Answer:
Magnesiums atomic number is 12
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convert 575.1 mmHg to atm
Answer:
= .7567105263
Explanation:
1 atm = 760 mmHg
575.1 mmHg (1 atm/760mmHg) = .7567105263 atm
If a sample of CO2 contains 48 g of oxygen, how many moles of CO2 are there?
Answer:
0.75 moles of CO2
Explanation:
molar mass of CO2⇒ 44.01 g/mol
molar mass Oxygen ⇒ 31.998 g/mol
divide the mass given by molar mass of oxygen so
48÷31.998= 1.50009376 moles of O
then you use the ration of oxygen to carbon to find the moles of CO2 which is one mole of CO2 for 2 moles of oxygen
1.50009376 moles of O ×[tex]\frac{1}{2}[/tex] = 0.75004688
with sig figs
0.75 moles of CO2
What is the approximate vapor pressure when the liquid water boils at about 50°C?
380 mmHg
760 mmHg
100 mmHg
5 mmHg
Answer:
380
Explanation:
Answer each of the following questions with increases, decreases, or does not change.
a) If the intermolecular forces in a liquid increase, the normal boiling point of the liquid increases .
b) If the intermolecular forces in a liquid increase, the vapor pressure of the liquid does not change .
c) If the surface area of a liquid increases, the vapor pressure increases .
d) If the temperature of a liquid decreases, the equilibrium vapor pressure increases .
Drag and drop your selection from the following list to complete the answer:
does not change decreases increases
Answer:
1) Increases
2) decreases
3) increases
4) decreases
Explanation:
When the intermolecular forces in a liquid increases, the greater vapour pressure of the liquid decreases accordingly.
Since the vapour pressure is proportional to temperature, as temperature increases, the vapour pressure increases alongside.
As intermolecular forces increases, the boiling point increases accordingly since more energy is required to break intermolecular bonds.
Lastly, the greater the surface area, tell greater the vapour pressure since more liquid surface area is now available.
The normal boiling point of benzene is 80.1°C. What is its enthalpy of vaporization if the vapor pressure at 26.1°C is 100 torr?
The heat of vaporization of benzene is required.
The heat of vaporization of benzene is 33009 J/kg.
[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K
[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K
[tex]R[/tex] = Gas constant = 8.314 J/mol K
[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr
[tex]\Delta H[/tex] = Heat of vaporization
From the Clausius–Clapeyron equation
[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]
The heat of vaporization of benzene is 33009 J/kg.
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Which of the following is a good definition of matter?
O A. Anything that is made up of light and gravity
O B. Anything that has mass and takes up space
O C. Anything that produces heat and mass
O D. Anything that has energy and creates heat
Answer:
B
Explanation:
I did the question before and got it right.
Label the parts of the electric circuit that best correspond to the heart, arteries, veins, and cells.
Answer:
1 ➡️ Cells
2 ➡️ Arteries
3 ➡️ Veins
4 ➡️ Heart
Explanation:
The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.
The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.
From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood
The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.
What is an electric circuit?The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.
In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.
In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.
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help me please loves
Answer:
K and ba
Explanation:
Answer: K and Ba
Explanation:
What is the volume of a substance that has a mass of 59 g and a density of 1.98 g/mL?
(show all work)
Answer:
29.8
Explanation:
The formula for volume is mass/ density, so 59/1.98. 29.8 is the answer.
A teaspoon of salt, NaCl has a mass of about
5.0 g. How many formula units are in a
teaspoon of salt?
Answer: The answer is 5.15x10^22
Explanation:
The formula unit present in a teaspoon of salt [tex]NaCl[/tex] having a mass of about 5.0 g is [tex]5.15 \times10^{22}[/tex] formula units.
Molar mass, also known as molecular weight, is the mass of one mole of a substance. It is calculated by summing up the atomic masses of all the atoms in a molecule. The unit of molar mass is grams per mole (g/mol).
Now, to determine the number of formula units in a teaspoon of salt (NaCl), we need to use Avogadro's number and the molar mass of NaCl.
Avogadro's number [tex](N_a)[/tex] is approximately. [tex]6.022 \times10^{23}[/tex] formula units per mole.
The molar mass of [tex]NaCl[/tex] is the sum of the atomic masses of sodium (Na) and chlorine ([tex]Cl[/tex]), which are approximately 22.99 g/mol and 35.45 g/mol, respectively.
To calculate the number of formula units in 5.0 g of [tex]NaCl[/tex], we can follow these steps:
Now, calculate the number of moles of [tex]NaCl[/tex] using its molar mass:
Moles = Mass / Molar mass
Moles = [tex]5.0 g[/tex] / [tex](22.99 g/mol + 35.45 g/mol)[/tex]
Calculate the number of formula units using Avogadro's number:
Formula units = [tex]Moles \times Avogadro's number[/tex]
Let's perform the calculation:
Molar mass of [tex]NaCl[/tex]= [tex]22.99 g/mol + 35.45 g/mol = 58.44 g/mol[/tex]
Moles of [tex]NaCl[/tex] = [tex]5.0 g[/tex] / [tex]58.44 g/mol[/tex] ≈ [tex]0.0856 mol[/tex]
Formula units = [tex]0.0856 mol \times (6.022 \times 10^{23})[/tex] formula units/mol ≈ [tex]5.15 \times10^{22}[/tex]formula units.
Therefore, there are approximately [tex]5.15 \times10^{22}[/tex] formula units in a teaspoon of salt ([tex]NaCl[/tex]) having mass [tex]5.0 g[/tex].
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Scientists are experimenting with pure samples of isotope X which is radioactive. The sample has a mass of 20. Grams. The half-life was measured to be 232 seconds. There is a second sample that weighs 80 grams. What is the half-life of the second sample
Answer:
Explanation:
Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .
Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .
what are the benefits of using supercritical fluids in EGS
Answer:
See detailed explanation.
Explanation:
Hello.
In this case, it is firstly necessary to cite that EGS accounts enhanced geothermal systems which are man-made reservoirs, placed where lots of hot rock is present but there is lack natural permeability, which requires a fluid to be injected into the subsurface to re-open it and therefore creating permeability.
Typically, water has been used for this purpose, but due to the current issue on saving water alternative methods such as supercritical fluids has been being implemented because they have better dynamic properties such as lower viscosities and therefore larger flow velocities, supercritical CO2 is easy and cheap to get as low temperatures are required to turn it in supercritical condition.
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What do you need to measure the amount of energy in food
Answer:
u measure how much power it has
Explanation:
for example u can power a light bulb woth it if u can it was 100eg energy
1.547 grams of hydrated MgSO4 is heated in a crucible. After heating, 0.7554 grams of anhydrous MgSO4 remains in the crucible. How many waters of hydration were attached to the MgSO4
Answer:
7
Explanation:
Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O
The following data were obtained from the question:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mole of H₂O = x =?
Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mass of H₂O =?
Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)
Mass of H₂O = 1.547 – 0.7554
Mass of H₂O = 0.7916 g
Finally, we shall determine the value of the x as illustrated below:
Mass of MgSO₄.xH₂O = 1.547 g
Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]
= 24 + 32 + 64 + x(2 + 16)
= 120 + 18x
Mass of H₂O = 0.7916 g
Molar mass of xH₂O = 18x
Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O
18x/ 120 + 18x = 0.7916/1.547
Cross multiply
0.7916 (120 + 18x) = 18x × 1.547
94.992 + 14.2488x = 27.846x
Collect like terms
94.992 = 27.846x – 14.2488x
94.992 = 13.5972x
Divide both side by 13.5972
x = 94.992 / 13.5972
x = 7
Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O
Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.
The number of moles of attached water molecules is 7.
Mass of hydrated MgSO4 = 1.547 grams
Mass of anhydrous MgSO4 = 0.7554 grams
Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x
Number of moles of anhydrous MgSO4 = 0.7554 grams /120
Number of moles of anhydrous salt = Number of moles of hydrated salt
0.7554 grams /120 = 1.547 grams/120 + 18x
0.7554(120 + 18x) = 1.547 × 120
90.6 + 13.6x = 185.6
185.6 - 90.6 /13.6 = x
x = 7
The number of moles of attached water molecules is 7.
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What occurs after cytokinesis is completed at the end of meiosis I?
O Four haploid cells are formed.
O Two diploid cells are formed.
OTwo haploid cells are formed.
O Four diploid cells are formed.
Answer. After cytokinesis is completed at end of meiosis - I two haploid cells are formed.on:
Answer:
C. TWO HAPLOID CELLS ARE FORMED
Explanation:
I TOOK THE EDGUNITY TEST AND I GOT IT CORRECT
What is the pressure if the height of a column of mercury is 0.20 m and the density of mercury is 13,600 kg/m3? (remember, gravity is 9.81 m/s2)
Answer:
[tex]p=26683.2Pa[/tex]
Explanation:
Hello,
In this case, since the pressure is computed via:
[tex]p=h*\rho*g[/tex]
Whereas h is the 0.520-m height, [tex]\rho[/tex] is the 13600-kg/m³ density and the g the 9.81-m/s² gravity. Thus, the pressure in Pa is:
[tex]p=0.20m*13,600 \frac{kg}{m^3} *9.81\frac{m}{s^2} \\\\p=26683.2\frac{kg*\frac{m}{s^2} }{m^2} =26683.2\frac{N}{m^2}\\ \\p=26683.2Pa[/tex]
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What does chemical equations and chemical formulas have in common?
Answer:
Chemical symbols refer to chemical elements only. They do not necessarily refer to atoms of that element, but also to ions.
Explanation:
why are copper pipes used in place of old lead pipes for plumbing systems?
Calcium reacts with sulfur forming calcium sulfide. What is the theoretical yield (g) of CaS(s) that could be prepared from 7.19 g of Ca(s) and 2.67 g of sulfur(s)? Enter your answer with two decimal places. Do not type units with your answer.
Answer:
The theoretical yield of CaS is 6.01 g.
Explanation:
The balanced reaction is given as:
[tex]Ca+S\rightarrow CaS[/tex]
The molar mass of Ca and S is 40.08 and 32.065 g/mol respectively.
Number of moles = [tex]\frac{Mass}{Molar Mass}[/tex]
So, 7.19 g of Ca contains [tex](\frac{7.19}{40.08})[/tex] mol of Ca or 0.179 mol of Ca
Also, 2.67 g of S contains [tex](\frac{2.67}{32.065})[/tex] mol of S or 0.0833 mol of S
According to the balanced equation:
1 mol of Ca produces 1 mol of CaS
So, 0.179 mol of Ca produces 0.179 mol of CaS
According to the balanced equation:
1 mol of S produces 1 mol of CaS
So, 0.0833 mol of S produces 0.0833 mol of CaS
As the least number of mol of CaS (product) is produced from S , therefore, S is the limiting reactant.
So, thoretically, 0.0833 mol of CaS is produced.
The molar mass of CaS is 72.143 g/mol.
So, the mass of 0.0833 mol of CaS is [tex](0.0833\times 72.143)[/tex] g or 6.01 g
Hence, the theoretical yield of CaS is 6.01 g.
How can a shoreline be destroyed by a Hurricane?
Answer:
costal erosion
Explanation:
can cause erosion
Answer:
The water piles up with nowhere to go but onto land when it gets to the coast. The rising water, called storm surge, can submerge low-lying areas and towns along the coast. Combined with the crashing waves of the storm, the storm surge can cause demolishing docks, houses, roads, and erode beaches.
Explanation:
A sample of an unknown gas weighs 0.419 grams and produced 5.00L of gas at 1.00atm (unknown gas only) and 298.15K, what is the molar mass (g/mole) of this unknown gas
Answer:
molar mass of unknown gas = 1.987 g/mol
Explanation:
First, the number of moles of the unknown gas is found
Using the ideal gas equation: PV = nRT
P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹
n = PV/RT
n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)
n = 0.2109 moles
Molar mass = mass/ number of moles
molar mass = 0.419 g/ 0.2109 mols
molar mass of unknown gas = 1.987 g/mol
The molar mass of unknown gas by using ideal gas equation = 1.987 g/mol.
Ideal gas equationThis equation gives the relation between pressure, volume, temperature as given below:
[tex]PV = nRT[/tex]
P = 1.00 atm, V = 5.00 L, T = 298.15 K, R = 0.082 L.atm.mol⁻¹K⁻¹
Substitute the above values in the above equation as follows:
n = (1.00 atm * 5.00 L)/(298.15 K *0.082 L.atm.mol⁻¹K⁻¹)
n = 0.2109 moles
Formula for molar mass[tex]Molar mass = mass/ number of moles[/tex]
Calculate molar mass by using the above equation,
molar mass = 0.419 g/ 0.2109 moles
The molar mass of unknown gas = 1.987 g/mol
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True or False: The exact location of an electron can be measured thanks to
modern science.
Answer:
false you can not get a exact location of electrons from just modern science
What happens when the elements in group 2 react with water?
Answer:
The Group 2 metals become more reactive towards the water as you go down the Group.
Explanation:
These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.
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True or False: Particles that are moving faster have a higher temperature
Answer:
true
Explanation:
I'm not sure why cause I dont know how to explain but it's TRUE
Answer:
True
Explanation:
The particles moving faster in a substance the hotter it gets.
Based on the visible cell structure, which of the following statements is true?
Answer:I think it would be the third one.
Explanation:
Determine the value of the equilibrium constant (report your answer to three significant figures) for the following reaction if an equilibrium mixture contains 0.010 mol of solid PbBr2, and is 0.0100 M in Pb2+ ions and 0.0250 M in Br1- ions. Use the notation 4.31e-5 to indicate a number such as 4.31 x 10-5.
Answer:
6.25e-6 is the value of the equilibrium constant
Explanation:
we have this equation
[tex]PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)[/tex]
When at a state of equilibrium,
we have the concentration of Pb^2+ to be 0.01
we have the concentration of Br^- to be 0.025
the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1
[PbBR2] = 1
such tht
Keq = [Pb^2+] x [Br-]^2
we already know the values of these from the above.
0.01x0.025^2
= 0.01 x 0.000625
= 0.00000625
= 6.25 x 10^-6
= 6.25e^-6
Calculate the density of the aluminum cylinder with a diameter 0f 1.3 cm weighing 18 grams. Height of the cylinder is 5.2 Cm. Find percent error (Actual density of Aluminum is 2.7 g/cm3). (Formula for volume of cylinder is πr2h)
Answer:
Percent error = 3.7%
Explanation:
Given data:
Density of Al cylinder = ?
Weight of cylinder = 18 g
Diameter = 1.3 cm
Height = 5.2 cm
Actual density of Al = 2.7 g/cm³
Percent error = ?
Solution:
First of all we will calculate the volume of cylinder through given formula.
V = πr²h
r = diameter /2
V = 22/7 × (0.65 cm)²× 5.2 cm
V = 22/7 × 0.4225cm²× 5.2 cm
V = 6.89 cm³
Now we will calculate the density.
d = m/v
d = 18 g/ 6.89 cm³
d = 2.6 g/cm³
Percent error:
Percent error = measured value - actual value /actual value × 100
Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³ × 100
Percent error = 3.7%
Negative sign shows that measured or experimental value is less than actual value.
What can the chemical formula tell us about a compound?
Answer:
A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.
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Question 11
4 pts
Using the formula 2H202 --> 2H2O + O2, if 7.30 moles of peroxide are
decomposed, how many moles of oxygen will be formed?
Answer:
3.65 mol O₂
Explanation:
Step 1: RxN
2H₂O₂ → 2H₂O + O₂
Step 2: Define
Given - 7.30 mol H₂O₂
Solve - x mol O₂
Step 3: Stoichiometry
[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 74.6 kJ of heat. Before the reaction, the volume of the system was 8.20 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ΔE, in kilojoules.
Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ