PLEASE HELP ITS A TESTT!!

Question 1:

Gwar [14]

Answer:

1.a

2.c

hope helps

Explanation:

mark

pa

8 0

3 years ago

What element is always present in an organic compound.

Andreyy89

Carbon is always present in an organic compound

3 0

3 years ago

Calculate the following question

lord [1]

Answer:

12.55 → pH

Explanation:

pH = - log [H⁺]

pH = 14 - pOH

pOH = - log [OH⁻]

pOH = - log 0.035 = 1.45

14 - 1.45 = 12.55 → pH

Remember that [OH⁻] . [H⁺] = 1×10⁻¹⁴ M

6 0

4 years ago

cooling a sample of matter from 70°c to 10°c at constant pressure causes its volume to decrease from 873.6 to 712.6 cm3. classif

jek_recluse [69]

Explanation:

Expression for the coefficient of thermal expansion is as follows.

$\alpha = \frac{1}{V}(\frac{\Delta V}{\Delta T})$

where, V = initial volume

$\Delta V$ = Final volume - initial volume

= (712.6 - 873.6) $cm^{3}$

= -161 $cm^{3}$

Now, we will calculate the change in temperature as follows.

$\Delta T$ = Final temperature - Initial temperature

= (10 + 273) K - (70 + 273) K

= 283 K - 343 K

= -60 K

Substituting these values into the equation as follows.

$\alpha = \frac{1}{873.6} \times (\frac{161}{60}) K^{-1}$

= 0.00307 $K^{-1}$

It is known that for non-ideal gases the value of alpha is 0.366% which is 0.00366 per Kelvin. As it is close to our result, hence the given sample of gas is a non-ideal gas.

3 0

3 years ago

How many grams of chlorine gas can be produced if 15 grams of FeCl3 reacts with 4 moles of O2? What is the limiting reactant? Wh

Mariana [72]

Answer:

The limiting reactant is FeCl3

The excess reactant is O2

The theoretical yield Cl2 is 9.84 grams

The % yield = 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles of O2 = 4.0 moles

Mass of Cl2 = 9.5 grams = actual yield

Step 2: The balanced equation

4 FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate the limiting reactant

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

FeCl3 has the smallest amount of moles, this is the limiting reactant. It will be completely consumed ( 0.0925 moles).

O2 is in excess. There will react 3/4 * 0.0925 = 0.0694 moles

There will remain 4.0 - 0.0694 = 3.3904 moles O2

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 we'll have 6/4 * 0.0925 = 0.13875 moles Cl2

Step 6: Calculate mass of Cl2

Mass Cl2 = moles Cl2 * molar mass Cl2

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams = theoretical yield

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) *100%

% yield = (9.5 grams / 9.84 grams) * 100%

% yield = 96.5 %

3 0

4 years ago