An atom is most stable when it has eight electrons in its outermost energy level.

If the solubility of a gas in water is 5.0g/L when the pressure of the gas above the water is 2.0 atm, what is the pressure of t

Serggg [28]

F the solubility of a gas in water is 5.0g/L when the pressure of the gas above the water is 2.0 atm, what is the pressure of the gas above the water when the solubility of the gas is 1.0 g/L

Here's how to solve this one.
The formula for solubility is
P1 / P2 = solubility1 / solubility2 
P1=2*1/5= .4 atm

So the correct answer is 0.4 atm.

5 0

3 years ago

Which has more mass: an armful of feathers or an armful of lead? which has more volume: a kilogram of feathers or a kilogram of

valina [46]

Mass is related to weight. It measures the resistance of the substance to acceleration when a net force is applied.
Therefore: armful of lead has more mass

volume measures the amount of 3-dimensional space that an object occupies.
Therefore: kilogram of feather has more volume as it is less dense.

density is the mass per unit volume. Since armful of lead has more mass than a handful of feathers, while kilogram of feather has more volume than a kilogram of lead.
Therefore: kilogram of lead has higher density.

8 0

3 years ago

Perform the calculation and record the answer with the correct number of significant figures.

kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

Answer:

For A: The answer becomes 0.1

For B: The answer becomes 0.4884

Explanation:

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

Rule applied for addition and subtraction:

The least precise number present after the decimal point determines the number of significant figures in the answer.

Rule applied for multiplication and division:

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

For A: (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

$\Rightarrow \frac{0.4}{3.19}=0.125$

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

For B: (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

6 0

3 years ago

Plesse help asap!!!

Naily [24]

Answer:

4.4×10² cm³

Explanation:

From the question given above, the following data were obtained:

Diameter (d) = 68.3 mm

Height (h) = 0.120 m

Volume (V) =?

Next, we shall convert the diameter (i.e 68.3 mm) to cm.

This can be obtained as follow:

10 mm = 1 cm

Therefore

68.3 mm = 68.3 mm / 10 mm × 1 cm

68.3 mm = 6.83 cm

Therefore, the diameter 68.3 mm is equivalent 6.83 cm.

Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:

1 m = 100 cm

Therefore,

0.120 m = 0.120 m/ 1 m × 100 cm

0.120 m = 12 cm

Therefore, the height 0.120 m is equivalent 12 cm.

Next, we shall determine the radius of the cylinder. This can be obtained as follow:

Radius (r) is simply half of a diameter i.e

Radius (r) = Diameter (d) /2

r = d/2

Diameter (d) = 6.83 cm

Radius (r) =?

r = d/2

r = 6.83/2

r = 3.415 cm

Finally, we shall determine the volume of the cylinder as follow:

Radius (r) = 3.415 cm

Height (h) = 12 cm

Volume (V) =?

Pi (π) = 3.14

V = πr²h

V = 3.14 × (3.415) ² × 12

V = 440 cm³

V = 4.4×10² cm³

Therefore, the volume of the cylinder is 4.4×10² cm³

8 0

2 years ago

He isotope 62(over28ni has the largest binding energy per nucleon of any isotope. calculate this value from the atomic mass of n

irina [24]

The atomic mass of the isotope Ni ( 62 over 28 ) = 61.928345 amu.
Mass of the electrons: 28 · 5.4584 · 10^(-4 ) amu = 0.0152838 amu ( g/mol )
Mass of the nuclei:
61.928345 amu - 0.0152838 amu = 61.913062 amu (g/mol)
The mass difference between a nucleus and its constituent nucleons is called the mass defect.
For Ni ( 62 over 28 ): Mass of the protons: 28 · 1.00728 amu = 28.20384 amu
Mass of the neutrons: 34 · 1.00866 amu = 34.299444 amu
In total : 62.49828 amu
The mass defect = 62.49828 - 61.913062 = 0.585218 amu
Nucleus binding energy:
E = Δm · c² ( the Einstein relationship )
E = 0.585218 · ( 2.9979 · 10^8 m/s )² · 1 / (6.022 · 10^23) · 1 kg / 1000 g =
= 0.585218 · 8.9874044 · 10 ^16 : (6.022 · 10^23) · 0.001 =
= ( 5.2595908 : 6.022 ) · 0.001 · 10^(-7 ) =
= 0.0008733 · 10^(-7) J = 8.733 · 10^(-11) J
The nucleus binding energy per nucleon:
8.733 · 10^(-11) J : 62 = 0.14085 · 10 ^(-11) =
= 1.4085 · 10^(-12) J per nucleon.

4 0

3 years ago