Determine how many liters 8.65 g of carbon dioxide gas would occupy at the following conditions.

Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

2 years ago

If the cylindrical pistons are 25.000 cm in diameter at 20.0 ∘c, what should be the minimum diameter of the cylinders at that te

attashe74 [19]

Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
$pV=nRT$
where
p = pressure
V = volume
n = moles
T = temperature
R = gas constant

State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
d₂ = piston diameter

Because V, n, and R remain the same between the two temperatures, therefore
$\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}$

If the supported load is W kg, then
$p_{1} = \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} = 20.3718W \, Pa$
Similarly,
$p_{2} = \frac{4W}{\pi d_{2}^{2}} \, Pa$

$\frac{p_{1}}{p_{2}} = \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}$

Because p₁/p₂ = T₁/T₂, therefore
$16d_{2}^{2} = \frac{293}{423} \\\\ d_{2}^{2} = \frac{0.6927}{16} \\\\ d_{2} = 0.2081 \, m$

The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter

8 0

3 years ago

g If a small amount of a strong acid is added to a buffer made up of a weak acid, HA, and the sodium salt of its conjugate base,

ikadub [295]

Answer:

Explanation is in the answer

Explanation:

The pH of the buffer solution does not change appreciably because the strong acid (free H⁺) reacts with conjugate base of buffer producing more weak acid. pH formula of buffers is (Henderson-Hasselbalch formula):

pH = pKa + log ( [A⁻] / [HA] )

The addition of strong acid decreases [A⁻] increasing [HA]. pH change just in the log of the ratio of [A⁻] with [HA], that is a real little effect over pH of the buffer solution.

5 0

2 years ago

How many liters of carbon dioxide will be produced when 128 L of ethane are burned?

Alja [10]

Answer:

i know the problem

2c2h6(g)+702(g)->4co2(G)+6g20(1) thats double replacement i dont know the first part

4 0

2 years ago

Draw a resonance structure, complete with all formal charges and lone (unshared) electron pairs, that shows the resonance intera

icang [17]

Answer:see the picture attached

Explanation:

3 0

2 years ago