Answer:
1) 2.69 * 10²³ PBr₅
2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁
Explanation:
Question 1)
We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.
Phosphorus pentabromide is given by PBr₅.
To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.
To convert from grams to moles, we will find the molar mass of PBr₅.
Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:
[tex](30.974)+5(79.904) = 430.494\text{ g/mol}[/tex]
And since we want to convert from grams to moles, we can write the following ratio:
[tex]\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}[/tex]
Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.
To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.
So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:
[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
In combination, starting with 192 grams of PBr₅, we will acquire:
[tex]\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
Cancel like units:
[tex]\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}[/tex]
Multiply. Hence:
[tex]=2.6858...\times 10^{23}\text{ PBr$_5$}[/tex]
Since the final answer should have three significant digits, our final answer is:
[tex]= 2.69\times 10^{23} \text{ PBr$_5$}[/tex]
So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.
Question 2)
We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.
3.42 kilograms is equivalent to 3420 grams of table sugar.
Again, we can convert from grams to moles, and then from moles to molecules.
First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:
[tex]12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}[/tex]
To cancel units, we can write our ratio as:
[tex]\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}[/tex]
With grams in the denominator.
And by definition:
[tex]\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Combining the two ratios and the starting value, we acquire:
[tex]3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Cancel like units:
[tex]=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}[/tex]
Multiply:
[tex]\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
Rewrite:
[tex]\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
The resulting answer should have three significant digits. Hence:
[tex]=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}[/tex]
So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.
Answer:
2.69×10²³ molecules of PBr₅
6.02×10²⁴ molecules of C₁₂H₂₂O₁₁
Explanation:
To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.
[tex]192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules[/tex]
Now, we know that there are about 2.69×10²³ molecules of PBr₅.
To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.
[tex]3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules[/tex]
Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.
A tank contains oxygen gas at 2.551 atm. What is the pressure in mmHg?
The force exerted on the container by the particles of the matter is called pressure. The tank containing oxygen gas at 2.551 atm will have a pressure of 1939 mm Hg.
What is pressure?Pressure is the property used to estimate the force experienced by the system due to the liquid or the gas held in it. The pressure of the gas can be calculated by the ideal gas and force and area.
The pressure is created due to the collision of the particles of the gases and liquids on the wall perpendicularly. It is estimated in Pascal (Pa) as the standard unit along with atm and mmHg.
It is known that 1 atm = 760 mm Hg
Given,
The pressure of oxygen gas = 2.551 atm
Using the conversion factor the pressure from atm to mm Hg is calculated as,
1 atm = 760 mm Hg
2.551 atm = (2.551 atm × 760 mm Hg) ÷ 1 atm
= 1938.76 mmHg
Therefore, 1939 mm Hg is the pressure of the oxygen gas.
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Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water . Suppose 2.73 g of methane is mixed with 6.7 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
3.8g of H2O are produced
Explanation:
The balanced reaction of the problem is:
CH4 + 2O2 → CO2 + 2H2O
Where 1 mole of CH4 reacts with 2 moles of O2
To solve this question we need to find, as first, the moles of each reactant in order to find limiting reactant. With limiting reactant we can find the moles of H2O produced and its mass as follows:
Moles CH4 - 16.04g/mol-
2.73g * (1mol/16.04g) = 0.170 moles CH4
Moles O2 -32g/mol-
6.7g (1mol/32g) = 0.209 moles O2
For a complete reaction of 0.170 moles of CH4 are needed:
0.170 moles CH4 * (2 mol O2 / 1mol CH4) = 0.340 moles O2
As there are just 0.209 moles of O2, oxygen is limiting reactant
The moles of water produced are:
0.209 moles O2 * (2mol H2O / 2mol O2) = 0.209 moles H2O
Mass water -Molar mass: 18.01g/mol-
0.209 moles H2O * (18.01g/mol) = 3.8g of H2O are produced
Evaluate the exponential expression (−2)6.
A general exponential expression is something like:
A^n
This means that we need to multiply the number A by itself n times.
Using that we will get (-2)^6 = 64
With that definition, we can rewrite:
(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)
So we just need to solve the above expression.
Also, remember the rule of signs:
(-)*(-) = (+)
We will get:
(-2)*(-2)*(-2)*(-2)*(-2)*(-2) = [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]
= 4*4*4 = 16*4 = 64
Then we got:
(-2)^6 = 64
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What is an example of an extensive property
A sentence using the word Compound
Answer:
The air smelled like a compound of diesel and gasoline fumes.
crassify the given quantities into scalar quantity and vetor quantity
Answer:
where is the quantities?
How many protons does Tin have?
A. 50
B. 68
C. 118
Hello There!
Tin has 50 protons.Hope that helps you!
~Just a felicitous girlie
#HaveASplendidDay
[tex]SilentNature[/tex]
In an ELISA, the compound 4-chloro-1-naphthol is used because:_______
a. it turns color in the presence of an enzyme that is bound to the secondary antibody
b. it helps the primary antibody bind to the protein
c. it helps the secondary antibody to bind to the protein
d. all of the choices
Answer:
a. It turns color in the presence of an enzyme that us bound to the secondary antibody.
Explanation:
The compound chloronapthenel is used in the reaction because it changes the color in the presence of an enzyme. It is strong organic compound which is used in biochemical processes.
what is the difference between Absorption and adsorption
Hydrogengasand oxygengas react to form water vapor. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .
The question is incomplete. The complete question is :
Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]
= 11.0 mol
g Identify the process in which the entropy increases. Group of answer choices a decrease in the number of moles of a gas during a chemical reaction the phase transition from a gas to a liquid the phase transition from a solid to a gas freezing water
Answer:
phase transition from a solid to a gas
Explanation:
Entropy refers to the degree of disorderliness in a system. The more disorderly a system is, the greater the entropy of the system.
Decrease in the number of moles of a gas decreases the entropy of the system. Similarly, the entropy of solids is less than that of liquids. The entropy of liquids is less than that of gases.
Therefore, a change of phase from solid to gas represents an increase in entropy of the system.
The average temperature at the South Pole In January is - 35.4 °C.
Convert this temperature to degrees Fahrenheit. Round your answer to 3 significant digits.
°F
Answer:
-31.72°F
Explanation:
(-35.4°C × 9/5) + 32 = -31.72°F
The average temperature at the South Pole In January is - 35.4 °C. This temperature in Fahrenheit is -31.72 °F
To convert Celsius to Fahrenheit, you can use the formula:
°F = (°C × 9/5) + 32
Let's calculate the temperature at the South Pole in degrees Fahrenheit:
°F = (-35.4 × 9/5) + 32
°F = (-63.72) + 32
°F = -31.72
Rounding to three significant digits, the temperature at the South Pole in degrees Fahrenheit is approximately -31.7 °F. The negative sign indicates that the temperature is below the freezing point in both Celsius and Fahrenheit scales. The South Pole experiences freezing temperatures, as it is located near the Earth's southernmost point and experiences long periods of darkness during January.
Hence, the temperature in Fahrenheit is -31.7 °F.
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Radon-220 undergoes alpha decay with a half-life of 55.6 s.?
Assume there are 16,000 atoms present initially and calculate how many atoms will be present at 0 s, 55.6 s, 111.2 s, 166.8 s, 222.4 s, and 278.0 s (all multiples of the half-life). Express your answers as integers separated by commas.
Calculate how many atoms are present at 50 s, 100 s, and 200 s (not multiples of the half-life).
The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.
The equation below gives the number of atoms present at time t
[tex]N=Noe^-kt[/tex]
N = Number of atoms present at time t
No = Number of atoms initially present
k = decay constant
t = time taken
Given that;
t1/2 = 0.693/k
where t1/2 = half life
k = 0.693/t1/2
k = 0.693/ 55.6 s
k = 0.0125 s-1
Substituting values;
N = 16,000 e^-0.0125(0)
N = 16,000 atoms
At 50 s
N = 16,000 e^-0.0125(50)
= 8564 atoms
At 100 s
N = 16,000 e^-0.0125(100)
= 4584 atoms
At 200 s
N = 16,000 e^-0.0125(200)
= 1313 atoms
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For a colligative property such as freezing point depression, :________
a) the charge on the particle affects the property.
b) only the molar mass of the particle matters, not the number of particles.
c) the size of the particle affects the property but not the charge.
d) the number of particles matter but not what they are.
what is food nutrients
Answer:
Nutrients arw compounds in foods essential to life and heath
Answer: In simple terms nutrients are the energy that you get from food certain foods give more nutrients and others give close to none. That is what nutrients in your food is
Explanation:
Consider an equilibrium (K1) that is established after 10 mL of compound A and 10 mL of compound B are mixed. Now, imagine the equilibrium (K2) where 1 mL of compound A is added to 100 mL of compound B. How are K1 and K2 related algebraically (read this question VERY carefully, at least one more time)?
K1 and K2 are related algebraically because once the values are inserted into the equilibrium equation, both equations will yield a denominator of 100.
Explain the general properties of aqueous solution based on the following support your answer with examples for each case
1. Electrolytes versus non-electrolyte
2. Precipitation reaction
3. Solubility
An aqueous solution, based on its name, is a water based solution, such that the solvent is water. In such solution, ionic compounds when dissolved, tend to dissociate into the constituent ions, for example, sodium chloride, NaCl forms an aqueous solution of sodium, Na⁺ and chloride, Cl⁻
1) Therefore, aqueous solutions are good electrolyte when ionic, and are therefore, good electrolytes which conduct electricity compared to solids that form non-electrolyte
2) A precipitation reaction is the insoluble product formed by the combination of cations and anions to form ionic solids that are insoluble
3) Aqueous solutions are made with substance that are soluble in water
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How is a light bulb related to radiation?
A. Light bulbs do not have anything to do with radiation.
B. A light bulb emits radiation when its filament is burned out and can longer emit light. This makes it important to remove dead lightbulbs as soon as they wear out.
C. As light is emitted from the filament, the energy in the metal is replaced as lightbulbs absorb background radiation. This prevents the filament from burning out quickly but the radiation cannot be used as power so electricity is still required for the lightbulb to work.
D. The light emitted by a light bulb is a form of radiation that occurs when the filament heats up and its thermal emission gains enough energy to move into the visible spectrum.
Answer:
As light is emitted from the filament, the energy in the metal is replaced as lightbulbs absorb background radiation. This prevents the filament from burning out quickly but the radiation cannot be used as power so electricity is still required for the lightbulb to work.
As discussed in class, the Fischer esterification reactants and products are at equilibrium. How was the equilibrium of the reaction that you performed shifted towards the products
Answer:
See explanation
Explanation:
The particular reactants in the Fischer esterification reaction were not stated.
Generally, a Fischer esterification is a reaction that proceeds as follows;
RCOOH + R'OH ⇄RCOOR' + H2O
This reaction occurs in the presence of an acid catalyst.
We can shift the equilibrium of this reaction towards the products side in two ways;
I) use of a large excess of either of the reactants
ii) removal of one of the products as it is formed.
Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.
Calculate the molality of each of the following solutions: (a) 36.2 g of sucrose (C12H22O11) in 323 g of water, m (b) 8.63 moles of ethylene glycol (C2H6O2) in 1889 g of water.
Answer:
(a) m = 0.327 m.
(b) m = 4.57 m.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly considering the fact that the molality is computed by dividing the moles of solute by the kilograms of solvent, in this case water; in such a way, we proceed as follows:
(a) We firstly calculate the moles of 36.2 grams of sucrose as its molar mass is 342.3 g/mol:
[tex]\frac{36.2g}{342.3g/mol} =0.106mol[/tex]
Next, the kilograms of water in this case are 0.323 kg so that the molality will be:
[tex]m=\frac{0.106mol}{0.323kg}\\\\m=0.327m[/tex]
(b) In this case, we directly realize that the kilograms of water are now 1.889 kg so that the molality will be:
[tex]m=\frac{8.63mol}{1.889kg}=4.57m[/tex]
Clearly, the both of them in molal, m, units.
Regards!
choose isomer of hexanoic acid?
A) - penthylformiate
B) _ izopropyleacetate
C) _ methylpropanoate
D) _ a- Methylbutanoic acid
Answer:
THR ANSWER IS C) _ methylpropanoate
What is the difference between elimination and substitution reaction
Identify the key factors that will determine if a reaction undergoes elimination or substitution mechanism.
Use the following reagents to determine the type of reaction pathway expected and determine the products in each reaction.
a. Tert BuO- in tertbutanol and chlorobutane
b. KOH in water and bromobutane
c. NaI in acetone and bromobutane
Write a conclusion of no more than two paragraphs to summarize your results
Answer:
a) E2
b) SN2
c) SN2
Explanation:
A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.
We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.
the ability for carbon to form long chain or rings is
1. Consider the following thermochemical reaction for kerosene:
2 C12H26(l) + 37 O2(g) 24 CO2(g) + 26 H2O(l) + 15,026 kJ
(a) When 21.3 g of CO2 are made, how much heat is released?
(b) If 500.00 kJ of heat are released by the reaction, how grams of C12H26 must have been consumed ?
(c) If this reaction were being used to generate heat, how many grams of C12H26 would have to be reacted to generate
enough heat to raise the temperature of 750g of liquid water from 10oC to 90oC?
2. Consider the reaction: NaNO3(s) + H2SO4(l) → NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ
How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
3. What is the enthalpy change when 49.4 mL of 0.430 M sulfuric acid reacts with 23.3 mL of 0.309 M potassium
hydroxide?
3.
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l) ΔH° = –111.6 kJ/mol
do you have the specific heat for part 2?
What volume (in liters) of a solution contains 0.14 mol of KCl?
1.8 M KCl
Express your answer using two significant figures.
Answer:
[tex]\boxed {\boxed {\sf 0.078 \ L }}[/tex]
Explanation:
We are asked to find the volume of a solution given the moles of solute and molarity.
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.
moles of solute = 0.14 mol KCl molarity= 1.8 mol KCl/ Lliters of solution=xSubstitute these values/variables into the formula.
[tex]1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}[/tex]
We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.
[tex]\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}[/tex]
[tex]1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl[/tex]
[tex]1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl[/tex]
Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.
[tex]\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]
[tex]x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]
The units of moles of potassium chloride cancel.
[tex]x= \frac{0.14 }{1.8 L}[/tex]
[tex]x=0.07777777778 \ L[/tex]
The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.
[tex]x \approx 0.078 \ L[/tex]
There are approximately 0.078 liters of solution.
If you add a solution of NaOH to a solution of H₂CO₃, two reactions occur, one after the other. Write the chemical equations for these two reactions. (Hint: NaOH dissociate into Na+ and OH-, and the hydroxide ion is the actual base).
We have a solution of NaOH and H₂CO₃
First, NaOH will dissociate into Na⁺ and OH⁻ ions
The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃
The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule
If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃
Consider the reaction below. How much heat is absorbed if 5.00 moles of nitrogen react
with excess oxygen?
2 N2 (8) + O2(g) → 2 N20 (8) AHrxn- +163.2 kJ
Explanation:
The given chemical reaction is:
[tex]2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ[/tex]
When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.
When 5.00 mol of nitrogen requires how much energy?
[tex]5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ[/tex]
Hence, the answer is 408 kJ of heat energy is required.
2. The reaction of a triglyceride with methanol in the presence of a strong base to form
methyl esters and glycerol is called
O A. transesterification.
O B. saponification.
O C. ester formation.
O D. dehydration condensation.
Answer:
The answer it's A. transesterification
Calculate the enthalpy change for the reaction of hydrogen and chlorine using the bond energies below.
"BOND." " BOND ENERGY "
H-H. 436
CL-CL. 242
H-CL. 431
Answer:
final-intial temperature= enthalpy change
What did Millikan discover
Answer:
Robert Millikan was a physicist who discovered the elementary charge of an electron using the oil-drop experiment
Answer:
the mass of an electron using the Oil-Drop experiment.
Explanation:
