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dem82 [27]
3 years ago
11

What volume of a 2.5 M stock solution of acetic acid (HC2H3O2) is required to prepare

Chemistry
1 answer:
Schach [20]3 years ago
8 0

<span> </span>

Answer is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH) = c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH) = 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.

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Ymorist [56]

Answer:

Iron: Fe

Co: Cobalt

Na: Sodium

Tin: Sn

P: Phosphorus

F: Flourine

Fe: Iron

Magnesium: Mg

Uranium: U

Ca: Calcium

Carbon: C

Lead: Pb

Ag: Silver

Zn: Zinc

Ni: Nickle

----------------

There is honestly no right answer for this but here is what I would put:

Atomic mass increases as you go from left to right. If you look at the periodic table, it would be between Sn and Sb. It would bet here because Sn is 118 and Sb is 121. Basing it off of Antimony and putting it in group 15, the properties are that it is metallic and is a poor conductor of heat. I would call it Stin, which would be shortened to St.

7 0
3 years ago
A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a
Semenov [28]

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

We are given:

Mass percentage of CH_4 = 20 %

So, mole fraction of CH_4 = 0.2

Mass percentage of C_2H_4 = 30 %

So, mole fraction of C_2H_4 = 0.3

Mass percentage of C_2H_2 = 35 %

So, mole fraction of C_2H_2 = 0.35

Mass percentage of C_2H_2O = 15 %

So, mole fraction of C_2H_2O = 0.15

We know that:

Molar mass of CH_4 = 16 g/mol

Molar mass of C_2H_4 = 28 g/mol

Molar mass of C_2H_2 = 26 g/mol

Molar mass of C_2H_2O = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}

where,

\chi_i = mole fractions of i-th species

m_i = molar masses of i-th species

n_i = number of observations

Putting values in above equation:

\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}

\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75

Hence, the correct answer is Option d.

3 0
3 years ago
Which part of the atom cannot have its location accurately determined and is modeled by a cloud around the center of the atom? A
Tju [1.3M]

Answer:

Electron

Explanation:

The answer would be the electron because it is constantly moving so its location cannot be accurately determined

3 0
3 years ago
Read 2 more answers
Qué tipo de reacciones pertenecen las combustiones?
navik [9.2K]

Answer:

Si una reacción química libera energía, se llama reacción exotérmica. El ejemplo más común es la combustión, en la cual la energía se manifiesta en forma de calor y luz. Si, por el contrario, la reacción química requiere energía del medio para efectuarse, recibe el nombre de reacción endotérmica.

5 0
3 years ago
For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)
aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0
3 years ago
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