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Masja [62]
3 years ago
5

What is the pH of a 2.5 x 10-5 M solution of HCI?

Chemistry
1 answer:
aalyn [17]3 years ago
7 0

Answer:

The pH of the solution is 4.60.

Explanation:

The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:

pH = -log (H30 +)

pH= -log (2,5 x 10-5)

<em>pH=4.60</em>

You might be interested in
a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what
jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

P\cdot V = n\cdot R_{u}\cdot T (1)

Where:

P - Pressure, measured in kilopascals.

V - Volume, measured in liters.

n - Molar quantity, measured in moles.

T - Temperature, measured in Kelvin.

R_{u} - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}} (2)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in liters.

T_{1}, T_{2} - Initial and final temperature, measured in Kelvin.

If we know that P_{1} = 121\,kPa, P_{2} = 202\,kPa, V_{1} = 2.7\,L, T_{1} = 288\,K and T_{2} = 303\,K, the final volume of the gas is:

V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}

V_{2} = 1.702\,L

The gas will occupy a volume of 1.702 liters.

6 0
3 years ago
Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

57.5% was the percent yield.

C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0
3 years ago
Sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). This makes it easy to isolat
zepelin [54]

Answer:

The correct answer is 0.12 grams.

Explanation:

The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.

Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.  

The number of moles or n can be determined by using the equation, mass/molar mass.  

R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).  

On putting the values we get:

n = PV/RT

= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)

= 0.0043447 mol

Therefore, mass of CO will be moles * molar mass of CO

= 0.0043447 mol * 28 g/mol

= 0.12 g

3 0
3 years ago
HELP ASAP!
Solnce55 [7]

Answer:

oxygen

Explanation:

3 0
3 years ago
Which part of the atom has to stay the same for it to be the same element? number of neutrons
kolezko [41]
For it to be the same element it must contain the same number of protons
5 0
3 years ago
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