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2 years ago

In what ways are final grade calculations and atomic mass calculations similar?

Chemistry

1 answer:

Gennadij [26K]2 years ago

3 0

Answer:

atomic mass calculations are similar to final grade calculations in how the two are calculated. In school, your given different kinds of work that has a different effect on your grade. Homework might be worth 20 percent while a tests are worth 80 percent. the percentage of these assignments are the weight they have on your grade. when you multiply the weight by the score, you get a weighted score that you can add with other weighted scores to reach a total, that total would be your grade. Atomic mass calculations are found the same way, first turn the percent abundance into a decimal (divide by 100, 97% becomes .97) multiply it by Mass (AMU) and you get that isotopes mass contribution. do this for the isotopes for whatever element, and add up the total. the total will be your element's atomic mass.

Explanation:

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a 20.4 g aluminum sphere and a 49.4 g iron sphere are both added to 75.2 mL of water contained in a graduated cylinder. what is

alexandr1967 [171]

The level in the beaker will increase because the volumes of the spheres will also be added to the volume of the water. First, we must determine the volume of each sphere. For this, we will use:

density = mass / volume
We can check the density of both aluminum and iron in literature, and given the mass, we may obtain the volume.

Aluminum:
Density = 2.70 g/ml
Mass = 20.4 g
Volume = 20.4 / 2.7 = 7.56 ml

Iron:
Density = 7.87 g/ml
Mass = 49.4 g
Volume = 49.4 / 7.87 = 6.28 ml

Now, we add these volumes to the volume of water present:
75.2 + 6.28 + 7.56 = 89.04

The new level will be 89.0 ml

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3 years ago

Read 2 more answers

Potential (P) energy or kinetic (K) energy

JulsSmile [24]

Potential energy, kinetic energy would be if they were already running

8 0

3 years ago

ASAP PLEASE

dimaraw [331]

The mass of Hydrogen is 2 g/mol
The mass of Helium is 4 g/mol

6 0

3 years ago

0.785 moles of N2,fill a balloon at 1.5 atm and 301 K.<br> What is the volume of the balloon?

Tcecarenko [31]

Answer:

V = 12.93 L

Explanation:

Given data:

Number of moles = 0.785 mol

Pressure of balloon = 1.5 atm

Temperature = 301 K

Volume of balloon = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will put the values.

V = nRT/P

V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm

V = 19.4 L /1.5

V = 12.93 L

7 0

3 years ago

2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to

evablogger [386]

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

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2 years ago