The mass of one cubic meter of gold is 19,300 kilograms.
We can use the specified density and the following formula to determine the mass of one cubic meter of gold:
Mass = Density × Volume
Given:
Density of gold = 19.3 g/cm³
Remember that when converting from cubic centimeters to cubic meters, 1 cubic meter is equal to 1,000,000 cubic centimeters.
Density = Mass / Volume
Mass = Density × Volume
Volume = 1 cubic meter = 1,000,000 cubic centimeters
Density = 19.3 g/cm³
Mass (grams) = Density × Volume = 19.3 g/cm³ × 1,000,000 cm³ = 19,300,000 g
For converting grams to kilograms, divide by 1000:
Mass (kilograms) = 19,300,000 g / 1000 = 19,300 kg
Therefore, the mass of one cubic meter of gold is 19,300 kilograms.
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Answer:
Explanation:
Question:
Gold has a density of
19.3
g
/
c
m
3
.
What is the mass, in kilograms, of one cubic meter of gold?
Density:
The density of an object is a quantity derived from the relation between mass (fundamental quantity) and volume (derived quantity). No matter the size of the object, its density will be the same because it is a constant property that is only varied when changing the mass or volume of the object. There are other factors that affect density such as temperature and pressure.
Data:
Volume:
V
=
1
m
3
Density:
ρ
=
19.3
g
c
m
3
Equation:
The equation to find the mass of the gold is given...
See full answer below.
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Density: Definition, Formula & Practice Problems
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Chapter 1 / Lesson 10
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In chemistry we like the number of atoms or molecules in moles.
If one mole is equal to 6.02*10^23 atoms and you have 9.00 *10^23 atoms then how many moles do you have?
Answer;
= 1.495 moles
Explanation;
-One mole contains 6.02 ×10^23 particles
Therefore;
1 mole = 6.02 ×10^23 atoms
Thus; 9.00 ×10^23 atoms will have;
= 9.00 ×10^23 atoms / 6.02 ×10^23 atoms
= 1.495 moles
Answer:
2.49
⋅
10
−
12
moles Pb
Explanation:
Before doing any calculations, it's worth noting that atoms do not contain moles, it's the other way around.
A mole is simply a collection of atoms. More specifically, you need to have exactly
6.022
⋅
10
23
atoms of an element in order to have one mole of that element - this is known as Avogadro's number.
In your case, you must determine how many moles of lead would contain
1.50
⋅
10
12
atoms of lead.
Well, if you know that one mole of lead must contain
6.022
⋅
10
23
atoms of lead, it follows that you get
1.50
⋅
10
12
atoms of lead in
1.50
⋅
10
12
atoms of Pb
⋅
1 mole Pb
6.022
⋅
10
23
atoms of Pb
=
2.49
⋅
10
−
12
moles Pb
Answer:
B
Explanation:
We are given that ammonia can be produced from hydrogen gas and nitrogen gas according to the equation:
We want to determine the mass of hydrogen gas that must have reacted if 0.575 g of NH₃ was produced.
To do so, we can convert from grams of NH₃ to moles of NH₃, moles of NH₃ to moles of H₂, and moles of H₂ to grams of H₂.
We are given that the molar masses of NH₃ and H₂ are 17.03 g/mol and 2.0158 g/mol, respectively.
From the equation, we can see that two moles of NH₃ is produced from every three moles of H₂.
With the initial value, perform dimensional analysis:
*Assuming 100% efficiency.
Our final answer should have three significant figures. (The first term has three, the second term has four (the one is exact), the third term is exact, and the fourth term has five. Hence, the product should have only three.)
In conclusion, our answer is B.
Answer:
0.01917 m^3/kg.
Explanation:
Given:
P = 15 MPa
= 1.5 × 10^4 kPa
T = 350 °C
= 350 + 273
= 623 K
Molar mass of water, m = (2 × 1) + 16
= 18 g/mol
= 0.018 kg/mol
R = 0.4615 kPa·m3/kg·K
Using ideal gas equation,
P × V = n × R × T
But n = mass/molar mass
V = (R × T)/P
V/M = (R × T)/P × m
= (0.4615 × 623)/1.5 × 10^4
= 0.01917 m^3/kg.
The specific volume of superheated water vapor under the conditions of 15 MPa pressure and 350°C temperature, using the ideal gas equation, is approximately 0.01919 cubic meter per kilogram.
The question is asking to calculate the specific volume of superheated water vapor using the ideal gas equation P = ρRT, where P is the pressure, ρ is the density (inverse of specific volume), R is the gas constant, and T is the temperature.
To find the specific volume (v), we need to rearrange the ideal gas equation to v = RT/P. Given that the pressure P = 15 MPa = 15000 kPa, the gas constant R = 0.4615 kPa.m³/kg.K, and the temperature T = 350°C = 623.15 K (adding 273 to convert °C to K), we can substitute these values into our rearranged equation balance to calculate for v.
The specific volume v = (0.4615 kPa.m³/kg.K * 623.15 K) / 15000 kPa = 0.01919 m³/kg. So, the specific volume of superheated water vapor under the given conditions is approximately 0.01919 cubic meter per kilogram.
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(2) HCl
(3) HBr
(4) HI
HCl is the most polar bond, as greater the difference in electronegativity, the more polar the bond. Here, Chlorine is more electronegative than Hydrogen, Bromine, and Iodine, making the HCl bond most polar.
The most polar bond in the list provided is HCl. Polarity in bonds is determined by the difference in electronegativity between the two atoms in the bond. The greater the electronegativity difference, the more polar the bond. Chlorine (Cl) is more electronegative than Hydrogen (H), Bromine (Br), and Iodine (I), and so the bond between Cl and H is the most polar. The other three options (Cl2, HBr, HI) are either bonds between atoms of the same element (and therefore nonpolar because the electrons are shared equally) or are less polar due to the lower electronegativity difference.
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