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3 years ago

6

What is the name of Pb(NO3)2? Explain how you determined the bond type and the steps you used to determine the naming convention

for the compound. Please be specific :)

1 answer:

Elena L [17]3 years ago

5 0

Answer: Lead(II) nitrate but idk the rest

Explanation:

You might be interested in

How many dm³ of hydrogen,measured at s.t.p.,would be needed to reduce 47.7g of copper(II) oxide to copper?

GaryK [48]

Answer:

Option D. 13.44

Explanation:

We'll begin by calculating the number of mole in 47.7g of copper(II) oxide, CuO.

This can be obtained as follow:

Mass of CuO = 47.7 g

Molar mass of CuO = 63.5 + 16 = 79.5 g/mol

Mole of CuO =.?

Mole = mass /Molar mass

Mole of CuO = 47.7/79.5

Mole of CuO = 0.6 mole

Next, we shall write the balanced equation for the reaction. This is given below:

CuO + H2 —> Cu + H2O

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2 to produce 1 mole of Cu and 1 mole of H2O.

Next, we shall determine the number of mole of H2 needed to react completely with 0.6 mole of CuO.

This can be obtained as follow:

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2.

Therefore, 0.6 mole of CuO will also react with 0.6 mole of H2.

Finally, we shall determine the volume occupied by 0.6 mole of H2 at STP.

This can be obtained as follow:

1 mole of H2 occupied 22.4 dm³ at STP.

Therefore, 0.6 mole of H2 will occupy = 0.6 × 22.4 = 13.44 dm³.

Therefore, 13.44 dm³ of H2 is needed for the reaction.

4 0

3 years ago

The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has

Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate law expression is:

$Rate=k[NO]^2[O_2]$

Now we have to determine the number of molecules of $NO\text{ and }O_2$

In mixture 1 : There are 5 $NO$ and 4 $O_2$ molecules.

In mixture 2 : There are 7 $NO$ and 2 $O_2$ molecules.

In mixture 3 : There are 3 $NO$ and 5 $O_2$ molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(5)^2\times (4)$

$Rate=k(100)$

The rate law expression for mixture 2 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(7)^2\times (2)$

$Rate=k(98)$

The rate law expression for mixture 3 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(3)^2\times (5)$

$Rate=k(45)$

Hence, the mixture 1 has the fastest initial rate.

4 0

3 years ago

Which expression can be used to calculate the density of a 129.5-gram sample of bronze that has a volume of 14.8 cubic centimete

PilotLPTM [1.2K]

Answer:

it would be like 30

Explanation:

just divide the gram by cubic

5 0

2 years ago

If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

Schach [20]

<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

$0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL$

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

4 0

3 years ago

What kind of sport do you have interest the most and explain why? <br>

lidiya [134]

Answer

Gymnastics

Explanation:

for the simple fact that they do cool stuff

7 0

3 years ago