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3 years ago

Balance and evaluate the following equation: AgNO3 (aq) + PbBr2 (s) yields AgBr (s) + Pb(NO3)2 (aq)

Chemistry

2 answers:

SOVA2 [1]3 years ago

8 0

Answer:

The balance equation is given as:

$2AgNO_3 (aq) + PbBr_2 (s) \rightarrow 2AgBr (s) + Pb(NO_3)_2 (aq)$

Explanation:

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

The balance equation is given as:

$2AgNO_3 (aq) + PbBr_2 (s) \rightarrow 2AgBr (s) + Pb(NO_3)_2 (aq)$

According to reaction,2 moles of silver nitrate react with 1 mole of lead (II) bromide to give 2 moles of silver bromide and 2 moles of lead nitrate.

When aqueous silver nitrate reacts with solid lead(II) bromide it gives solid precipitate of silver bromide and aqueous solution of lead(II) nitrate.

4vir4ik [10]3 years ago

6 0

Balance the bromine (Br) in the equation by multiplying AgBr by 2. Further, multiply AgNO3 by 2 in order to balance NO3 in the equation. With this, the elements in each side become balance. This gives an answer of,
2AgNO3 (aq) + PbBr2 (s) ---> 2AgBr (s) + Pb(NO3)2 (aq)

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What is the volume of 3.66 x 1032 molecules of fluorine gas at STP?

kipiarov [429]

Answer:

1.36x10^10L

Explanation:

Step 1:

Determination of the mole of fluorine that contains 3.66x10^32 molecules. This is shown below:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This implies that 1 mole of fluorine also contains 6.02x10^23 molecules.

Now if 1 mole of fluorine contains 6.02x10^23 molecules,

Therefore, Xmol of fluorine will contain 3.66x10^32 molecules i.e

Xmol of fluorine = 3.66x10^32/6.02x10^23

Xmol of fluorine = 6.08x10^8 moles

Step 2:

Determination of the volume occupied by 6.08x10^8 moles of fluorine.

1 mole of any gas occupy 22.4L at stp. This means that 1 mole of fluorine also occupy 22.4L at stp.

Now if 1 mole of fluorine occupies 22.4L at stp,

Then 6.08x10^8 moles of fluorine will occupy = 6.08x10^8 x 22.4 = 1.36x10^10L

6 0

3 years ago

Select the correct answer.

elena-s [515]

Answer:

B. CO + H₂O → H₂ + CO₂

Explanation:

Redox reaction -

It is the type of reaction , where the process of oxidation and reduction takes place simultaneously , is known as a redox reaction .

Oxidation refers to the addition of oxygen atom or the removal of hydrogen atom .

And ,

Reduction refers to the addition of hydrogen atom or the removal of the oxygen atom .

From the options given in the question , the only reaction , which have both the process of oxidation as well as reduction , is -

CO + H₂O → H₂ + CO₂

In the above reaction ,

Oxidation process - CO → CO₂ ,

The addition of oxygen atom .

Reduction process - H₂O → H₂ ,

The removal of oxygen atom .

5 0

3 years ago

A compound is composed of 85.6% Carbon and the rest is Hydrogen. The molecular mass of the compound is 42.078g/mol. What is the

s344n2d4d5 [400]

The molecular formula for the compound is $C_{3} H_{6}$

Explanation:

As with all of these problems, we assume 100 g of an unknown compound.

And thus, we determine the elemental composition by the given percentages.

Moles of carbon = 85.64 / 12.011

= 7.13 mol.

Moles of hydrogen = 14.36 / 1.00794

= 14.25 mol.

There are 2 moles of hydrogen per mole of carbon. And thus the empirical formula is CH $_{2}$ .

And molecular formula = n × (empirical formula)

Thus, 42.08 = n × (12.011 + 2 × 1.00794)

And thus n = 3, and molecular formula = $C_{3} H_{6}$

3 0

3 years ago

Based on difference in electro negatives between atoms, which is the most polar bond pair? A. H — S B. H — CI C. H — O D. H — F

pickupchik [31]

I think it’s C
Good luck

3 0

3 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0

3 years ago