Answer:
I believe it is boron.
Explanation:
The vapor from elemental mercury metal is a neurotoxin. Most dentists have levels of mercury in their urine of 10 micrograms per liter (1 microgram is 10^-6 grams), which is about twice that
of their patients. The legal safe maximum exposure limit for mercury in urine is 4.053 x 10^17 atoms of mercury per liter of urine. Does this level mean that the average dentist is at risk for
mercury poisoning? In other words, is this amount, 10 micrograms per liter of urine, above or below the legal safe limit? Show your work and justify your answer.
Answer: 10 micrograms per liter of urine is below the legal safe limit and the dentist is not at risk for mercury poisoning.
Explanation:
To calculate the number of moles, we use the formula:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Given mass of mercury = [tex]10\mu g=10^{-5}g[/tex] (Conversion factor: [tex]1\mu g=10^{-6}g[/tex] )
Molar mass of mercury = 200.6 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of mercury}=\frac{10^{-5}g}{200.6g/mol}\\\\\text{Moles of mercury}=4.9\times 10^{-8}mol[/tex]
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms
So, [tex]4.9\times 10^{-8}[/tex] moles of mercury will contain = [tex]\frac{6.022\times 10^{23}}{1}\times 4.9\times 10^{-8}=2.95\times 10^{16}[/tex] number of atoms.
We are given:
Legal safe limit for mercury in urine = [tex]4.053\times 10^{17}[/tex] atoms
Calculated amount of mercury in urine = [tex]2.95\times 10^{16}[/tex] atoms
As, the calculated amount of mercury in urine is less than the legal safe limit of mercury in urine. So, the average dentist is not at risk for mercury poisoning.
Hence, 10 micrograms per liter of urine is below the legal safe limit and the dentist is not at risk for mercury poisoning.
Identify the following as a weak or strong acid or base: (a) NH3, (b) H3PO4, (c) LiOH, (d) HCOOH (formic acid), (e) H2SO4, (f) HF, (g) Ba(OH)2.
Answer:
Explained
Explanation:
a) NH3- Weak Base
b) H3PO4- Weak Acid
c) LiOH- Weak Base
d) HCOOH- Weak Acid
e) H2SO4- Strong Acid
f) HF- Weak Acid
g) Ba(OH)2- Strong Base
Which two elements have similar physical and chemical properties?
Explanation:
Elements in the same group have same number of valence electrons. And we know, the elements which have same number of valence electrons, have similar physical and chemical properties. Hence, the elements in the same group have similar physical and chemical properties.
Which landform is created by moving water?
A- Cirques
B- Dunes
C- Fjord
D- Valleys
10 points!
What does Ca stand for in Italy?
Give two examples. How does predation affect an ecosystem?
Answer:
It affects the amount of a species that may be hunted by said predator, controlled you can say.
It can cause evolution in prey that causes them to not be caught, Darwins theory in a way.
Explanation:
Explanation:
At the level of the community, predation reduces the number of individuals in the prey population. The best example of predation involve carnivorous interactions, in which one animal consumes another, Think of wolves hunting moose, owls hunting mice, or shrews hunting worms and insects.
Jill has a gel pen the gel pen has a mass of 8g and a valine of 2cm3 what is the density
Answer:
The answer is 4 g/cm³Explanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\ [/tex]
From the question
mass = 8 g
volume = 2 cm³
We have
[tex]density = \frac{8}{2} \\ [/tex]
We have the final answer as
4 g/cm³Hope this helps you
Two common materials we encounter in our daily lives are radiator coolant in automobiles and dry ice. The coolant in an automobile radiator, a solution of antifreeze (a glycol such as ethylene or propylene glycol) and water, freezes at 39°C and boils at 110°C. The temperature of the dry ice (solid carbon dioxide) used in some ice cream vending carts is –78°C. Convert the boiling point of radiator coolant (110°C) to degrees Fahrenheit. Convert the temperature of dry ice (–78°C) to the Kelvin scale.
Answer:
See explanation
Explanation:
Using the formula
°C = (F-32) × 5/9
Where;
°C = temperature in degrees centigrade
F= temperature in Fahrenheit
F= (9/5 ×°C) +32
F= (9/5 × 110) + 32
F= 230°F
To convert -78°C to Kelvin
-78°C + 273 = 195 K
How many moles of H2 can be formed if a 4.71 g sample of Mg reacts with excess HCl?
Answer:
0.2 mole of H2.
Explanation:
We'll begin by calculating the number of mole in 4.71 g of magnesium (Mg). This can be obtained as follow:
Mass of Mg = 4.71 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /Molar mass
Mole of Mg = 4.71 /24
Mole of Mg = 0.2 mole
Next, we shall write the balanced equation for the reaction.
This is illustrated below:
Mg + 2HCl —> MgCl2 + H2
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Finally, we shall determine the number of mole of H2 produced from the reaction. This can be obtained as illustrated below:
From the balanced equation above,
1 mole of Mg reacted to produce 1 mole of H2.
Therefore, 0.2 mole of Mg will also react to produce 0.2 mole of H2.
Thus, 0.2 mole of H2 was obtained from the reaction.
The number of moles of H₂ that can be formed if a 4.71 g sample of Mg reacts with excess HCl is 0.19625 moles
Let's represent the reaction with a chemical equation
Mg + HCl → MgCl₂ + H₂
Mg + 2HCl → MgCl₂ + H₂
HCl is in excess according to the question. This means Mg is the limiting reagent and therefore determine the amount of product.
Therefore,
24 g of Mg gives 2g of H₂
4.71 g of Mg will give ? H₂
cross multiply
mass of H₂ = 4.71 × 2 / 24
mass of H₂ = 9.42 / 24 = 0.3925 g
moles of H₂ = mass / molar mass
moles of H₂ = 0.3925 / 2
moles of H₂ = 0.19625 moles
read more: https://brainly.com/question/18757768?referrer=searchResults
Select all statements that are correct:____
A. The vapor pressure of a solution containing a nonvolatile solute is directly proportional to the mole fraction of the solvent.
B. The vapor pressure of a solution containing a nonvolatile solute is directly proportional to the mole fraction of the solute.
C. For molecules of similar size, the stronger the intermolecular forces, the lower the vapor pressure at room temperature.
D. For molecules of similar size, the stronger the intermolecular forces, the lower the vapor pressure and the higher the boiling point.
Answer:
A, C and D are correct.
Explanation:
Hello.
In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:
[tex]P_{vap}^{solution}=x_{solvent}P_{solvent}[/tex]
Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.
Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.
Best regards.
calculate the density of an object with a mass of 25g and a volume of 10ml
Answer:
The answer is 2.50 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} \\[/tex]
From the question
mass = 25 g
volume = 10 mL
We have
[tex]density = \frac{25}{10} = \frac{5}{2} \\ [/tex]
We have the final answer as
2.50 g/mLHope this helps you
A sample of an ideal gas has a volume of 3.25 L at 12.80 °C and 1.50 atm. What is the volume of the gas at 24.40 °C and
0.996 atm?
Answer:
You have got the Combined Gas Equation for an Ideal Gas, which holds that
P
1
V
1
T
1
=
P
2
V
2
T
2
V
2
=
P
1
×
V
1
×
T
2
T
1
×
P
2
...
Explanation:
Ling built a circuit with three light bulbs. She observed that all three light bulbs glowed.
Then she removed one light bulb, and the others stopped glowing. She explained that the
other light bulbs had burned out. How could Ling test her explanation? *
(1 Point)
Replace the battery and observe whether the two lights come back on.
Take another light bulb out of the circuit and observe whether the last one comes on.
Wait another 10 min to see whether the light bulbs suddenly start glowing again.
Put the light bulb that was removed back into the circuit and observe whether the other lights come on.
21
Answer:
Put the light bulb that was removed back into the circuit and observe whether the other lights come on.
Explanation:
If all three bulbs initially glowed and she supposes that after removing the first bulb, the others stopped glowing because they burned out, then she must replace the original bulb and see if the other two bulbs will glow.
If all three bulbs don't glow after replacing the first one, then all the bulbs may have probably burned out.
The amount of space a material occupies is its
Answer:
—MASS is the amount of matter that makes up something. - VOLUME _ is the amount of space that a material takes up.
Explanation:
what is rpxiaiteon uncrambled
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabatically and reversibly. If the decrease in pressure is also measured, we can use it to infer the value of γ = Cp/Cv and hence, by combining the two values, deduce the constant-pressure heat capacity. A fluorocarbon gas was allowed to expand reversibly and adiabatically to twice its volume; as a result, the temperature fell from 298.15 K to 248.44 K and its pressure fell from 202.94 kPa to 81.840 kPa. Evaluate Cp
Answer:
The value is [tex]C_p = 42. 8 J/K\cdot mol[/tex]
Explanation:
From the question we are told that
[tex]\gamma = \frac{C_p }{C_v}[/tex]
The initial volume of the fluorocarbon gas is [tex]V_1 = V[/tex]
The final volume of the fluorocarbon gas is[tex]V_2 = 2V[/tex]
The initial temperature of the fluorocarbon gas is [tex]T_1 = 298.15 K[/tex]
The final temperature of the fluorocarbon gas is [tex]T_2 = 248.44 K[/tex]
The initial pressure is [tex]P_1 = 202.94\ kPa[/tex]
The final pressure is [tex]P_2 = 81.840\ kPa[/tex]
Generally the equation for adiabatically reversible expansion is mathematically represented as
[tex]T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }[/tex]
Here R is the ideal gas constant with the value
[tex]R = 8.314\ J/K \cdot mol[/tex]
So
[tex]248.44 = 298.15 * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }[/tex]
=> [tex]C_v = 31.54 J/K\cdot mol[/tex]
Generally adiabatic reversible expansion can also be mathematically expressed as
[tex]P_2 V_2^{\gamma} = P_1 V_1^{\gamma}[/tex]
=>[tex] [ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}[/tex]
=> [tex]2^{\gamma} = 2.56[/tex]
=> [tex]\gamma = 1.356[/tex]
So
[tex]\gamma = \frac{C_p}{C_v} \equiv 1.356 = \frac{C_p}{31.54}[/tex]
=> [tex]C_p = 42. 8 J/K\cdot mol[/tex]
Propose a structure for an aromatic hydrocarbon, C9H12, that can form two C9H11Cl products on substitution of a hydrogen on the aromatic ring with chlorine. You do not have to consider stereochemistry. In cases where there is more than one answer, just give one. Ignore the ortho, para-directing effects of the alkyl groups in answering this question. Consider only the number of nonequivalent hydrogens on the aromatic ring.
Answer:
Explanation:
From the given study,
the molecular formula of the given aromatic compound = C9H12
the first process we suppose to carry out is to calculate the double bond equivalence.
So, the double bond equivalence [tex]DBE = {N_C+1-(\dfrac{(N_H+N_{Cl})-N_H}{2}) }[/tex]
[tex]DBE = {9+1-(\dfrac{(12+0)-0}{2}) }[/tex]
DBE = 4
Hence, the aromatic compound possesses 4 double bonds, this signifies that the might be a presence of benzene ring.
The substitutional products of the aromatic hydrocarbon C9H12 can be seen in the attached file below.
In the anaerobic fermentation of grain, the yeast Saccharomyces cerevisiae digests glucose from plants to form the products ethanol and propenoic acid by the following reactions: Reaction 1: C6H12O6 2 C2H5OH + 2 CO2 Reaction 2: C6H12O6 2 C2H3CO2H + 2 H2O In an open flow reactor 4000 kg of a 12% glucose-water solution flows in. During fermentation, 120 kg of carbon dioxide is produced together with 90 kg of unreacted glucose. What are the weight percents of ethyl alcohol and propenoic acid that exit the broth? Assume that none of the glucose is assimilated into the bacteria.
Answer:
Explanation:
The first step in order to solve this particular question is to make sure that the two reactions given in the question is balanced. Therefore, we have;
Reaction 1: C6H12O6 -----------------------------> 2 C2H5OH + 2 CO2.
Reaction 2: C6H12O6 ------------------------------> 2 C2H3CO2H + 2 H2O.
Next, we determine the number of moles of water and that of glucose. Recall that we are given from the question that the open flow reactor = 4000 kg of a 12% glucose-water solution flows in that is to say the percentage for water is [100% - 12% = 88%]. Also, the molar mass of water = H₂O = 18 kg/kmol and that for glucose =180 kg/kmol.
Number of moles of water = (4000 kg × 88%) ÷ 18 = 195.6 kmol.
Number of moles of glucose= (4000kg × 12%) ÷ 180 = 2.67 kmol.
Next thing to do is to determine the number of moles in the unreacted glucose . Therefore, the 90 kg of unreacted glucose ÷ 180kg/kmol = 0.5 kmol.
So, we have that During fermentation, 120 kg of carbon dioxide is produced. Thus, the number of kmol = 120kg÷ 44kg/mol = 2.73kmol.
For reaction 1, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.73kmol / 2 moles of CO₂ = 1.365 kmol.
For reaction 2, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.67kmol - 0.5 kmol - 1.365kmol = 0.805 kmol.
For both reaction, the kmol for outflow of glucose = 2.73 kmol.
Also, 2 × 0.805 + 1.365 × 0 = 1.61kmol.
Hence, 195.6 kmol + 1.61 =197.21 kmol.
The mass of ethanol = 46.1 kg/kmol × 2.73 kmol = 125.853 kg.
The weight percent of ethanol has been 29.72%, and the percent mass of propionic acid has been 47.79%.
The balanced chemical equations of the reactions have been:
[tex]\rm C_6H_1_2O_6\;\rightarrow\;2\;C_2H_5OH\;+\;2\;CO_2[/tex]
[tex]\rm C_6H_1_2O_6\;\rightarrow\;2\;C_2H_3CO_2H\;+\;2\;H_2O[/tex]
The solution has consisted of 12% glucose.
Water in the solution = 88%
The mass of the solution = 4000 kg.
The moles of glucose = 12% of 4000 kg
Moles of glucose = [tex]\rm \dfrac{12}{100}\;\times\;4000\;\times\;\dfrac{1}{180\;g/mol}[/tex]
Moles of glucose = 2.67 kmol.
Moles of water = 88 % of 4000 kg
Moles of water = [tex]\rm \dfrac{88}{100}\;\times\;4000\;\times\;\dfrac{1}{18\;g/mol}[/tex]
Moles of water = 195.6 kmol.
The unreacted glucose in the mixture = 90 kg
Moles of unreacted glucose:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of unreacted glucose = [tex]\rm \dfrac{90\;\times\;1000\;g}{180\;g/mol}[/tex]
Moles of unreacted glucose = 0.5 kmol
The mass of carbon dioxide produced = 120 kg.
Moles of carbon dioxide produced = [tex]\rm \dfrac{120\;\times\;1000\;g}{44\;g/mol}[/tex]
Moles of carbon dioxide produced = 2.73 kmol
Since 1 mole of glucose produces 2 moles of carbon dioxide.
2.73 kmol of carbon dioxide has been produced from 1.365 kmol of glucose.
The moles of ethanol produced by reaction 1 = 2 moles/ mole glucose.
The glucose present has been 2.67 kmol.
The ethanol produced = 5.34 kmol.
Moles of propionic acid produced = 5.34 kmol.
The mass of 5.34 kmol ethanol = Moles × molecular weight
The mass of ethanol produced = 5.34 × 1000 × 46.07g/mol
The mass of ethanol produced = 246.0138 kg.
The mass of propionic acid produced = 5.34 × 1000 × 74.08 g/mol
The mass of propionic acid produced = 395.5872 kg.
The mass of water produced = 5.34 × 1000 × 18 g/mol
The mass of water produced = 96.12 kg.
The remained glucose = 90 kg
The total mass in the reactor:
= Mass of glucose + water + propionic acid + ethanol
= 90 + 96.12 + 395.5872 + 246.0138 kg
= 827.721 kg.
% Mass of ethanol = [tex]\rm \dfrac{Mass\;of\;ethanol}{Total\;mass}\;\times\;100[/tex]
% Mass of ethanol = [tex]\rm \dfrac{246.0138}{827.721}\;\times\;100[/tex]
% Mass of ethanol = 29.72%
% Mass of Propionic acid = [tex]\rm \dfrac{Mass\;of\;propionic\;acid}{Total\;mass}\;\times\;100[/tex]
% Mass of Propionic acid = [tex]\rm \dfrac{395.5872}{827.721}\;\times\;100[/tex]
% Mass of Propionic acid = 47.79 %.
The weight percent of ethanol has been 29.72%, and the percent mass of propionic acid has been 47.79%.
For more information about weight percent, refer to the link:
https://brainly.com/question/18204076
Which ONE of the following has the most mass (assume they are all of the same volume)?
A block of styrofoam.
b. A block of copper.
c. A box of feathers.
d. An empty cardboard box.
Answer:
the answer is b
Explanation:
the Na2O valence is
This is something easy can someone please finish this one. i'll give brainliest.
Answer:
1. C maybe
2. C
3. B
4. A
5. A
6. B
7. C
8. A
9. C
10. A
You have two pieces of metal. Metal A has a higher temperature than metal B. What does this tell you about them?
Group of answer choices
There is more light energy in metal B.
The molecules in metal A are vibrating faster.
Metal B has more heat energy.
Metal A has more gravitational potential energy. Bl
Answer:
The molecules in metal A are vibrating faster
Which of the following correctly describes
a solid?
a. The particles do not move at all.
b. The particles are closely locked in pigsition and can vibrate in place.
c. The particles are free to
d. move about independently. colliding frequently
The particles are closely packed but have enough energy to slide past each
other.
Answer:A
Explanation: solids are locked into place B would be a liquid C would be a gas
An igneous rock has a coarse texture and is dark in color. How else can this rock be accurately described?
small crystals
cooled quickly
produced from magma
formed at Earth’s surface
Answer:
igneous rocks are produced from magma
Explanation:
Answer:
produced form magma
Explanation:
I took the test and got it right
4. Subatomic particles that are located in the nucleus and have no charge (neutral) are called
(10 Points)
protons
electrons
neutrons
nucleus
When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay: HCOOH(g) →CO2(g) + H2 (g) The rate of reaction is monitored by measuring the total pressure in the reaction container. Time (s) . . . P (torr) 0 . . . . . . . . . 220 50 . . . . . . . . 324 100 . . . . . . . 379 150 . . . . . . . 408 200 . . . . . . . 423 250 . . . . . . . 431 300 . . . . . . . 435 At the start of the reaction (time = 0), only formic acid is present. What is the formic acid pressure (in torr) when the total pressure is 364? Hint: use Dalton's law of partial pressure and the reaction stoichiometry.
Answer:
[tex]p_{HCOOH}=76torr[/tex]
Explanation:
Hello.
In this case, since the total pressure at any point of the experiment is written via the Dalton's law:
[tex]p=p_{HCOOH}+p_{CO_2}+p_{H_2}[/tex]
We can also write the partial pressure of formic acid in terms of its initial pressure (220 torr) and the change [tex]x[/tex] as the time goes by:
[tex]p_{HCOOH}=220-x[/tex]
Thus, based on the stoichiometry, since it is a first-order decay and all the stoichiometric coefficients are 1, we can infer that the decrease in the partial pressure of formic acid equals the increase in the partial pressure of both carbon dioxide and hydrogen, therefore we can write:
[tex]p_{CO_2}=p_{H_2}=x[/tex]
In such a way, we write the Dalton's law as shown below:
[tex]p=220torr-x+x+x\\\\p=220torr+x[/tex]
Thus, at the point in which the total pressure is 364 torr, the change is:
[tex]x=364torr-220torr=144torr[/tex]
It means, that the partial pressure of formic acid is:
[tex]p_{HCOOH}=220torr-144torr\\\\p_{HCOOH}=76torr[/tex]
Best regards.
35. How many atoms are in 39.2 moles? (1 mole = 6.02 x 10 23 atoms)
A. 2.36 x 10 25atoms
B.
6.5 x 10 23 atoms
C.
6.5 x 10-23 atoms
D.
236 x 10 23 atoms
STOP
How many significant figures
are in this number?
23,479
KE and PE Math
Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts of potential and kinetic energy the bowling ball has:
as it sits on top of a building that is 40 meters tall.
as it is half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second.
as it is just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second.
Questions
1. Does the bowling ball have more potential energy or kinetic energy as it sit on top of the building? Why?
2. Does the bowling ball have more potential energy or kinetic energy as it is half way through its fall? Why?
3. Does the bowling ball have more potential energy or kinetic energy just before it hits the ground? Why?
4. What is the potential energy of the bowling ball as it sits on top of the building?
5. What is the potential energy of the ball as it is half way through the fall, 20 meters high?
6. What is the kinetic energy of the ball as it is half way through the fall?
What is the kinetic energy of the ball just before it hits the ground?
Answer:
1. Answer: The bowling ball has more potential energy as it sits on top of the building. It does not have any kinetic energy because it is not moving.
2. Answer: The bowling ball has equal amounts of potential and kinetic energy half way through the fall. At the half way point, half of the potential energy has been converted to kinetic energy.
3. Answer: Just before the ball hits the ground, it has more kinetic energy. As it hits the ground the potential energy becomes zero.
4. Answer:
PE=784 J
5. Answer:
PE = 392 J
6. Answer:
KE= 392 J
Also, since the PE and KE are equal at the half way point and PE =392 J, KE = 392 J.
7. What is the kinetic energy of the ball just before it hits the ground?
Answer:
KE=784 J
At first I answered in the comments, but I am able to answer now. I hope this can help
Answer:
The person above has everything, except for 2 I put " kinetic energy since it is in motion, and is picking up speed. It has already used up all of the potential energy " I don't think the potential energy and kinetic energy is equal
Explanation:
In order to listen to the radio you need to use ______________ and _________ waves.
Answer:
ears and hands and listen the quistion
Answer:
horizotal , varitical