Of the following is not soluble in water KOH НСІ NaBr CaCO3 KI

What are the physical properties of liquid bleach

Archy [21]

I think it is Sodium,hypochlorite

6 0

3 years ago

For the following reaction, 4.64 grams of oxygen gas are mixed with excess benzene (C6H6). The reaction yields 3.95 grams of car

Reil [10]

Answer:

Theoretical yield for CO₂ is 5.10g

Explanation:

Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)

We convert the mass of oxygen to moles:

4.64 g /32 g/mol = 0.145 moles of O₂

Let's find out the 100% yield reaction of CO₂ (theoretical yield)

Ratio is 15:12. So let's make this rule of three:

15 moles of O₂ can produce 12 moles of CO₂

Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles

We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g

6 0

3 years ago

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ANSWER ASAP PLEASE The Florida panther is one of the most endangered animals on earth. Where is the air that Florida panthers br

lutik1710 [3]

A. in the atmosphere

7 0

3 years ago

Hydrogen and oxygen react to form water. At an atomic level, how do we know that a chemical reaction has occurred

makkiz [27]

Answer: Bubbles are evidence that a chemical reaction took place because cause formation of a precipitate. Bubbles are evidence that a chemical reaction took place because they indicate the formation of gas as a new product.

Explanation:

3 0

3 years ago

what is the pH of a solution that results when 0.010mol HNO3 is added to 500.ml of a solution that is 0.10M in aqueous ammonia a

WARRIOR [948]

Answer : The

pH of a solution is, 8.56

Explanation : Given,

$K_b=1.8\times 10^{-5}$

Concentration of ammonia (base) = 0.10 M

Concentration of ammonium nitrate (salt) = 0.55 M

First we have to calculate the value of $pK_b$ .

The expression used for the calculation of $pK_b$ is,

$pK_b=-\log (K_b)$

Now put the value of $K_b$ in this expression, we get:

$pK_b=-\log (1.8\times 10^{-5})$

$pK_b=5-\log (1.8)$

$pK_b=4.7$

Now we have to calculate the pOH of buffer.

Using Henderson Hesselbach equation :

$pOH=pK_b+\log \frac{[Salt]}{[Base]}$

Now put all the given values in this expression, we get:

$pOH=4.7+\log (\frac{0.55}{0.10})$

$pOH=5.44$

The pOH of buffer is 5.44

Now we have to calculate the pH of a solution.

$pH+pOH=14\\\\pH+5.44=14\\\\pH=14-5.44\\\\pH=8.56$

Thus, the pH of a solution is, 8.56

8 0

3 years ago