A compound and an oxidant react to produce heat and a new product in a _________

Indicate the FULL NAME and TOTAL NUMBER of each element in the chemical formula provided. *

REY [17]

Answer:

4 lead = Pb

2 nitrogen = N

6 oxygen = O

Explanation:

Know the rules of multiplying wth perentheses.

- Hope that helped! Please let me know if you need further explanation.

6 0

2 years ago

How many grams are in 2.45E24 formula units of Fp3BZ2? The molar mass of Fp3Bz2. is 97.05 g/mol.

dlinn [17]

Answer:

394.99g

Explanation:

The number of moles of a substance can be calculated by dividing the number of atoms of such substance by Avagadro's number (6.02 × 10^23)

n = nA ÷ 6.02 × 10^23

The number of atoms of Fp3BZ2 in this question is 2.45E24 formula units i.e. 2.45 × 10^24

n = 2.45 × 10^24 ÷ 6.02 × 10^23

n = 2.45/6.02 × 10^(24-23)

n = 0.407 × 10¹

n = 4.07moles

Using mole = mass/molar mass

Where; molar mass of Fp3Bz2. is 97.05

g/mol.

mass = molar mass × mole

mass = 97.05 × 4.07

mass = 394.99g

4 0

2 years ago

What is the ph of a solution with h+ = 7.0* 10

Mekhanik [1.2K]

Do you mean h=7.0+10? If so your answer is 70.

8 0

2 years ago

g When aqueous solutions of and are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M is mixed

solong [7]

The question is incomplete, the complete question is:

When aqueous solutions of NaCl and $Pb(NO_3)_2$ are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M NaCl is mixed with an excess of an aqueous solution of

Answer: The mass of lead chloride produced is 1.96 g

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{ \text{Volume of solution (mL)}}$ .....(1)

Given values:

Molarity of NaCl = 0.1000 M

Volume of the solution = 140.7 mL

Putting values in equation 1, we get:

$0.1000=\frac{\text{Moles of NaCl}\times 1000}{140.7}\\\\\text{Moles of NaCl}=\frac{0.1000\times 140.7}{1000}=0.01407mol$

The chemical equation for the reaction of NaCl and lead nitrate follows:

$Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)$

By the stoichiometry of the reaction:

If 2 moles of NaCl produces 1 mole of lead chloride

So, 0.01407 moles of NaCl will produce = $\frac{1}{2}\times 0.01407=0.007035mol$ of lead chloride

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Molar mass of lead chloride = 278.1 g/mol

Plugging values in equation 2:

$\text{Mass of lead chloride}=(0.007035mol\times 278.1g/mol)=1.96g$

Hence, the mass of lead chloride produced is 1.96 g

8 0

2 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

3 years ago

A compound and an oxidant react to produce heat and a new product in a __________ reaction. [1]