Hi . I need help on this problem please help.

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0

3 years ago

How do you think the amount of a material affects its tendency to sink or float? <br> ASAP

Ksivusya [100]

Answer:

it dependes on the material

Explanation:

what is the material

3 0

3 years ago

1. Review the information in the table below. Use the information to calculate rate of how many sandwiches are made in a 10-minu

PilotLPTM [1.2K]

Answer:

The answer to your question is given below.

Explanation:

Rate is simply defined as quantity per unit time. Mathematically it is represented as:

Rate = Quantity /time

Thus, we can obtain the rate as follow:

1. Quantity = 20 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 20/10

Rate = 2 sandwich per mins

2. Quantity = 30 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 30/10

Rate = 3 sandwich per mins

3. Quantity = 40 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 40/10

Rate = 4 sandwich per mins

4. Quantity = 50 sandwich

Time = 10 mins

Rate =?

Rate = Quantity /time

Rate = 50/10

Rate = 5 sandwich per mins

Thus, the complete table is given as follow:

Quantity >> Unit of measure >> Rate

10 >>>>>>> 10 min >>>>>>>>>> 1

20 >>>>>>> 10 min >>>>>>>>>> 2

30 >>>>>>> 10 min >>>>>>>>>> 3

40 >>>>>>> 10 min >>>>>>>>>> 4

50 >>>>>>> 10 min >>>>>>>>>> 5

4 0

3 years ago

A salad is an example of a mixture that is ... Mixed

melomori [17]

Heterogeneous Mixture

8 0

3 years ago

What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution

denis-greek [22]

Answer:

89.04 g of NaNO₃.

Explanation:

We'll begin by converting 838 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

838 mL = 838 mL × 1 L / 1000 mL

838 mL = 0.838 L

Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:

Volume = 0.838 L

Molarity = 1.25 M

Mole of NaNO₃ =?

Mole = Molarity × volume

Mole of NaNO₃ = 1.25 × 0.838

Mole of NaNO₃ = 1.0475 mole

Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:

Mole of NaNO₃ = 1.0475 mole

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ =?

Mass = mole × molar mass

Mass of NaNO₃ = 1.0475 × 85

Mass of NaNO₃ = 89.04 g

Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.

4 0

3 years ago