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Nikitich [7]
3 years ago
9

24. According to the author, the term avant-garde

Chemistry
1 answer:
REY [17]3 years ago
5 0
F.a particular art movement founded intentionally in the early 1900s
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Water carries helpful nutrients to new areas,
den301095 [7]

Answer:

Positive Flood I believe

8 0
3 years ago
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What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso
Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247

The pH of the buffer can be known as

pH = pK_{a} + log[\frac{[A-]}{[HA]}}]

The concentration of [A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\

Similarly, the concentration of [HA] = \frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236

So, the pH of the buffer is 6.1236.

5 0
3 years ago
I will give brainliest and 76 points
katen-ka-za [31]

Answer:

number 5 is the answer

Explanation:

8 0
2 years ago
If an object has a density of 0.00018 g/cm cubed, what density would it be if it had a mass of 10kg
Vladimir79 [104]

Answer:

Explanation:

first to get the density of some thing you have to devide the mass by the volume so 0.00018 (divided) by 10 kg and that gives you ur answer

3 0
2 years ago
A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0
3 years ago
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