Answer:
C6H1206
Explanation:
C6H12O6 is a monomer of carbohydrates also known as glucose, so it is not an electrolyte at all.
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
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]
When 435 J of heat is added to 3.4 g of olive oil that's at 21 Deg C, it's
temperature increases to 85 Deg C. Calculate the specific heat of Olive oil? Show work
Answer:
k Nishant
Explanation:
i don't know sorry but u can search in google
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
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
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
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:
In which atmospheric layer is the ozone layer?
A.troposphere
B.mesosphere
C.stratosphere
D.thermosphere
Answer:
stratosphere
Explanation:
contains a high concentration of ozone in relation to other parts of the atmosphere, although still small in relation to other gases in the stratosphere.
Answer:
stratosphere
Explanation: Most atmospheric ozone is concentrated in a layer in the stratosphere, about 9 to 18 miles (15 to 30 km) above the Earth's surface. Ozone is a molecule that contains three oxygen atoms.
A tree frog uses plants or trees for protection from the rain. The frog is protected from the rain and the tree is neither helped nor harmed
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
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:
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’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
for the following reaction, provide the missing information
Answer:
19. Option B. ⁰₋₁B
20. Option D. ²¹⁰₈₄Po
Explanation:
19. ²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ
Thus, we can determine ʸₓZ as follow:
228 = 228 + y
Collect like terms
228 – 228 = y
y = 0
88 = 89 + x
Collect like terms
88 – 89 = x
x = –1
Thus,
ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B
²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ
²²⁸₈₈Ra —> ²²⁸₈₉Ac + ⁰₋₁B
20. ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He
Thus, we can determine ᵘᵥX as follow:
u = 206 + 4
u = 210
v = 82 + 2
v = 84
Thus,
ᵘᵥX => ²¹⁰₈₄X => ²¹⁰₈₄Po
ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He
²¹⁰₈₄Po —> ²⁰⁶₈₂Pb + ⁴₂He
PLSSS HELP Match each image to the
correct step of meiosis. PLS HELP
Answer:
Explanation:
Here you go!! Hope this helps.