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3 years ago

Barium chlorate (Ba(ClO3)2) breaks down to form barium chloride and oxygen. What is the balanced equation for this reaction?

Chemistry

2 answers:

Allushta [10]3 years ago

7 0

The answer is Option 4:) have a good day

Scrat [10]3 years ago

6 0

Ba(ClO₃)₂ ----> BaCl₂ + 3O₂

:)

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How many grams of sodium chloride are present in a 0.75 m solution with a volume of 500.0 milliliters?

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3 years ago

If 115grams of a substance reacts with 84grams of another substance, what will be the mass of the products after the reaction?

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3 years ago

After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is

Tanzania [10]

Answer:

Ksp = 8.8x10⁻⁵

Explanation:

<em>Full question is:</em>

<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>

<em />

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

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3 years ago

"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.

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3 years ago