The pOH of an aqueous solution of 0.480 M trimethylamine (a weak base with the formula (CH3)3N) is .
Answer:
Explanation:
Kb of (CH₃)₃N is 7.4 x 10⁻⁵
initial concentration of (CH₃)₃N a is .48 M
(CH₃)₃N + H₂O = (CH₃)₃NH⁺ + OH⁻
a - x x x
x² / (a - x ) = Kb
x is far less than a so a - x can be replaced by a .
x² / a = Kb
x² = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶
x = 5.96 x 10⁻³
pOH = - log ( 5.96 x 10⁻³ )
= 3 - log 5.96
= 3 - .775
= 2.225
What happens when the elements in group 2 react with water?
Answer:
The Group 2 metals become more reactive towards the water as you go down the Group.
Explanation:
These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.
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Question 11
4 pts
Using the formula 2H202 --> 2H2O + O2, if 7.30 moles of peroxide are
decomposed, how many moles of oxygen will be formed?
Answer:
3.65 mol O₂
Explanation:
Step 1: RxN
2H₂O₂ → 2H₂O + O₂
Step 2: Define
Given - 7.30 mol H₂O₂
Solve - x mol O₂
Step 3: Stoichiometry
[tex]7.30 \hspace{3} mol \hspace{3} H_2O_2(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} H_2O_2} )[/tex] = 3.65 mol O₂
The empirical formula of CBr2 has a molar mass of 515.46 g/mol. What is the molecular formula of this
compound
Answer:
C3Br6
Explanation:
C= (1 X 12.011) = 12.011
Br= (2 X 79.904)= 159.808
159.808+12.011 = 171.819
515.46 divided by 171.819 = 3.00
so you mulitpy CBr2 by 3 which gives you C3Br6
SOMEONE PLEASE HELPPP
Which of the following is an Elementary compound?
A. CO2
B. N2
C. SO2
D. H2S
heeeeeeeeeelp please please please
Answer:
Explanation:
In my opinion the answer should be SO2
Answer:
a should be answer i think.
Which is an example of a current research focus in chemistry?
A. applying gene therapy to treat certain diseases
B. using hook-and-loop tape in the clothing industry
C. developing smoke detectors for common use
D. studying coal combustion as an energy source
Answer:
b is the correct answer
do not trust answer one
Explanation:
what’s the most abundant isotope of lawrencium
Answer:
266Lr
Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.
Explanation:
hopefully that helps you
The smallest form of matter that still retains the properties of an element
Answer:
atom
Explanation:
the atom is the smallest form.
A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain
Answer:
[tex]Total = 50.6\ moles[/tex]
Explanation:
Given
[tex]Propane = C_3H_8[/tex]
Represent Carbon with C and Hydrogen with H
[tex]C = 13.8[/tex]
Required
Determine the total moles
First, we need to represent propane as a ratio
[tex]C_3H_8[/tex] implies
[tex]C:H = 3:8[/tex]
So, we're to first solve for H when [tex]C = 13.8[/tex]
Substitute 13.8 for C
[tex]13.8 : H = 3 : 8[/tex]
Convert to fraction
[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]
Cross Multiply
[tex]3 * H = 13.8 * 8[/tex]
[tex]3 H = 110.4[/tex]
Solve for H
[tex]H = 110.4/3[/tex]
[tex]H = 36.8[/tex]
So, when
[tex]C = 13.8[/tex]
[tex]H = 36.8[/tex]
[tex]Total = C + H[/tex]
[tex]Total = 13.8 + 36.8[/tex]
[tex]Total = 50.6\ moles[/tex]
A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound.
a. 1.245 g Ni, 5.381 g I,
b. 2.677 g Ba, 3.115 g Br,
c. 2.128 g Be, 7.557 g S, 15.107 g
Answer:
you can see the empirical formula at the pic
The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.
What is empirical formula?
Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.
(a) 1.245 g Ni : 5.381 g I
Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2
So the formula is NiI2
(b) 2.677 g Ba : 3.115 g Br
Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038
= 0.02 : 0.04 = 1:2
So the formula is BaBr2
(c) 2.128 g Be : 7.557 g S
Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1
So the formula is BeS
Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.
To learn more about empirical formula, refer to the link below:
https://brainly.com/question/11588623
#SPJ2
Which of the following is a good definition of matter?
O A. Anything that is made up of light and gravity
O B. Anything that has mass and takes up space
O C. Anything that produces heat and mass
O D. Anything that has energy and creates heat
Answer:
B
Explanation:
I did the question before and got it right.
When does carbon dioxide absorb the most heat energy?
during freezing
during deposition
during sublimation
during condensation
Answer:
during sublimation
Explanation:
just took the test
What can the chemical formula tell us about a compound?
Answer:
A chemical formula tells us the number of atoms of each element that is in a compound. It contains the symbols of the atoms for the elements present in the compound as well as how many there are for each element in the form of subscripts.
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How to separate given mixture?
Answer:
Chromatography involves solvent separation on a solid medium.
Distillation takes advantage of differences in boiling points.
Evaporation removes a liquid from a solution to leave a solid material.
Filtration separates solids of different sizes.
Explanation:
If 25.6 mL isopropyl alcohol fully decomposes, what mass of H2 is formed? The density of isopropyl alcohol is 0.785 g/mL. g
Answer:
The correct answer is 0.67 g H₂
Explanation:
Isopropyl alcohol (C₃H₇OH) can decompose to give acetone (C₂H₆OH) and hydrogen gas (H₂) according to the following chemical equation:
C₃H₇OH (g) ⇒ C₂H₆CO(g) + H₂(g)
We can calculate the initial mass of isopropyl alcohol from the density and volume data:
density = m/V = 0.785 g/mL
⇒ m = density x V = 0.785 g/mL x 25.6 mL = 20.096 g C₃H₇OH
According to the chemical equation 1 mol of C₃H₇OH gives 1 mol H₂. The molar mass of C₃H₇OH is:
molar mass C₃H₇OH = (12 g/mol x 3) + (1 g/mol x 7) + 16 g/mol + 1 g/mol = 60 g/mol
molar mass H₂ = 1 g/mol x 2 = 2 g/mol
So, we obtain: 2 g H₂ from 60 g C₃H₇OH. We multiply this stoichiometric ratio (2 g H₂/60 g C₃H₇OH) by the initial mass of C₃H₇OH to obtain the mass of H₂ is formed:
20.096 g C₃H₇OH x (2 g H₂/60 g C₃H₇OH) = 0.6698 g ≅ 0.67 g H₂
True or False: Particles that are moving faster have a higher temperature
Answer:
true
Explanation:
I'm not sure why cause I dont know how to explain but it's TRUE
Answer:
True
Explanation:
The particles moving faster in a substance the hotter it gets.
How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The test variables (independent variables) and outcome variables (dependent variables) are the same things. B. The test variable (independent variable) controls the outcome variable (dependent variable). C. The test variable (independent variable) and outcome variable (dependent variable) have no affect on each other. D. The outcome variable (dependent variable) controls the test variable (independent variable).
Answer:
I'm on the exact same queston
Answer:
The test variable (independent variable) controls the outcome variable (dependent variable)
Explanation:
its right on study island
What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)
Answer:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
Explanation:
Hello.
In this case, for the undergoing chemical reaction:
[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]
We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
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Scientists are experimenting with pure samples of isotope X which is radioactive. The sample has a mass of 20. Grams. The half-life was measured to be 232 seconds. There is a second sample that weighs 80 grams. What is the half-life of the second sample
Answer:
Explanation:
Half life of radioactive materials do not depend upon the mass of the material . It only depends upon the nature of radioactive materials . The half life of 20 g is 232 seconds . That means 20 gram will be reduced to 10 gram in 232 seconds .
Half life of 80 gram is also 232 seconds . So , 80 gram will be reduced to 40 gram in 232 second .