Answer:
[tex]P_{2} = 3.259\,atm[/tex]
Explanation:
Let suppose that air inside the tire behaves ideally. The equation of state for ideal gases is:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]
As tire can be modelled as a closed and rigid container, there are no changes in volume and number of moles. Hence, the following relationship is constructed:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]
The final pressure is:
[tex]P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}[/tex]
[tex]P_{2} = \frac{298.15\,K}{278.15\,K}\cdot (3.04\,atm)[/tex]
[tex]P_{2} = 3.259\,atm[/tex]
Answer:
P2=3,26 atm
Explanation:
Considering that this gas behaves like an ideal gas, the equation used in noble gases is:
PxV = nxrxT
p is pressure, v is volume, n is number of moles, r is a constant of noble gases whose value is 0.082x10exp23 as long as you use it in the equations it has the same value (that is why it is called constant) and t would be the temperature.
Because v, r and n are values that did not change throughout the reaction of the gas, we are only going to take into account the p and t that did change throughout the reaction and that is why there is an initial and final pressure and a final and initial temperature.
So:
P1 / T1 = P2 / T2 (where value 1 refers to start, and 2 to end)
Regarding this equation, since we want to know the value of P2, which is the FINAL pressure of the gas, that is, what pressure did it achieve with the reaction, the equation that we would use would be solving for P2:
P2 = (T2 / T1) xP1 = 3.26 atm
Percent yield for sodium bromide if 3.87 moles of iron 3 bromide yielded 10.4 moles of sodium bromide
Answer:
89.57 %
Explanation:
Don't have one just thought id help <3
Answer: 89.6
Explanation:
A balloon containing helium gas expands from 230
mL to 860 mL as more helium is added. What was
the initial quantity of helium present if the
expanded balloon contains 3.8 x 10-4 mol,
assuming constant temperature and pressure?
Which of the variables are known?
Answer:
1.0 *10^(-4) mol
Explanation:
For gases:
n1/n2 = V1/V2
n1/3.8*10^(-4) mol = 230 mL/ 860 mL
n1 = 3.8*10^(-4)*230/860 = 1.0 *10^(-4) mol
A balloon containing gas expands from 230 mL to 860 mL as more helium is added. 1.0 × 10⁻⁴ mole was the initial quantity of helium present if the expanded balloon contains 3.8 × 10⁻⁴ mole assuming constant temperature and pressure.
Ideal Gas LawThe ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.
PV = nRT
where,
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant
T = Temperature
Now, calculating the ratio between the initial and the final numbers of moles of gas
PV = nRT
or, [tex]\frac{V}{n} = \frac{(RT)}{P}[/tex]
or, [tex]\frac{V}{n} = k[/tex]
or, [tex]\frac{V_1}{n_1} = \frac{V_2}{n_2}[/tex]
or, [tex]\frac{n_1}{n_2} = \frac{V_1}{V_2}[/tex] [Avogadro's Law]
Here, Volume and number of moles are the variables which are known.
Now put the values in above formula we get
⇒ [tex]\frac{n_1}{n_2} = \frac{V_1}{V_2}[/tex]
⇒ [tex]\frac{n_1}{3.8 \times 10^{-4}} = \frac{230}{860}[/tex]
⇒ [tex]n_{1} = \frac{3.8 \times 10^{-4} \times 230}{860}[/tex]
⇒ n₁ = 1.0 × 10⁻⁴ mole
Thus, we can say that 1.0 × 10⁻⁴ mole was the initial quantity of helium present if the expanded balloon contains 3.8 × 10⁻⁴ mole assuming constant temperature and pressure. Volume and number of moles are the variables which are known.
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Stoichiometry!
Please note:
- Use 6.022x1023 for avogadro’s number
- Ignore sig figs and do not round the final answer.
- Keep it to 1 decimal place.
Hey there!
C₃H₆O(l) + O₂(g) => CO₂(g) + H₂O(g)
First thing I want to do is balance the equation. Balance C:
C₃H₆O(l) + O₂(g) => 3CO₂(g) + H₂O(g)
Balance H:
C₃H₆O(l) + O₂(g) => 3CO₂(g) + 3H₂O(g)
Balance O:
C₃H₆O(l) + 4O₂(g) => 3CO₂(g) + 3H₂O(g)
Now we can properly solve the problem.
a.)
Density of acetone is 0.800 g/mL, we need to form 67.2 L of CO₂ at STP.
For every one mole of acetone reacted, 3 moles of CO₂ is produced.
Let's convert 67.2 L to moles: At STP, one mole of a gas takes up 22.4 liters.
67.2 ÷ 22.4 = 3 moles
So turns out we want to produce 3 moles of CO₂, which means we need one mole of acetone.
The molar mass of acetone is 58.08 g/mol. The density of acetone is 0.800 g/mL. We need the volume of one mole.
58.08 ÷ 0.800 = 72.6
72.6 mL of acetone is needed.
b.)
For every one molecule of acetone reacted, 3 molecules of water is produced.
We are combusting 3.011 x 10²² molecules of acetone.
3.011 x 10²² x 3 = 9.033 x 10²²
Find the number of moles:
(9.033 x 10²²) ÷ (6.022 x 10²³) = 0.15 moles
The molar mass of water is 18.015 g/mol.
0.15 x 18.015 = 2.7 grams
2.7 grams of water vapor is formed.
Hope this helps!
El succinato de butilo es un repelente de insecto utilizado en la casa contra hormigas y cucarachas. Su composicion es de 65.28 por ciento de C, 9,63 por ciento de H y de 27,79 por ciento de O. Su masa molecular determinada experimentalmente es de 230 u, ¿cuales son las formulas empiricas y molecular del succinato de butilo?
Answer:
what is the English translation, id like to help
Explanation:
17. A gas tank holds 2785 L of propane, C3H8, at 830. mm
Hg. What is the
volume of the propane at standard pressure?
Answer:
3041.5L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 2785 L
P1 (initial pressure) = 830mmHg
P2 (final pressure) = stp = 760mmHg
V2 (final volume) =?
Applying the Boyle's law equation P1V1 = P2V2, the final volume propane can be achieved by as illustrated below:
P1V1 = P2V2
830 x 2785 = 760 x V2
Divide both side by 760
V2 = (830 x 2785)/760
V2 = 3041.5L
Therefore, the volume of the propane at standard pressure is 3041.5L
By applying Boyle's Law and using the given initial conditions, the volume of propane at standard pressure is calculated to be 3041.71 L.
The question involves applying the gas law to determine the volume of propane at standard pressure. The given conditions are 2785 L of propane at 830 mmHg, and we need to find the volume at standard pressure (760 mmHg). Using Boyle's Law, which states that P1V1 = P2V2 where P is pressure and V is volume, we can calculate the new volume at standard pressure.
First, convert the initial pressure to the same units as the standard pressure, which is already in mmHg. Then, set up the equation with P1 = 830 mmHg, V1 = 2785 L, P2 = 760 mmHg, and solve for V2:
V2 = (P1 × V1) / P2 = (830 mmHg × 2785 L) / 760 mmHg
V2 = 3041.71 L
So, the volume of propane at standard pressure is 3041.71 L.
Michelle has a cup of hot coffee. The liquid coffee weighs 236 grams. Michelle adds a few teaspoons sugar and 25 grams of milk to the coffee. Michelle stirs the mixture until everything is combined. The mixture now weighs 271 grams. How many grams of sugar did Michelle add to the coffee?
Answer:
10grams
Explanation:
So it weighs 236 grams added with 25 grams. So it now weighs 261 grams so 10 grams of sugar remains in it.
Final answer:
Michelle added 10 grams of sugar to her coffee. The calculation involved subtracting the initial coffee weight and the weight of added milk from the final mixture weight.
Explanation:
To determine how many grams of sugar Michelle added to her coffee, we need to analyze the before and after weights of the mixture. The initial weight of the coffee was 236 grams, and after adding milk, the total weight became 271 grams. Given that Michelle also added 25 grams of milk, we can calculate the weight of the sugar.
The calculation to find the weight of the sugar added is as follows:
Subtract the initial weight of the coffee (236 grams) from the final weight of the mixture (271 grams) to account for the added ingredients: 271 grams - 236 grams = 35 grams.
Subtract the weight of the milk (25 grams) from the 35 grams to find the weight of the sugar: 35 grams - 25 grams = 10 grams.
Therefore, Michelle added 10 grams of sugar to her coffee.
Nitrifiers are bacteria
a. capable of living in symbiosis in the roots of some plants.
b. capable of oxidizing NH3 using NO.
c. that convert N2 to NH3.
d. that generate nitrites or nitrates.
e. capable of using NH3 as an electron donor.
Answer:
(D.) Nitrifiers are bacteria that generate nitrites or nitrates.
Explanation:
In the nitrogen cycle which occurs in nature, ammonia and ammonium compounds in the soil from organic sources and are converted to nitrites and nitrates by aerobic microorganisms.
Nitrifiers, as the name implies, are these such aerobic bacteria which oxidize inorganic constituents in the soil to generate energy. Examples of these nitrifiers are nitrobacter and nitrosomonas.
A 2.425 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.91 ∘ C to 30.28 ∘ C. The heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ / ∘ C, what is the heat of combustion per gram of the material?
Answer:
-63.79 kJ/g
Explanation:
According to the law of conservation of energy, the sum of the heat released by the combustion of the new organic material (Qcomb) and the heat absorbed by the bomb calorimeter (Qbc) is zero.
Qcomb + Qbc = 0
Qcomb = -Qbc [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Qbc = C × ΔT
where,
C: calorimeter constantΔT: change in the temperatureQbc = C × ΔT
Qbc = 28.81 kJ/°C × (30.28°C - 24.91°C) = 154.7 kJ
From [1],
Qcomb = -154.7 kJ
The heat of combustion per gram of the material is:
-154.7 kJ / 2.425 g = -63.79 kJ/g
Answer:
The heat of combustion per gram of the material is -63.8 kJ/ gram
Explanation:
Step 1: Data given
Mass of a ew organic material = 2.425 grams
The initial temperature of the calorimeter = 24.91 °C
The final temperature of the calorimeter = 30.28 °C
The heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ /°C
Step 2: Calculate heat
Q = c*ΔT
⇒with c = the heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ /°C
⇒with ΔT = The change of temperature = T2 - T1 = 30.28 - 24.91 = 5.37 °C
Q = 28.81 kJ/ °C * 5.37 °C
Q = 154.7 kJ
Step 3: Calculate the heat of combustion per gram of the material
heat of combustion per gram = -Q / mass (negative since it's exothermic)
heat of combustion per gram = -154.7 kJ / 2.425 grams
heat of combustion per gram = -63.8 kJ/ gram
The heat of combustion per gram of the material is -63.8 kJ/ gram
6) At constant temperature, the pressure of a gas is 780 mmHg and
volume of the gas is 2.0 L. Calculate the volume of the gas when
pressure decreases to 620 mmHg.
Answer:
The volume of the gas when pressure decreases to 620 mmHg is 2.52 L
Explanation:
Boyle's law is a law related to gases that establish a relationship between the pressure and the volume of a given amount of gas, without variations in temperature, that is, at constant temperature. So, this law establishes that at constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
o P * V = k
Having a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment, if you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be true:
P1 * V1 = P2 * V2
In this case:
P1: 780 mmHgV1: 2 LP2: 620 mmHgV2: ?Replacing:
780 mmHg* 2 L= 620 mmHg* V2
Solving:
[tex]V2=\frac{780 mmHg* 2L}{620 mmHg}[/tex]
V2= 2.52 L
The volume of the gas when pressure decreases to 620 mmHg is 2.52 L
Answer:
2.52 L
Explanation:
Step 1:
The following data were obtained from the question:
Initial pressure (P1) = 780 mmHg
Initial volume (V1) = 2.0 L.
Final pressure (P2) = 620 mmHg.
Final volume (V2) =?
Step 2:
Determination of the final volume of the gas.
From the question given, we discovered that the temperature is constant. Since the temperature is constant, the gas is simply obeying Boyle's law.
Applying the Boyle's law equation, the final volume of the gas can be obtained as follow:
P1V1 = P2V2
780 x 2 = 620 x V2
Divide both side by 620
V2 = (780 x 2) /620
V2 = 2.52 L
Therefore, the volume of the gas when the pressure decreases to 620 mmHg is 2.52 L
An atom has the following electron configuration.
1s22s22p63s23p3
How many valence electrons does this atom have?
3
5
8
15
Answer:
5 electrons
Explanation:
Valence electrons are electrons located in the outermost energy shell, so you want to count the number of electrons in the last energy shell.
You can divide the configuration into 1s2 / 2s22p6 / 3s23p3 to see the energy shells in this atom. There are 3 shells occupied by the atom's electrons, so you need to count the electrons in the third shell as those are its valence electrons.
2 + 3 = 5 valence electrons total
Note: you don't count the 3 before the letter because that only indicates the shell level, not the number of electrons. Count only the exponents.
An atom that has the following electron configuration.
1s22s22p63s23p3 contains only 5 valence electrons in its outermost shell.
Electronic configuration of chemical elements is the classification and ordering of electrons around the nucleus of an atom in reference to their energy levels.
Electrons can be ordered and classified using Aufbau's principle.
Aufbau's principle posits that in the ground state of an atom, electrons are ordered to the subshell of the lowest energy levels before filling the higher energy level subshell.
The valence electron of a chemical element is the outermost electron participating actively in a chemical bonding.
From the given electronic configuration, we will first determine the total atomic number (total electrons) before we determine the valence electrons in the outermost shell.
Given that
1s² 2s² 2p⁶ 3s² 3p³
There are (2 + 2 + 6 + 2 + 3) total electrons in the elctronic configuration.
= 15 total electrons
The maximum number of electrons in the energy levels is determined by using the formula is [tex]\mathbf{2n^2}[/tex]
where;
n = 1, 2, 3 ....
In the first energy level, we have the maximum number of electrons to be:
= 2(1)²
= 2 electrons
In the second energy level, we have:
= 2(2)²
= 2(4) electrons
= 8 electrons
Since we have only 15 electrons, the number of the valence electrons in the atom is:
= 15- (2 + 8 ) electrons
= 15 electrons - 10 electrons
= 5 valence electrons left in the outermost shell.
In conclusion, an atom that has the following electron configuration 1s22s22p63s23p3 contains only 5 valence electrons in its outermost shell.
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Consider a mixture of two gases A and B, confined to a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the total pressure in the vessel?
Answer:
Total pressure increased
Explanation:
When gas C is added in the vessel then number of mole increases and number of collision depends on the number of molecules present in the vessel and on adding gas C ,mole also increases hence number of collision increases therefore pressure also increases because number of collision increases.
Total pressure increases.
How many grams of CO2 gas would occupy 25.7 L of volume at a temperature of –22.2oC and a pressure of 997 mm Hg? (***Must convert moles into gram at the end of this problem)
Answer:
72.028g
Explanation:
Data obtained from the question. This includes:
Volume (V) = 25.7 L
Temperature (T) = –22.2°C
Pressure (P) = 997 mmHg
Mass of CO2 =..?
Mass of CO2 can be obtained as follow:
Step 1:
Conversion to appropriate unit:
It is important to express each variables in their appropriate units in order to obtain the desired result in the right unit.
For temperature:
We shall be converting from celsius to Kelvin. This is illustrated below:
Temperature (Kelvin) = temperature (celsius) + 263
temperature (celsius) = –22.2°C
Temperature (Kelvin) = –22.2°C + 263
Temperature (Kelvin) = 250.8K
For Pressure:
We shall be converting from mmHg to atm. This is illustrated below:
760mmHg = 1 atm
Therefore, 997 mmHg = 997/760 = 1.31 atm
Step 2:
Determination of the number of mole of CO2.
With the ideal gas equation, we can obtain the number of mole of CO2 as follow:
Volume (V) = 25.7 L
Temperature (T) = 250.8K
Pressure (P) = 1.31 atm
Gas constant (R) = 0.082atm.L/Kmol
Number of mole (n) =...?
PV = nRT
n = PV/RT
n = (1.31 x 25.7)/(0.082 x 250.8)
n = 1.637 mole
Step 3:
Converting 1.637 mole of CO2 to grams. This is illustrated below:
Number of mole of CO2 = 1.637 mole
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 =...?
Mass = number of mole x molar Mass
Mass of CO2 = 1.637 x 44
Mass of CO2 = 72.028g.
What do methanol (CH2O), ethanoic acid (C2H4O2), and glucose (C3H6O3) all have in common?
a. They are all carbohydrates
b. They all have the same empirical formula
c. They all have the same % composition
d. All of the above are true
Answer:
d
Explanation:
Carbohydrates are compounds containing carbon, hydrogen, and oxygen. Therefore, a is true.
An empirical formula is the simplest ratio of atoms present in a compound. Therefore, C2H4O2 and C3H6O3, (if you simplified them like you would a fraction) would be CH2O. Therefore b is correct,
They also have the same % composition, with a ratio of 1 carbon : 2 hydrogen : 1 oxygen. Therefore, c is correct.
Since a, b and c are all correct, the answer is d, all of the above are true.
Final answer:
Methanol, ethanoic acid, and glucose all share the same empirical formula CH2O, despite having different molecular structures and functions.
Explanation:
What methanol (CH2O), ethanoic acid (C2H4O2), and glucose (C6H12O6) all have in common is that they all have the same empirical formula, which is CH2O. This simplified empirical formula indicates a molar ratio between carbon, hydrogen, and oxygen, which is 1:2:1. However, these compounds have different molecular formulas, meaning they contain different numbers of atoms, but when simplified to the lowest whole number ratio, they share the same empirical formula.
Methanol, a simple alcohol with the molecular formula CH4O, ethanoic acid (also known as acetic acid, found in vinegar), with a molecular formula of C2H4O2, and glucose, a simple sugar with the molecular formula C6H12O6, while all different compounds, share the empirical formula CH2O. This showcases an interesting aspect of organic chemistry where different compounds can have identical empirical formulas but vary greatly in structure and function.
The middle ear converts sound from what into vibrations that can be sent through what?
Answer:
.
Explanation:
Eardrum or tympanic membrane changes sound to vibration and pass through malleus, incus and stapes.
What
was the initial volume of the hydrogen in cm3?
Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
Answer:
Explanation:
To get the initial volume of the hydrogen in cm³, we apply the equation generated from Charles' law.
V1/T1=V2/T2
Where;
V1=initial volume of gas
T1=initial temperature of gas
V2=final volume of gas
T2=final temperature of gas
We make V1 subject of formula
V1=(V2×T1)/T2
Given;
V2=212cm³
T1=50°C=(50+273)K=323K
T2=-5.0°C=(-5.0+273)K=268K
V1=(212×323)/268
V1=255.5cm³
Therefore, the initial volume of the hydrogen was 255.5cm³
what mass of helium is in a 2.00 L balloon at STP? pls help me lol
Once upon a time there was a lovely
princess. But she had an enchantment
upon her of a fearful sort which could
only be broken by love's first kiss.
She was locked away in a castle guarded
by a terrible fire-breathing dragon.
Many brave knights had attempted to
free her from this dreadful prison,
but non prevailed. She waited in the
dragon's keep in the highest room of
the tallest tower for her true love
and true love's first kiss. (laughs)
Like that's ever gonna happen. What
a load of - (toilet flush)
Allstar - by Smashmouth begins to play. Shrek goes about his
day. While in a nearby town, the villagers get together to go
after the ogre.
What are the 3 advantages and uses of polyester fabric
Answer:
Synthetic fiber fabric has good heat resistance and thermoplasticity.
Good light resistance, light resistance is second only to acrylic.
Good chemical resistance. ...
High strength and elastic recovery. ...
Good water absorption. ...
you can choose any three from here
hope that was helpful.Thank you!!!
Final answer:
Polyester fabric has three main advantages: it is highly stain-resistant, has excellent wrinkle resistance, and is used in various industrial applications.
Explanation:
Advantages and Uses of Polyester Fabric:
Polyester fabric is highly stain-resistant, making it a durable option for apparel and home furnishings. It can resist spills and stains, making it easier to clean and maintain.Polyester fabric has excellent wrinkle resistance due to the cross-linking of polymer strands. This makes it a popular choice for permanent-press clothing that requires minimal ironing or maintenance.Industrial polyester fibers, yarns, and ropes are used in various applications such as car tire reinforcements, conveyor belts, safety belts, and coated fabrics. Polyester's high-energy absorption properties make it suitable for these industrial uses.Calculate the wavelength of yellow light produced by a sodium lamp if the frequency of radiation is 3.34 x 10^14 Hz
Answer : The wavelength of yellow light produced by a sodium lamp is, [tex]8.98\times 10^{-7}m[/tex]
Explanation : Given,
Frequency of radiation = [tex]3.34\times 10^{14}Hz=3.34\times 10^{14}s^{-1}[/tex]
conversion used : [tex]Hz=s^{-1}[/tex]
Formula used :
[tex]\nu=\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency of radiation
[tex]\lambda[/tex] = wavelength of radiation
c = speed of light = [tex]3\times 10^8m/s[/tex]
Now put all the given values in the above formula, we get:
[tex]3.34\times 10^{14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}[/tex]
[tex]\lambda=8.98\times 10^{-7}m[/tex]
Therefore, the wavelength of yellow light produced by a sodium lamp is, [tex]8.98\times 10^{-7}m[/tex]
Which of the following is an example of maintaining homeostasis? A. Learning B. Jumping C. Shivering D. Smiling
Answer:
Shivering.
Explanation:
Homeostasis is the bodies need to keep things at a normal. When you are cold you shiver to produce body heat and get back to normal
Consider the balanced equation for the decomposition of ozone:
2O3(g) 3O2(g)
Can the rate law for this chemical reaction be deduced from its balanced equation?
A)Yes; since the reactant and product are both oxygen gases, the reaction takes 0 seconds.
B)Yes; the coefficients of the balanced equation determine the order of the reaction.
C)No; the rate law must be determined experimentally.
D)No; the rate will change randomly.
Answer:
C)No; the rate law must be determined experimentally.
Explanation:
Rate Laws are empirical relationships defining rates of reaction. For any given reaction The Empirical Rate Law is the product of the concentration of the reactants raised to the power of their order of reaction. That is, for the hypothetical reaction aA + bB => Products its Empirical Rate Law is ...
Rate = k[A]ᵃ[B]ᵇ where k = the rate constant and a & b are orders of reaction.
Order of reaction is a 'rate trend' when one changes a rate factor such as concentration. This can only be determined by experimental observation by physically increasing or decreasing concentration and observing the change in reaction rate relative to a reference reaction of interest. Typically they are whole numbers but can be decimal fractions.
By graphing experimental outcomes of Rxn Rate vs Change in Concentration one can define the order of reaction. Observations give ...
0-order reactions => change concentration => no change in rate
1st order reactions => change concentration => proportional linear change in rate
2nd order reactions => change concentration => exponential change in rate.
Answer:
The answer is C
Explanation:
I had the question on edg
2. What is the percentage by mass of sodium (Na) in a formula unit of sodium
hydrogen carbonate (NaHCO3)?
The percentage by mass of sodium in sodium hydrogen carbonate (NaHCO3) is 27.4%, calculated by dividing the atomic mass of sodium by the molar mass of NaHCO3 and multiplying by 100%.
Explanation:
To calculate the percentage by mass of sodium (Na) in a unit of sodium hydrogen carbonate (NaHCO3), you first need to know the atomic masses of the elements involved. The atomic mass of Na is approximately 23 g/mol, H is about 1 g/mol, C is approximately 12 g/mol and O is approximately 16 g/mol.
Using these numbers, the molar mass of NaHCO3 is about 84 g/mol (23 for Na, 1 for H, 12 for C, and 48 for the three O atoms). To find the percentage of Na in NaHCO3, divide the atomic mass of Na by the molar mass of NaHCO3 and multiply by 100%:
(23 g/mol ÷ 84 g/mol) x 100% = 27.4%.
Therefore, the percentage by mass of sodium in a formula unit of sodium hydrogen carbonate is 27.4%.
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The mass percentage of sodium (Na) in a formula unit of sodium hydrogen carbonate (NaHCO₃) is 27.36%. The calculated molar mass of NaHCO₃ is 84.01 g/mol.
To determine the percentage by mass of sodium (Na) in a formula unit of sodium hydrogen carbonate (NaHCO₃), we need to follow these steps:
1. Find the atomic masses of each element:
Sodium (Na) = 22.99 g/mol
Hydrogen (H) = 1.01 g/mol
Carbon (C) = 12.01 g/mol
Oxygen (O) = 16.00 g/mol
2. Calculate the molar mass of NaHCO₃:
NaHCO₃ = 22.99 g/mol (Na) + 1.01 g/mol (H) + 12.01 g/mol (C) + 3 x 16.00 g/mol (O)
NaHCO₃ = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol
3. Determine the mass percentage of sodium (Na):
Mass % Na = (22.99 g / 84.01 g) x 100
Mass % Na = 27.36%
Therefore, the mass percentage of sodium (Na) in sodium hydrogen carbonate (NaHCO₃) is 27.36%.
When 189.6 g of ethylene (C2H4) burns in oxygen to give carbon dioxide and water, how many grams of CO2 are formed? C2H4(g) + O2(g) → CO2(g) + H2O(g) (unbalanced)
Answer:
596 g of CO₂ is the mass formed
Explanation:
Combustion reaction:
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)
We determine moles of ethylene that has reacted:
189.6 g . 1mol / 28g = 6.77 moles
We assume the oxygen is in excess so the limiting reagent will be the ethylene.
1 mol of ethylene produce 2 moles of CO₂ then,
6.77 moles will produce the double of CO₂, 13.5 moles.
We convert the moles to mass: 13.5 mol . 44 g /1mol = 596 g
For the vaporization of a liquid at a given pressure: A.ΔG is positive at all temperatures. B.ΔG is negative at all temperatures. C.ΔG is positive at low temperatures, but negative at high temperatures (and zero at some temperature). D.ΔG is negative at low temperatures, but positive at high temperatures (and zero at some temperature). E.None of these.
Answer:
C. ΔG is positive at low temperatures, but negative at high temperatures (and zero at some temperature).
Explanation:
Since we need to give energy in the form of heat to vaporize a liquid, the enthalpy is positive. In a gas, molecules are more separated than in a liquid, therefore the entropy is positive as well.
Considering the Gibbs free energy equation:
ΔG= ΔH - TΔS
+ +
When both the enthalpy and entropy are positive, the reaction proceeds spontaneously (ΔG is negative) at high temperatures. At low temperatures, the reaction is spontaneous in the reverse direction (ΔG is positive).
Why is stoichiometry is used in cooking?
Answer:
Explanation:
stoichiometry is used in cooking because it helps you determine the amount or proportion of compounds you will need in a chemical reaction. Stoichiometry is present in daily life, even in the cooking recipes we make at home. The reactions depend on the compounds involved and how much of each compound is needed to determine the product that will result.
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Answer:
Ammonia
Explanation:
Answer:
ammonia NH3
Explanation:
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Se toman 200 ml de una disolución de un frasco de ácido clorhídrico ClH en cuya etiqueta se lee: Pureza: 35 % en peso; densidad = 1,15 g/ml. ¿Cuál es su concentración en g/l?
Answer:
Concentration of solution is 402g/mL
Explanation:
The HCl solution has a purity of 35% w/w. That means there are 35g of HCl per 100g of solution. As density of solution is 1.15g/mL, volume of the solution is:
100g solution × (1mL / 1.15g) = 86.96mL. Thus concentration of solution in g/mL is:
35g of HCl / 86.96mL = 0.402g/mL. As 1L contains 1000mL:
[tex]0.402\frac{g}{mL} \frac{1000mL}{1L}[/tex] = 402g/mL
Concentration of solution is 402g/mL
If 16 grams of sugar, C12H22O11, are in a spoonful, how many molecules of sugar are in the spoonful?
Answer:
If 16 grams of sugar are in a spoonful, 2.83*10²² molecules of sugar are in the spoonful
Explanation:
Avogadro's Number or Avogadro's Constant is the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of that substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.
So you need to know how many moles are 16 grams of sugar, C₁₂H₂₂O₁₁. For that you must know the molar mass of sugar. Being:
C: 12 g/moleH: 1 g/moleO: 16 g/molethe molar mass of sugar is:
C₁₂H₂₂O₁₁= 12 *12 g/mole + 22 * 1 g/mole + 11* 16 g/mole= 342 g/mole
Then the following rule of three applies: if 342 grams of sugar are present in 1 mole of substance, 16 grams in how many moles will they be?
[tex]moles=\frac{16 grams*1 mole}{342 grams}[/tex]
moles= 0.047
Now, knowing Avogadro's number, you can apply the following rule of three: if 1 mole of sugar contains 6.023 * 10²³ molecules, 0.047 moles how many molecules does it have?
[tex]molecules=\frac{0.047 moles*6.023*10^{23} }{1 mole}[/tex]
molecules= 2.83*10²²
If 16 grams of sugar are in a spoonful, 2.83*10²² molecules of sugar are in the spoonful
Answer:
2.82x10^22 molecules.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules.
The above hypothesis implies that 1 mole of sugar, C12H22O11 also contains 6.02x10^23 molecules.
1 mole of C12H22O11 = (12x12) + (22x1) + (16x11) = 144 + 22 + 176 = 342g
Now, if 342g of C12H22O11 contains 6.02x10^23 molecules,
Therefore 16g of C12H22O11 will contain = (16x6.02x10^23)/342 = 2.82x10^22 molecules.
Therefore, 2.82x10^22 molecules of the sugar are in the spoon
The ingredient label for a fruit drink reads: water, sugar, orange juice, high fructose corn syrup, vitamin C, sodium benzoate (preservative), Red 40, Blue 1.
What can you conclude about the fruit drink’s contents?
A) The product contains no preservatives
B) The product contains more juice than water.
C) The product contains more sugar than juice.
D) The product contains more sugar than any other ingredient.
Answer:
C
Explanation:
C is the only one that is true. Because sugar comes before juice in the ingredients listed, there is more of it.
A flexible vessel contains 43L of gas where the pressure is 1.3 atm. What will the volume be when the pressure is 0.61 atm, the temperature remaining constant?
Answer:
91.64L
Explanation:
Step 1:
Data obtained from the question.
This includes the following:
Initial volume (V1) = 43L
Initial pressure (P1) = 1.3 atm
Final volume (V2) =..?
Final pressure (P2) = 0.61 atm
Step 2:
Determination of the final volume.
To solve for the final volume, we'll apply the Boyle's law equation since the temperature is constant. The final volume is obtained as follow:
P1V1 = P2V2
1.3 x 43 = 0.61 x V2
Divide both side by 0.61
V2 = (1.3 x 43)/0.61
V2 = 91.64L
Therefore, the new volume is 91.64L
The volume will be 91.64 L when the pressure is 0.61 atm
From Boyle's law equation, we have that
P₁V₁ = P₂V₂
Where P₁ is the initial pressure
V₁ is the initial volume
P₂ is the final pressure
and V₂ is the final volume
From the given information
P₁ = 1.3 atm
V₁ = 43 L
P₂ = 0.61 atm
V₂ = ?
Putting the parameters into the formula, we get
1.3 × 43 = 0.61 × V₂
∴ [tex]V_{2} = \frac{1.3 \times 43}{0.61}[/tex]
V₂ = 91.64 L
Hence, when the pressure is 0.61 atm, the volume will be 91.64 L
Learn more here: https://brainly.com/question/20277275
A balloon of air occupies 10.0L(V2)at25.0°C(T2)and1.00atm(P2). What temperature (T1) was it initially, if it occupied 9.40 L (V1) and was in a freezer with a pressure of 0.939 atm (P1)?
The initial temperature (T₁) was 263.0145 K.
Given Conditions:
Final volume (V₂) = 10.0 L
Final temperature (T₂) = 25.0°C = 298.15 K
Final pressure (P₂) = 1.00 atm
Initial volume (V₁) = 9.40 L
Initial pressure (P₁) = 0.939 atm
Using the Combined Gas Law:
[tex]\[ \dfrac{P_1 \times V_1}{T_1} = \dfrac{P_2 \times V_2}{T_2} \][/tex]
Rearrange to Solve for T₁:
[tex]\[ T_1 = \dfrac{P_1 \times V_1 \times T_2}{P_2 \times V_2} \][/tex]
Plugging in the Values:
[tex]\[ T_1 = \dfrac{0.939 \, \text{atm} \times 9.40 \, \text{L} \times 298.15 \, \text{K}}{1.00 \, \text{atm} \times 10.0 \, \text{L}} \][/tex]
Calculate T₁:
[tex]\[ T_1= \dfrac{2630.145}{10} = 263.0145 \, \text{K} \][/tex]