If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction
Answer:
7.71 × 10⁻⁴ M/s
Explanation:
The initial rate of the reaction can be expressed by using the formula:
[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]
where;
Pressue P = 1.00 atm
Volume V =5.74mL = (5.74 /1000) L
Rate R = 0.082 L atm/mol.K
Temperature = 298 K
[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]
= 2.35 × 10⁻⁴ mol
Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]
Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]
Δ[O₂] = 0.04626 M
The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]
= [tex]\dfrac{0.04626}{60}[/tex]
= 7.71 × 10⁻⁴ M/s
A bicycle rider is applying a force of 20 N while heading south against a wind blowing from the south with a force of 5 N
Answer:
the bike will go down 5 N
An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant. B) an increase in reaction rate due to a decrease in the kinetic energy of the reactants. C) an increase in the rate of reaction because reactant molecules collide with greater energy. D) an increase in both the reaction rate and activation energy due to increased kinetic energy.
Answer:
C) an increase in rate of reaction because reactant molecules collide with greater energy
Explanation:
Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.
The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.
Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:_______.
Mass % W Mass % Cl
72.17 27.83
56.45 43.55
50.91 49.09
46.36 53.64
If a molecule of each compound contains only one tungsten atom, what are the formulas for the four compounds?
Answer:
WCl₂, WCl₄, WCl₅, WCl₆
Explanation:
Molar Mass of Tungsten = 184 g/mol
Mass of Chlorine = 35.5 g/mol
In the first compound;
Percentage of tungsten = 72.17 %
Upon solving;
72.17 % = 184
100 % = Total mass
Total mass of compound = 254.95g
Mass of chlorine = 254.95 - 184 = 70.95 (Dividing by 35.35; This is approximately 2 Chlorine atoms.
The Formular is WCl₂
In the second compound;
Percentage of tungsten = 56.45 %
Upon solving;
56.45 % = 184
100 % = Total mass
Total mass of compound = 325.95 g
Mass of chlorine = 325.95 - 184 = 141.95g (Dividing by 35.35; This is approximately 4 Chlorine atoms.
The Formular is WCl₄
In the third compound;
Percentage of tungsten = 50.91 %
Upon solving;
50.91 % = 184
100 % = Total mass
Total mass of compound = 361.42 g
Mass of chlorine = 361.42 - 184 = 177.42 (Dividing by 35.35; This is approximately 5 Chlorine atoms.
The Formular is WCl₅
In the fourth compound;
Percentage of tungsten = 46.39 %
Upon solving;
46.39 % = 184
100 % = Total mass
Total mass of compound = 396.64 g
Mass of chlorine = 396.64 - 184 = 212.64 (Dividing by 35.35; This is approximately 6 Chlorine atoms.
The Formular is WCl₆
Which element will gain three electrons to form an anion?
A. aluminum
B. chromium
C. iron
D. nitrogen
Answer:
D represents the element nitrogen which will gain three valence electrons forming a 3 ion.
Answer:
answer is
Explanation:
D
Which of the following elements has the largest atomic radius? Bromine, Barium, Magnsium, Zinc
Answer:
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
Explanation:
HELP FAST PLZ!!!!! Which phase change allows a substance to transform from a liquid to a
gas?
melting
Ofreezing
O ionization
condensation
deionization
O
evaporation
sublimation
Answer:evaporation
Explanation:
Answer:
Pretty sure its evaporation, if its not I'm very sorry.
A piece of metal has a mass of 0.650 kilograms, has a width of 0.136 meters, and has a length of 0.0451 meters.Part A: If the metal’s volume is 291 cm3, what is the height of the metal in centimeters? (The width & length values given above are in a different unit!)
Part B: What is the density of this piece of metal?
Answer:
height = 4.74 cm
density = 2.23 g/ cm³
Explanation:
Mass of metal = 0.650 kg (650 g)
Width = 0.136 m
Length = 0.0451 m
Volume of metal = 291 cm³
Height in cm = ?
density of metal =?
Solution:
Width = 0.136 m (0.136 m×100 cm/1m = 13.6 cm)
Length = 0.0451 m (0.0451 m×100 cm/1m = 4.51 cm)
First of all we will calculate the height:
Volume = height× width× length
291 cm³ = h × 13.6 cm × 4.51 cm
291 cm³ = h × 61.34 cm²
h = 291 cm³ / 61.34 cm²
h = 4.74 cm
Density:
d = m/v
d = 650 g/291 cm³
d = 2.23 g/ cm³
1.
Light near the middle of the ultraviolet region of the electromagnetic spectrum
has a frequency of 2.73 X 1016 s-1
a.
What is the wavelength of this radiation in meters (m)?
b.
What is the energy associated with this radiation in kcal?
Explanation:
Given that,
The frequency of electromagnetic spectrum is [tex]2.73\times 10^{16}\ Hz[/tex]
(A) Let the wavelength of this radiation is [tex]\lambda[/tex]. We know that,
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{2.73\times 10^{16}}\\\\\lambda=1.09\times 10^{-8}\ m[/tex]
So, the wavelength of this radiation is [tex]1.09\times 10^{-8}\ m[/tex].
(B) Let E is the energy associated with this radiation. Energy of an electromagnetic radiation is given by :
[tex]E=hf[/tex]
h is Planck's constant
[tex]E=6.63\times 10^{-34}\times 2.73\times 10^{16}\\\\E=1.8\times 10^{-17}\ J[/tex]
1 kcal = 4184 J
It means,
[tex]1.8\times 10^{-17}\ J=\dfrac{1}{4184}\times 1.8\times 10^{-17}\\\\=4.3\times 10^{-21}\ \text{kcal}[/tex]
Hence, this is the required solution.
On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential energy of two O atoms versus the distance between their nuclei.
Answer:
Explanation:
We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.
From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.
Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.
What is the mass in grams of 2.5 moles of Al?
Answer:
One mole of Al weighs 27g.
2.5 moles of Al weigh 67.5g.
what kind of bonds are there in H2O?
Answer: it is covalent and there are 2 hydrogen molecules and 1 oxygen molecule.
Explanation: it just is
Three colorless solutions in test tubes, with no labels, are in a test tube rack on the laboratory bench. Lying beside the tests tubes are three labels : 0.10 M Na2CO3, 0.10 M HCL, and 0.10 M KOH. You are to place the labels on the test tubes using only the three solutions present. Here are your tests:
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm.
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Identify the labels for test tubes 1, 2, and 3
Answer:
Test tube 1 0.10 M HCL
Test tube 2 0.10 M KOH
Test tube 3 0.10 M Na2CO3
Explanation:
From the question we are told that
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm
Generally this warmth is as a result of a reaction between an acid and a base and the acid is 0.10 M HCL and the base is 0.10 M KOH , the heat generated is know as the heat of neutralization,
The reaction is
[tex]HCl_{(aq)} + KOH_{(aq)} \rightarrow KCl_{(aq)} + H_2O_{(l)} + \Delta H[/tex]
We are also told from the reaction that
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Generally carbon dioxide gas is produced is as a result of a reaction between the acid HCl and Na2CO3.
The reaction is
[tex]2HCl -{(aq)} + Na_2 CO_3_{(aq)} \rightarrow 2 NaCl _{(aq)} + CO_2_{(g)} + H_2O_{(l)}[/tex]
Hence from this explanation above we see that the solution in test tube 1 is 0.10 M HCL while solution in test tube 2 is 0.10 M KOH and then solution in test tube three is 0.10 M Na2CO3
What does the term mass mean in science?
A. The measure of the force of gravity on an object.
B. The amount of matter in an object.
C. The Amount of space that matter takes up.
Answer:
B
Explanation:
Mass is defined as the quantity of matter in an object.
which angles are right
Answer:
a right is 90 degrees
2. (6 pts) In a reaction, 235 mL of 1.50 M HCl solution reacts completely with an excess amount of
aluminum. If the hydrogen gas is collected over water in a container with a volume of 3.60 L and at a
temperature of 25.0 °C, calculate the pressure in the container. The vapor pressure of water is 23.78
mmHg (Table 6.4, page 232).
2Al(s) + 6HCl(aq) + 3H2(g) + 2AlCl3(aq)
Answer:
[tex]P=1.23atm[/tex]
Explanation:
Hello.
In this case, since the total pressure in the container includes the pressures of both hydrogen and water:
[tex]P=P_{H_2}+P_{H_2O}[/tex]
For the reacting solution of HCl, based on the 6:3 mole ratio with hydrogen in the chemical reaction, we can next compute the yielded moles o hydrogen:
[tex]n_{H_2}=0.235L*1.50\frac{molHCl}{L}*\frac{3molH_2}{6molHCl} =0.176molH_2[/tex]
Then, by using the ideal gas equation we compute the pressure of hydrogen for the collected 3.60 L at 25.0 °C (298.15 K):
[tex]P_{H_2}=\frac{n_{H_2}RT}{V} =\frac{0.176mol*0.082\frac{atm*L}{mol*K}*298.15K}{3.60L}=1.20atm[/tex]
Finally, since the vapor pressure of water in at is 0.03129, the total pressure is then:
[tex]P=1.20atm+0.03129atm\\\\P=1.23atm[/tex]
Best regards!
Why can we use a water-ethanol mixture but not a water-hexane mixture as a recrystallization solvent? because water and hexane are ___________________.
Answer:
immiscible liquids
Explanation:
When two liquids are to be used for the purpose of recrystallization the both liquids must be miscible with each other in all proportions.
This implies that, if two liquids are immiscible, it will be difficult for the mixture to be useful in recrystallization.
Water and hexane are immiscible because water is a polar solvent and hexane is a non polar solvent.
Water is miscible with ethanol because ethanol contains polar -OH groups that interact effectively with water leading to miscibility of ethanol with water in all proportions.
How many carbon atoms are in vitamin c?
Answer:
molecules can be much bigger. one molecule of vitamin c is made up of 20 atoms (6 carbons, 8 hydrogens, and 6 oxygens
27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?
Answer:
8.69 is the pH at the equivalence point
Explanation:
Formic acid, HCHO₂, reacts with NaOH as follows:
HCHO₂ + NaOH → NaCHO₂ + H₂O.
At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:
Moles: 0.0278L * 0.797mol/L = 0.02216moles
To reach the equivalence point it is necessary to add:
0.02216mol * (1L / 0.928mol) = 0.0239L
Total volume in the equivalence point:
0.0278L + 0.0239L = 0.0517L
Concentration: 0.02216moles / 0.0517L = 0.429M
The equilibrium of NaCHO₂, CHO₂⁻, in water is:
CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)
Where Kb, 5.56x10⁻¹¹ is defined as:
5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]
In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:
5.56x10⁻¹¹ = [X] [X] / [0.429M]
2.383x10⁻¹¹ = X²
4.88x10⁻⁶ = X = [OH⁻]
As pOH = -log [OH⁻]
pOH = 5.31
And pH = 14 - pH
pH = 8.69 is the pH at the equivalence point
How long would it take a bus traveling 52 km/h to travel 130 km
Answer:
2 and a half hours
Explanation:
Time is equal to distance over speed
9 What 11 letter word describes the chemical balance within an organism?
___________________________________
10 What 7 letter word describes the waterproof outer layer for soft-bodies creatures?
___________________________________
11 What word starting with the letter "C" is the liquid material found within cells?
___________________________________
12 What word starting with the letter "C" is a process of organizing things into groups scientifically?
___________________________________
Answer:
9.) Homeostasis
10.) Epidermis
11.) Cytoplasim
12.) Classification
The wood in my house is crumbling. *
Problem
Hypothesis
Law
Theory
Answer:
Problem
Explanation:
The given statement is a problem. It states the problem that the house of the speaker has been undergoing with. This problem gives rise to the Hypothesis in which the 'why' question is asked. The reason of the crumbling of the wood is stated in the hypothesis. Any reason placed of the happening of the event is stated to be hypothesis.
The ion with the smallest diameter is ________. The ion with the smallest diameter is ________. Be2 Sr2 Ca2 Ba2 Mg2
Answer:
Be2^+
Explanation:
Ionic diameter increases down the group. This implies that Be2^+ will have the smallest diameter.
This extremely small diameter makes Be2^+ to differ considerably from other ions of group 2 elements.
For instance, the compounds of beryllium are mostly covalent in nature.
Rahul and Manav each were given a mixture of iron filings and sulphur powder. Rahul heated the mixture strongly and a new substance was formed. Write three points of difference between the two.Required to answer.
Answer:
sure
Explanation:
The substance formed after heating the mixture of that of Rahul is caleed a compound. Whereas, Manav's mixture still remains in its current stae that is a heterogeneous mixture.
The compound formed is in black in color whereas the mixture is a mix of brownish-red and yellow.
The compound is a homogeneous mixture whereas the mixture is a heterogenous mixture because of its uneven distribution.
A sample of N2O effuses from a container in 47s .
Required:
How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?
Answer:
112.92 s.
Explanation:
Let M₁ be the molar mass of N₂O
Let t₁ be the time taken for N₂O to effuse.
Let M₂ be the molar mass of I₂
Let t₂ be the time taken for I₂ to effuse.
Molar mass (M₁) of N₂O = (14×2) + 16 = 28 + 16 = 44 g/mol
Time (t₁) of effusion of N₂O = 47 s
Molar mass (M₂) of I₂ = 127 × 2 = 254 g/mol
Time (t₂ ) of effusion of I₂ =?
The time take for the same amount of I₂ to effuse can be obtained as follow:
t₂/t₁ = √(M₂/M₁)
t₂/47 = √(254 / 44)
Cross multiply
t₂ = 47 × √(254 / 44)
t₂ = 112.92 s
Therefore, it will take 112.92 s for the same amount of I₂ to effuse.
When a substance changes its size, shape, or state, this is called a __________.
a
change of atoms
b
physical change
c
chemical change
d
change of state
Answer:
its a physical change
Explanation:
its b
Research Hypatia's achievements in the world
of science.
What is she most known for?
Write down three interesting facts about her
life.
Answer:
See explanation
Explanation:
Hypatia is popular for her work in mathematics. She also did some work in the area of astronomy. Her well know work in mathematics is her ideas about conic sections.
She was born the Theon of Alexandria and she was a professional mathematician in her life time.
She was the greatest mathematician of her time and she was telling leader of the Neoplatonist school of philosophy in Alexandria. By so doing, she conquered the culture of sexism in her time.
She was trained by her father in mathematics and eventually replaced him. She was the last major mathematician in the Alexandrian tradition.
In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.
Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?
Answer:
a) - 0.2 M
b) - 0.2 M
c)- 0
Explanation:
The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:
MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol
a). Molarity = moles CuBr₂/1 L solution
moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol
Volume in L = 375 mL x 1 L/1000 mL = 0.375 L
M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M
b). When is added to water, CuBr₂ dissociates into ions as follows:
CuBr₂ ⇒ Cu²⁺ + 2 Br⁻
We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:
0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M
c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).
Fossils in the gobi desert helped scientists learn?
Answer:
dinosaurs laid eggs,sabretooths were carnivores, bats lived millions of years ago ,fossils of cynobacteria exist?
Explanation:
What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.
Answer:
i
[tex]J_{m} = 20 [/tex]
ii
[tex]J_{m} = 22.5 [/tex]
Explanation:
From the question we are told that
The first temperatures is [tex]T_1 = 25^oC = 25 +273 =298 \ K[/tex]
The second temperature is [tex]T_2 = 100^oC = 100 +273 = 373 \ K[/tex]
Generally the equation for the most highly populated rotational energy level is mathematically represented as
[tex]J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}[/tex]
Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]
Also
B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]
Generally the energy require per mole to move 1 cm is 12 J /mole
So [tex]0.244 \ cm^{-1}[/tex] will require x J/mole
[tex]x = 0.244 * 12[/tex]
=> [tex]x = 2.928 \ J/mol [/tex]
So at the first temperature
[tex]J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 20 [/tex]
So at the second temperature
[tex]J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 22.5 [/tex]