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kari74 [83]
3 years ago
15

Determine the [h3o+] of a 0.210 m solution of formic acid.

Chemistry
1 answer:
Nataly [62]3 years ago
3 0
When the Pka for formic acid = 3.77
and Pka = -㏒ Ka 
   3.77 = -㏒ Ka
∴Ka = 1.7x10^-4 

when Ka = [H+][HCOO-}/[HCOOH]

when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2      by solving this equation for X

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
       = -㏒ 0.0059
       = 2.23 

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What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?
Vadim26 [7]

Answer:

158 L.

Explanation:

What is given?

Pressure (P) = 1 atm.

Temperature (T) = 112 °C + 273 = 385 K.

Mass of methane CH4 (g) = 80.0 g.

Molar mass of methane CH4 = 16 g/mol.

R constant = 0.0821 L*atm/mol*K.

What do we need? Volume (V).

Step-by-step solution:

To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:

PV=nRT.

Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.

So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:

80.0g\text{ CH}_4\cdot\frac{1\text{ mol CH}_4}{16\text{ g CH}_4}=5\text{ moles CH}_4.

So, in this case, n=5.

Now, let's solve for 'V' and replace the given values in the ideal gas law equation:

V=\frac{nRT}{P}=\frac{5\text{ moles }\cdot0.0821\frac{L\cdot atm}{mol\cdot K}\cdot385K}{1\text{ atm}}=158.04\text{ L}\approx158\text{ L.}

The volume would be 158 L.

6 0
1 year ago
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Answer:

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Explanation:

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Answer:

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Explanation:

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What is the answers and pls show work if possible!!
taurus [48]

D = m / V


It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...


V = L x W x H

Volume = Length x Width x Height


start by converting 200.0 mg into grams

1000 mg = 1 g

200. mg x (1 g / 10^3 mg) = 0.200 g


V = m / D

V = 0.200 g / (19.32 g/cm^3)

V = 0.01035 cm^3


Convert 2.4 ft and 1 ft to cm

2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm

1 ft = 30.48 cm


Compute the height (thickness)

V = LxWxH

H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm

H = 4.64 x 10^-6 cm


Convert to nanometers

4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm


Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.


Atomic radius gold = 174 pm

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4 0
3 years ago
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Elena L [17]

Answer:

The answer to your question is: letter B

Explanation:

Reaction

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From the information given and the reaction, we can conclude that:

Green solid = Cr2O3 (s)     "s" means solid

Colorless liquid = CCl4 (l)    "l" means liquid   and is the other reactant

Purple solid = CrCl3(s)        CrCl3 is purple and "s" solid

Then, as a green specks remains it means that the excess reactant is Cr2O3, so, CCl4 is the limiting reactant.

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