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
The smallest form of matter that still retains the properties of an element
Answer:
atom
Explanation:
the atom is the smallest form.
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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A liquid has a volume of 62.7 mL and a density of 2.59 g/mL. What is its mass? (show all work)
Answer:
162.4
Explanation:
The formula for mass is density* volume so 62.7 multiplied by 2.59 equals 162.393. then you round so your anwser would be 162.4
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.
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 .
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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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
SOMEONE PLEASE HELPPP
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:
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
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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The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its effects on human health if the receiving water is later used as a drinking water source. One way in which chromate can be removed from solution is by its reaction with ferrous ions (Fe2+) to form a mixture of chromic hydroxide and ferric hydroxide solids [Cr(OH)3(s) and Fe(OH)3(s), respectively], which can then be filtered out of the water. The overall reaction can be represented as
CrO42- + 3 Fe2+ + 8 H2O --> Cr(OH)3(s) + 3 Fe(OH)3(s) + 4 H+
How much particulate matter would be generated daily by this process at a facility that treats 60 m3/h of a waste stream containing 4.0 mg/L Cr, if the treatment reduces the Cr concentration enough to meet a discharge limit of 0.1 mg/L?
Answer:
45727g
Explanation:
So, have the overall ionic equation given as the following;
CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.
So, we have (from the question) that the amount or quantity of the waste stream daily = 60m^3/h, and the waste stream daily contains waste stream containing = 4.0 mg/L Cr, and the discharge limit = 0.1 mg/L.
Step one: convert m^3/ h to L/h. Therefore, 60 m^3/h × 1000dm^3 = 60000 L/h .
Step two: Determine or calculate the the value of Cr used up.
The value of Car used up ={ 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.
Step three: Determine or calculate the mass of Cr(OH)3 and the mass of Fe(OH)3.
The number of moles of Cr = 234/52 = 4.5 moles.
Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.
Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.
And the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.
Hence, the total = 463.5 g + 1441.8 g = 1905.3 g.
Step four: Determine or calculate the How much particulate matter would be generated daily.
The amount of the particulate that would be generated daily = 24 × 1905.3 = 45727g.
Which of the following represents a species with 16 protons and 18 electrons? A) Ar B) S C) S²⁻ D) Si⁴⁻ E) S²⁺ Question 5 of 40
Answer:S²⁻
Explanation:
Sulphur is in group 16. The atomic number of sulphur is 16.
However, sulphur can accept two electrons to form the sulphide ion S²⁻ which contains 18 electrons, hence the answer provided above.
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.
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
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]
Which element contains one set of paired and three unpaired electrons in its fourth and outer main energy level?
Explanation:
Phosphorus (P) because of the 5 valence electrons total, 3 of them are in the 3p sublevel, and according to Hund's rule, they "single-fill" each orbital first.
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:
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