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miv72 [106K]
3 years ago
8

What is 1 Oz equal to in grams?

Chemistry
2 answers:
lakkis [162]3 years ago
4 0
1 Oz is 28.3495 grams

hope this helps!
pentagon [3]3 years ago
3 0

Conversion factor for ounces and grams.

<em>1 ounce is equal to approximately 28.4 grams. </em>

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Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices
lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

  • P= 0.980 arm
  • V= 1.20 L
  • T= 287 K
  • R= 0.082 \frac{atmL}{molK}
  • n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082\frac{atmL}{molK}× 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082\frac{atmL}{molK}× 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

molar mass=\frac{0.458 grams}{0.04997 moles}

Solving:

<u><em>molar mass= 9.17 </em></u>\frac{g}{mol}

Finally, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

Learn more about

molar mass:

brainly.com/question/5216907

brainly.com/question/11209783

brainly.com/question/7132033

brainly.com/question/17249726

ideal gas law:

brainly.com/question/4147359

#SPJ1

6 0
2 years ago
What is the part of the cartridge that actually flies outward (the projectile)?
zhannawk [14.2K]

Answer:

the culotte, the fulminant capsule and the sheath or cap together with remains of powder

Explanation:

A bullet has a nose, sheath or casing, a short propellant charge, and a capsule.

All these components are part of the vouchers. The classification of these is in metallic and semi-metallic.

3 0
2 years ago
A jet airplane reaches 750 km/h on a certain flight. What distance does it cover in <br> 10 min?
malfutka [58]

The distance that jet covered in 10 min is 125 Km

calculation

distance = speed  x time

speed=750 km/h

time= 10/60hrs

=750 x10/60=125 Km

3 0
3 years ago
Read 2 more answers
A 35.40 gram hydrate of sodium carbonate, Na2CO3•nH2O, is heated to a constant mass. Its final weight is 30.2 g. What is formula
Sever21 [200]

Answer:

Na₂CO₃•H₂O

Explanation:

After it is heated, the remaining mass is the mass of sodium carbonate.

30.2 g Na₂CO₃

Mass is conserved, so the difference is the mass of the water:

35.4 g − 30.2 g = 5.2 g H₂O

Convert masses to moles:

30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃

5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O

Normalize by dividing by the smallest:

0.285 / 0.285 = 1.00 mol Na₂CO₃

0.289 / 0.285 = 1.01 mol H₂O

The ratio is approximately 1:1.  So the formula of the hydrate is Na₂CO₃•H₂O.

3 0
3 years ago
Read 2 more answers
What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
7 0
3 years ago
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