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pentagon [3]
4 years ago
5

What is the total ionic equation for the following reaction?

Chemistry
2 answers:
schepotkina [342]4 years ago
5 0

Answer: Option (d) is the correct answer.

Explanation:

An equation in which electrolytes are represented in the form of ions is known as an ionic equation.

Strong electrolytes easily dissociate into their corresponding ions. Hence, they form ionic equation.

H_{2}CrO_{4} is a strong acid and Ba(OH)_{2} is a strong bases, therefore, both of them will dissociate into ions.

Thus, total ionic equation will be as follows.

2H^{+} + CrO_{4}^{-} + Ba^{2+} + 2OH^{-} \rightarrow Ba^{2+} + CrO_{4}^{-} + 2H_{2}O

maxonik [38]4 years ago
5 0

Answer:

2H+ + CrO4– + Ba2+ + 2OH– > Ba2+ + CrO4– + 2H2O

Explanation:

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The chemical formula for magnesium oxide is 2MgO. The number 2 represents the number of ___.
Korvikt [17]

Answer:

magnesium oxide molecules

Explanation:

this is because the two is before the formula of MgO

7 0
3 years ago
Which of the following has the greatest mass, in grams? 1 atom of lead (Pb) 0.5 mol silver (Ag) atoms 0.3 mol gold (Au) atoms 25
Strike441 [17]

Answer : Option C) 0.3 mol of Gold.

Explanation : Amongst the options given in the question, 0.3 mol of Au is the greatest in mass in grams.

As 0.3 mol X atomic weight of Au (196.966) = 59.088 grams;

Silver has 0.5 mol X atomic weight of Ag (107.86) = 53.93 grams;

The other options are not relevant as they deal in the atomic range which has mass value very less as compared to the moles of elements.

Therefore, it is clear that Au has the greatest mass amongst the given choices.

8 0
4 years ago
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mrs_skeptik [129]

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4 years ago
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A 29.4 ml sample of 0.347 m triethylamine, (c2h5)3n, is titrated with 0.375 m hydrobromic acid. at the equivalence point, the ph
Debora [2.8K]

A 29.4 ml sample of 0.347 m triethylamine, (c2h5)3n, is titrated with 0.375 m hydrobromic acid. at the equivalence point, the pH is 5.81.

Given,

Volume of Triethylamine (C₂H₅)₃N = 29.4 ml = 0.0294 L

Molarity of (C₂H₅)₃N = 0.275 M

Molarity of HBr = 0.375M

Solution:

(C₂H₅)₃N + HBr ⇔ (C₂H₅)₃ NH⁺  +  Br⁻

⇒ Volume of HBr = mole of (C₂H₅)₃N / Molarity of HBr

⇒ Volume of HBr = 0.00572 mol / 0.375 mol/l

⇒ Volume of HBr = 0.0152 L

∴ Total Volume = 0.0294 + 0.0152

⇒ Total Volume = 0.0446 L

Concentration of (C₂H₅)₃NH⁺ = \frac{0.00572}{0.0446} ⇒ 0.128 M,

(C₂H₅)₃NH⁺ + H₃O⁺ ⇄ (C₂H₅)₃N + H₃O⁺

let's assume, (C₂H₅)₃N = x,

⇒ H₃O⁺ = x, and

⇒ (C₂H₅)₃NH⁺ = 0.128 - x

∴ Now, k = kw / kb

⇒ k = 1.9 × 10⁻¹¹

and, k = [(C₂H₅)₃N][H₃O⁺] / (C₂H₅)₃NH⁺

⇒ 1.9 × 10⁻¹¹ = x^{2} / (0.128 - x)

Thus,  x = 1.55 × 10⁻⁶ M,

Hence, pH = - log[x]

⇒ - log [1.55 × 10⁻⁶ ]

⇒ - log (1.55) - log (10⁻⁶)

⇒ - log (1.55) + 6log10

⇒ - 0.19 + 6

⇒ 5.81 = pH

Therefore, pH = 5.81.

To learn more about equivalence point here

brainly.com/question/14782315

#SPJ4

7 0
2 years ago
The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a
mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0
3 years ago
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