A pharmacy technician measures the mass of a pill. the mass of one pill is 461 mg. what is the mass in kilograms of 100 pills? w

Problem 2

You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.

Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.

Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.

#days              Amount in micrograms

0                              216

3                               108

6                                54

9                                27

Problem One

Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.

Table

Bond               Energy Kj/Mol               Bond Length pico meters

N - N                 167                                                145

N=N                  418                                                125

N≡N                  942                                               110

Rules

As the number of bonds INCREASES, the energy within the bond also INCREASES

As the number of bonds INCREASES, the distance of the bond DECREASES.

The answer is a total of 74,844 calories.

Answer:

9.69g

Explanation:

To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.

1 mole of gas takes up 22.4L at STP.

Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning

X moles of N2 = 7.744/22.4 = 0.346 moles

Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:

Molar Mass of N2 = 2x14 = 28g/mol

Number of moles N2 = 0.346 moles

Find the mass of N2 =?

Mass = number of moles × molar mass

Mass of N2 = 0.346 × 28

Mass of N2 = 9.69g

Hence, 7.744L of N2 consists of 9.69g of N2

Answer:

The correct options include choice 2, 3, and 6.

Explanation:

Density is identified as the mass of a substance per unit volume occupied by that substance.

The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.

2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.

3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.

6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.

The same metals in both instances will yield consistent densities due to the fixed density of the metal.

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol / 2 mol CH3COOH)

= 0.005 mol

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm