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Vinvika [58]
3 years ago
12

How many grams of Carbon are in sample of 400 grams of Co2?​

Chemistry
1 answer:
RoseWind [281]3 years ago
3 0

Answer:

Mass of carbon = 109.1 g

Explanation:

Given data:

Mass of carbondioxide = 400 g

Mass of carbon = ?

Solution:

Molar mass of carbon = 12 g/mol

Molar mass of CO₂ = 44 g/mol

Mass of carbon in 400g of  CO₂:

Mass of carbon = 12 g/mol/44 g/mol × 400 g

Mass of carbon = 109.1 g

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100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
2 years ago
Which of the following measure would remain consistent no matter it's location?
elena55 [62]
Mass using grams because of the balance scale is evenly weighted not from springs and gravity like a normal scale.
4 0
3 years ago
Read 2 more answers
Is sulfur dioxide a compound, mixture or element? Thanks!
horsena [70]
It is a compound because a compound is two or more different elements chemically combined.
5 0
2 years ago
How many PO43? ions are in a mole of K3PO4?
adoni [48]

K₃PO₄ → 3K⁺ (aq) + PO₄³⁻(aq)

One mole of PO₄³⁻ ion gets dissociated from one mole of K₃PO₄

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ ions

One mole of PO₄³⁻ ions x  (6.022 x 10²³ ions/ 1 mole of PO₄³⁻ ions )

= 6.022 x 10²³ ions

Therefore , there are 6.022 x 10²³ PO₄³⁻ ions in a mole of K₃PO₄.

7 0
2 years ago
If a 45.6 g piece of aluminum has a density of 2.1 g/mL, what volume would it have?
Mashcka [7]

Answer:

\boxed {\tt 21.7142857 \ mL}

Explanation:

The density formula is:

d=\frac{m}{v}

Rearrange this formula for v, volume. Multiply by v, then divide by d.

d*v=\frac{m}{v}

d*v=m

\frac{d*v}{d} =\frac{m}{d}

v=\frac{m}{d}

Volume can be found by dividing the mass by the density. The mass of the aluminum is 45.6 grams and the density is 2.1 grams per milliliter.

v=\frac{45.6 \ g }{ 2.1 \ g/mL}

Divide. Note the grams, or g, will cancel out.

v=\frac{45.6 }{2.1 \ g}

v= 21.7142857 \ mL

The volume of the aluminum is 21.7142857 milliliters.

3 0
3 years ago
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