

Explanation:

At equivalence point :

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :

pH = 1/2 [ pKw + pKa + log c ]

Ka for acetic acid = 1.75 X 10⁻⁵

so pKa = -log (1.75 X 10⁻⁵) = 4.74

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169

so log c = log 0.02169 = -1.66

putting the values....

pH = 1/2 [14 + 4.74 - 1.66 ]

pH = 1/2 [ 17.08] = 8.54

6 0

3 years ago

Read 2 more answers

a technician mixes 80 ml of a 5% solution with 10 ml of water. what is the final percentage strength of the solution prepared?

Vinvika [58]

A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.

given that :

8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL

therefore, 80 × 5 = 10 × x %

x % = 40 %

Therefore, the final percentage strength of the solution is 40 %

Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.

To learn more about percentage strength here

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5 0

1 year ago

The word science is derived from latin word..........which means............

stepan [7]

Scientia, meaning knowledge and skill

8 0

3 years ago