Answer:
1116g
Explanation:
We'll convert moles O2 -> moles of Na2O -> grams of Na2O.
Based on our balanced equation, we have 1 mole of O2 for every 2 moles of Na2O. This is our mole to mole ratio.
9 mol O2 x [tex]\frac{2 mol Na2O}{1 mol O2}[/tex] = 18 mol Na2O
We can convert mols -> grams using the molar mass of Na2O- 62g.
18 mol Na2O x [tex]\frac{62g Na2O}{1 mol}[/tex] = 1116g
Thorium-236 has a half life of 10.0 minutes. If I have 250 grams of Thorium-236, How many half lives will have taken place after 50.0 minutes?
What happens to the entropy when a solution is made?
A. The entropy increases.
B. The entropy decreases.
C. The entropy goes to zero.
D. The entropy is unaffected.
Answer:
The entropy increases
Explanation:
Just took the quiz
A metal forms two oxides X and Y when contains 28.9% and 21.0% oxygen respectively.
a) calculate the ratio by mass of metal in the two oxides
b) what chemical law does it illustrate
Please help no Link please
15.
A tank containing 173 grams of methane, CH (9), registers 15.1 atmospheres at 298 Kelvin. What is the volume of the tank (assuming the entire volume is available to the gas)?
A 3,410L
B. 213
C.O 175L
D. 280.
Chemistry 4/28 5454
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Answer: The volume of tank is 17.5 L.
Explanation:
Given: Mass of methane = 173 g
Pressure = 15.1 atm
Temperature = 298 K
Molar mass of methane is 16.04 g/mol.
Therefore, moles of methane are calculated as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{173 g}{16.04 g/mol}\\= 10.78 mol[/tex]
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\15.1 atm \times V = 10.78 mol \times 0.0821 L atm/mol K \times 298 K\\V = \frac{10.78 mol \times 0.0821 L atm/mol K \times 298 K}{15.1 atm}\\= 17.5 L[/tex]
Thus, we can conclude that volume of the tank is 17.5 L.
3. How many times does earth rotate on its axis in one year?
Answer:
There are approximately 366.25 sidereal days in a year so that the Earth spins 366.25 times with respect to distant stars in a year
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Explanation:
Consider the balanced equation Zn + 2HCl ZnCl2 + H2 How many
moles of ZnCl2 will be produced if 2 moles of HCl are used?
Answer:
1 mole of ZnCl₂
Explanation:
Just from the stoichiometric equation/ balanced equation:
Zn(s) + 2HCl(aq) → ZnCl₂(s) + H₂(g)
1 mole 2 moles 1 mole 1 mole
Therefore: 2 moles of 2HCl produce 1 mole of ZnCl₂
Bears stop coming to a river ecosystem where they have been eating many fish each day. The fish the bears eat normally eat smaller fish, which eat plants along the river bottom.
What happens to the ecosystem?
Both the larger and the smaller fish populations grow quickly but then die out because the plant life is insufficient for them all to eat.
The larger fish population will drop first, and the smaller fish population will grow quickly. The plants will die off because too many of the smaller fish are eating them.
The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.
The smaller fish population begins to eat more plants and to grow. The larger fish have more food to eat so their population is able to grow, too.
Answer:
The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.
Explanation:
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1) How many moles of gaseous arsine (AsH3) occupy 0.372 L at STP?
In sig fig 4
2) What is the density of gaseous arsine?
In sig fig 4
Thanks!
Answer: (1). There are 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.
(2). The density of gaseous arsine is 3.45 g/L.
Explanation:
1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol[/tex]
2). As number of moles are also equal to mass of a substance divided by its molar mass.
So, number of moles of Arsine [tex](AsH_{3})[/tex] (molar mass = 77.95 g/mol) is as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g[/tex]
Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.
[tex]Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L[/tex]
Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.
At 47c a gas has a pressure of 140kpa. The gas is cooled until the pressure decreases to 105kpa. If the volume remains constant, what will the final temperature be in kelvin’s? In degrees Celsius
Answer:
The final temperature is equal to 240 K or -33.15°C
Explanation:
Given that,
Initial temperature of the gas, T₁ = 47°C = 320 K
Initial pressure, P₁ = 140 kpa
Final pressure, P₂ = 105 kpa
We need to find the final temperature if the volume remains constant. The relation between temperature and pressure is given by :
[tex]P\propto T[/tex]
or
[tex]\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{105\times 320}{140}\\\\T_2=240\ K\\\\T_2=-33.15^{\circ} C[/tex]
So, the final temperature is equal to 240 K or -33.15°C.
Jerry is trying to classify cells by their physical characteristics. He discovers a multicellular organism containing cells that have a nucleus and a cell wall as well as the ability to conduct photosynthesis. Into which of the three domains would this organism most likely fit? A. Archaea B. bacteria C. Eukarya D. Viral
Domains eukarya would this organism most likely fit.
The domain eukarya comprised of eukaryotes or organisms whose cells contain true nucleus.
What is a domain?It is the largest of all groups in the classification of life. There are three domains:-
Archaea domainBacteria domainEukarya domainWhat is Eukarya?It is the domain of organism called eukaryotes. These are the organism who have a well defined nucleus and membrane bound organelles.
Hence, C) option is correct.
To know more about domain here
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What volume will 28 grams of nitrogen gas occupy at 27 Celsius and a
pressure of 785 mm Hg?
Answer: [tex]2.49\ m^3[/tex]
Explanation:
Given
Mass of nitrogen present [tex]m=28\ g[/tex]
Temperature [tex]T=27^{\circ}C\equiv 300\ K[/tex]
Pressure [tex]P=785\ mm\ \text{of}\ Hg\ \text{or}\ 1.032\ atm[/tex]
The molar mass of Nitrogen [tex]M=28\ g/mol[/tex]
No of moles of nitrogen present
[tex]n=\dfrac{m}{M}\\\\n=\dfrac{28}{28}\\\\n=1[/tex]
Using [tex]PV=nRT[/tex]
[tex]\Rightarrow 1.032\times V=1\times 8.314\times 300\\\\\Rightarrow V=\dfrac{2494.2}{1.032}\\\\\Rightarrow V=2494.2\ L\ \text{or}\ 2.49\ m^3[/tex]
T/F Adsorption is a real problem in gravimetry, especially when the particle size is large
Answer:
False
Explanation:
During the workup portion of the reaction of alkenes with HBr as described in the experiment provided, a student transferred the reaction mixture to a separatory funnel, rinsed the reaction flask with diethyl ether, and added the ether rinses to the separatory funnel. The student then added sodium bicarbonate to the separatory funnel. Extremely vigorous bubbling occurred. What did the student do wrong
Answer:
Explanation:
Because of the acid-base reaction, as sodium bicarbonate is introduced to the separatory funnel, the additional or unreacted HBr reacts vigorously to yield CO2 gas, which exits the separatory funnel together with any dissolved compound(s) in the ether layer. This is due to a wrong and incorrect selection of the solvent mixture and the addition of sodium bicarbonate to an acidic solution.
Nothing to do with it until it has leaked out of the separatory funnel. Even then, the student may separate the components from the remaining reaction mixture by washing the ether coating layer several times with brine water, then running it into a dry sodium sulfate bed and evaporating solvent ether under decreased pressure.
7. Which object would be the least dense in a tub of
water?
A. Golf ball
B. Glass marble
C. Rubber ball
D. Beach ball
Why do we have a leap year in our calendar every four years?
Answer:
every 4 years we add an extra day, February 29th, to our calendars these extra days called leave days help synchronize our human created calendars with us orbit around the sun and the actual passing of our seasons
How many stoms of oxygen in 4 molecules of HNO,?
Answer: the answer is a
How many grams of potassium carbonate are needed to make 300ml of a 4.5M solution?
Answer:
186.3g
Explanation:
4.5moles of K₂CO₃ is in 1000ml
? moles of K₂CO₃ is in 300 ml
(4.5 × 300)/ 1000 = 1.35 moles of K₂CO₃
1 mole of K₂CO₃ = (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138g
1.35 moles of K₂CO₃ = ?
= (1.35 × 138)/1 = 186.3g
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
Answer:
[tex][SO_3]=0.25M[/tex]
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
[tex][SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}[/tex]
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
[tex][SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\[/tex]
[tex][SO_3]=0.25M[/tex]
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State the methods of preparing salts
Answer:
Preparation of Salts by The Action of An Acid Upon a Metal.
Preparation of Salts by Double Decomposition.
Preparation of Salts by Neutralization.
Using A Soluble Base (alkali)
(b). Using An Insoluble Base.
Preparation of Salts by the Action of An Acid on The Trioxocarbonate (IV) of A Metal.
Explanation:
How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?
Answer:25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
Explanation:
24.3 2 An artifact classified as seeds, found in a site at Newlands Cross, Ireland, is found to have a 14C radioactivity of 0.103 counts per second per gram of carbon. If living carbon-containing objects have an activity of 0.255 counts per second per gram of carbon, estimate the age of the artifact?
The half-life of 14C is 5730 years.
______ years
Answer:
Age ≅ 7500 years
Explanation:
All radioactive decay is 1st order kinetics and described by the expression
A = A₀e^-kt => t = ln(A/A₀) / -k
k = 0.693 / t(half life) = (0.693 / 5730)yrs⁻¹ = 1.21 x 10⁻⁴ yrs⁻¹
t = Age = [ln(0.103/0.255) / - 1.21 x 10⁻⁴] yrs = 7500 years
Genes influence an organism's traits by coding for:
A
Cells
Answer:
First, the protein may be a structural protein, contributing to the physical properties of cells or organisms. ... Second, the protein may be an enzyme that catalyzes one of the chemical reactions of the cell. Therefore, by coding for proteins, genes determine two important facets of biological structure and function.
Explanation:
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is scandium a transition metal?
Answer:no
Explanation:
Answer:
Scandium is a transition metal
Explanation:
Rubbing alcohol and vegetable oil soluble on insoluble?
How many grams are there in 1.8055 x 10^25 molecules of sodium sulfate? Hint: Convert to moles first
Answer:
4258.82 g of Na₂SO₄
Explanation:
From the question given above, the following data were obtained;
Number of molecules of Na₂SO₄ = 1.8055x10²⁵ molecules.
Number of mole of Na₂SO₄ =?
From Avogadro's hypothesis,
6.02×10²³ molecules = 1 mole
Therefore,
6.02×10²³ molecules = 1 mole of Na₂SO₄
Next, we shall determine the mass of 1 mole of Na₂SO₄. This can be obtained as follow:
1 mole of Na₂SO₄ = (23×2) + 32 + (16×4)
= 46 + 32 + 64
= 142 g
Thus,
6.02×10²³ molecules = 142 g of Na₂SO₄
Finally, we shall determine the mass of Na₂SO₄ that contains 1.8055x10²⁵ molecules. This can be obtained as follow:
6.02×10²³ molecules = 142 g of Na₂SO₄
Therefore,
1.8055x10²⁵ molecules
= (1.8055x10²⁵ × 142) / 6.02×10²³
= 4258.82 g of Na₂SO₄
Thus, 4258.82 g of Na₂SO₄ contains 1.8055x10²⁵ molecules
How will an increase in wind speed affect soil erosion?
O A Soil erosion will increase.
O B. Soil erosion will decrease.
O C. Soil erosion will completely stop.
OD Soil erosion will remain the same.
Answer:
A. Soil erosion will increase.
Option A is correct. Increase in wind speed will increase the rate of erosion.
The process of eroding, transferring, and depositing tiny particles as well as nutrients from the top layer of soil is known as soil erosion by wind.
In arid and semiarid regions, wind-driven soil erosion is a serious issue that impedes the environmental sustainability of animal husbandry and agriculture and threatens ecological security.
A wind that is blowing at 30 mph will erode at a rate that is more than 3 times faster than a wind that is blowing at 20 mph. As soil moisture rises, wind erosion declines. For instance, dry soil erodes approximately 1.3 times more quickly than soil with just enough moisture to support plant life.
Learn more about soil erosion;
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8. Which of the following elemental gases would have properties closest to an ideal gas?
a. hydrogen
argon
b. helium
d. fluorine
Answer:
Helium
Explanation:
Please tell me if it was right
How much does REAL carbon fiber cost ? lets say as big as a piece of paper
Answer:
Today, the average total production cost of “standard modulus” carbon fiber is in the range of $7-9 per pound.
When 16.35 moles of SI reacts with 11.26 moles of N2, how many moles of SI3N4 are formed
Answer:
5.450 mol Si₃N₄
Explanation:
Step 1: Write the balanced equation
3 Si + 2 N₂ ⇒ Si₃N₄
Step 2: Establish the theoretical molar ratio between the reactants
The theoretical molar ratio of Si to N₂ is 3:2 = 1.5:1.
Step 3: Establish the experimental molar ratio between the reactants
The experimental molar ratio of Si to N₂ is 16.35:11.26 = 1.45:1. Comparing both molar ratios, we can see that Si is the limiting reactant.
Step 4: Calculate the moles of Si₃N₄ produced from 16.35 moles of Si
The molar ratio of Si to Si₃N₄ is 3:1.
16.35 mol Si × 1 mol Si₃N₄/3 mol Si = 5.450 mol Si₃N₄
a) How do you prepare %3 (w/v) Na2CO3 solution from Na2CO3⸱2H2O? (15p) Na2CO3 MW=106 g/mol
Answer:
4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL
Explanation:
A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.
Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:
106g/mol + 2*MW H2O
106g/mol + 2*18g/mol = 142g/mol
That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:
3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =
4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL