Answer:
[tex]5H_2O+C_5H_{10}O_5\rightarrow 5CO_2+20H^++20e^-[/tex]
Explanation:
Hello.
In this case, when ribose (C₅H₁₀O₅) yields carbon dioxide (CO₂) we write:
[tex]C_5H_{10}O_5\rightarrow CO_2[/tex]
Which needs to be balanced by adding water and hydrogen ions:
[tex]5H_2O+C_5H_{10}O_5\rightarrow 5CO_2+20H^++20e^-[/tex]
You can also see that there are 20 transferred electrons, since the carbon atoms in the ribose have 0 as their oxidation state and the carbon atoms in the carbon dioxide have +4 as the oxidation state, thus, each carbon transfers 4 electrons, a five carbon atoms transfer 20 electrons overall.
In such a way, since the carbon is increasing its oxidation state, such half reaction is an oxidation half reaction.
Best regards.
Identifying the particles of an atom
Answer:
6 protons 8 neutrons 5 electrons
Explanation:
protons are blue neutrons are red electrons are green
Match the following terms and definitions.
1. major ecological grouping of plants and animals
food chain
biome
environmental factor
decomposers
2. bacteria and fungi that break down dead matter
a state of change in which the end result is equal or
3.
balanced
the basic relationships that show how a community of plants,
animals, and bacteria live and grow and how these living
4. things are dependent on each other as well as the Sun, soil,
and other nonliving parts of their environment; a cycle of
relationships
line of plants and animals that shows the order in which
5.
organisms are eaten
a condition or conditions of the nonliving surroundings, such
6.
as light, temperature, water, and so on
ecosystem
dynamic equilibrium
Answer:
major ecological grouping of plants and animals
food chain
*biome*
bacteria and fungi that break down dead matter
*decomposers*
a state of change in which the end result is equal or
balanced
*dynamic equilibrium*
the basic relationships that show how a community of plants,
animals, and bacteria live and grow and how these living
*ecosystem*
things are dependent on each other as well as the Sun, soil,
and other nonliving parts of their environment; a cycle of
relationships
line of plants and animals that shows the order in which
*Food chain*
organisms are eaten
a condition or conditions of the nonliving surroundings, such
*environmental factor*
Explanation:
which element Shows very similar chemical properties to barium?
Calculate the density of the aluminum cylinder with a diameter 0f 1.3 cm weighing 18 grams. Height of the cylinder is 5.2 Cm. Find percent error (Actual density of Aluminum is 2.7 g/cm3). (Formula for volume of cylinder is πr2h)
Answer:
Percent error = 3.7%
Explanation:
Given data:
Density of Al cylinder = ?
Weight of cylinder = 18 g
Diameter = 1.3 cm
Height = 5.2 cm
Actual density of Al = 2.7 g/cm³
Percent error = ?
Solution:
First of all we will calculate the volume of cylinder through given formula.
V = πr²h
r = diameter /2
V = 22/7 × (0.65 cm)²× 5.2 cm
V = 22/7 × 0.4225cm²× 5.2 cm
V = 6.89 cm³
Now we will calculate the density.
d = m/v
d = 18 g/ 6.89 cm³
d = 2.6 g/cm³
Percent error:
Percent error = measured value - actual value /actual value × 100
Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³ × 100
Percent error = 3.7%
Negative sign shows that measured or experimental value is less than actual value.
What is the mass/volume percent of a solution with 7.5g of Na2CO3 dissolved in 20.0 g of water?
9.3 %
27%
7.5%
Answer:
7.5%
Explanation:
I just know it's that
help me please i’ll give u a good rating
Answer:
d
Explanation:
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
Answer each of the following questions with increases, decreases, or does not change.
a) If the intermolecular forces in a liquid increase, the normal boiling point of the liquid increases .
b) If the intermolecular forces in a liquid increase, the vapor pressure of the liquid does not change .
c) If the surface area of a liquid increases, the vapor pressure increases .
d) If the temperature of a liquid decreases, the equilibrium vapor pressure increases .
Drag and drop your selection from the following list to complete the answer:
does not change decreases increases
Answer:
1) Increases
2) decreases
3) increases
4) decreases
Explanation:
When the intermolecular forces in a liquid increases, the greater vapour pressure of the liquid decreases accordingly.
Since the vapour pressure is proportional to temperature, as temperature increases, the vapour pressure increases alongside.
As intermolecular forces increases, the boiling point increases accordingly since more energy is required to break intermolecular bonds.
Lastly, the greater the surface area, tell greater the vapour pressure since more liquid surface area is now available.
Determine the mass (in grams) of NaCl in 294 grams of a 24.1% (m/m) NaCl solution. Be sure to report to the correct number of significant figures with no units.
I NEED HELP WITH THIS URGENTLY!!!!!
Answer:
70.9 grams 3 sig figs
Explanation:
24.1% of 294 grams = 0.241(294 gram) = 70.854 grams ≅ 70.9 grams 3 sig figs
Determine the number of moles of oxygen atoms in each of the following.
1) 4.93 mol H2O2
2) 2.01 mol N2O
Answer :
Part 1: 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.
Part 2: 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.
Explanation :
Part 1: 4.93 mol [tex]H_2O_2[/tex]
In 1 mole of [tex]H_2O_2[/tex], there are 2 atoms of hydrogen and 2 atoms of oxygen.
As, 1 mole of [tex]H_2O_2[/tex] contains 2 moles of oxygen atoms.
So, 4.93 moles of [tex]H_2O_2[/tex] contains [tex]4.93\times 2=9.86[/tex] moles of oxygen atoms.
Thus, 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.
Part 2: 2.01 mol [tex]N_2O[/tex]
In 1 mole of [tex]N_2O[/tex], there are 2 atoms of nitrogen and 1 atom of oxygen.
As, 1 mole of [tex]N_2O[/tex] contains 1 mole of oxygen atoms.
So, 2.01 moles of [tex]N_2O[/tex] contains [tex]2.01\times 1=2.01[/tex] moles of oxygen atoms.
Thus, 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.
How many grams of H2SO4 are needed to prepare 5.0 L of a 2 M H2SO4 solution? You must show your work in order to receive credit.
why is steel a mixture
Answer:
Steel is a mixture because iron does not mix with the carbon and the other components.
Answer:
Steel is a mixture because it does not have a set chemical compound formula. It's a mixture of iron and carbon fused together with other metals or nonmetals to create a mixture.
Explanation:
Steel is an alloy , a solid solution
Steel is a homogenous mixture. It is not a pure substance because it is made form more than one component.
2.
(3x – 4y = -10
(6x + 3y = –42
SOLUTION:
Answer:
Is like for solving for the solution for both equations??
Which item has the most thermal energy?
A. 1 kg boiling water
B. 1 kg ice
C. 1 kg hot water just below the boiling point
D. 1 kg cold water
Answer:
1L of hot water just below the Boling point
Explanation:
asking questions is best to learn please ask more questions
Answer:
its 1kg of boiling water
Explanation:
When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?
C2H2(g) + 2H2(g) → C2H6(g)
Answer:
[tex]Y=75.6\%[/tex]
Explanation:
Hello.
In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:
[tex]m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2} *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6[/tex]
Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:
[tex]Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%[/tex]
Best regards.
Compute the values of the diffusion coefficients for the interdiffusion of carbon in both α-iron (BCC) and γ-iron (FCC) at 900°C. Which is larger? Explain why this is the case.
Answer:
α-iron (BCC) has faster diffusion rate because of lower values in activation energy and pre-exponential value.
Explanation:
Taking each parameters or data at a time, we can determine the values/a constant for each parameters in the diffusion coefficient equation.
For α-iron (BCC), the diffusion coefficient = pre-exponential value,Ao × e^( -Activation energy,AE)/gas constant,R × Temperature.
Converting the given Temperature, that is 900°C to Kelvin which is equals to 1173.15K.
For α-iron (BCC), the pre-exponential value, Ao = 1.1 × 10^-6, and the activation energy, AE = 87400.
Thus, we have that the diffusion coefficient = 1.1 × 10^-6 × e(-87400)/1173.15 × 8.31.
Diffusion coefficient for α-iron (BCC) = 1.41 × 10^-10 m^2/s.
Also, For the γ-iron (FCC), the pre-exponential value, Ao = 2.3 × 10^-5 and the activation energy, AE = 148,00.
From these values we can see that both the exponential value, Ao and the activation energy for γ-iron (FCC) are higher than that of α-iron (BCC).
Thus, the diffusion coefficient for the γ-iron (FCC) = 2.3 × 10^-5 × e ^-(14800)/8.31 × 1173.15.
Then, the diffusion coefficient for the γ-iron (FCC) = 5.87 × 10^-12 m2/s.
Therefore, there will be faster diffusion in α-iron (BCC) because of lower activation energy and vice versa.
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Answer:
I would say it is true
Explanation:
Determine the molar mass of CaO
The natural abundance for boron isotopes is 19.9% 10B and 80.1%
11B Calculate boron's atomic mass.
Answer:
10.801 amu
Explanation:
From the question given above, the following data were obtained:
Isotope A (¹⁰B):
Mass of A = 10
Abundance (A%) = 19.9%
Isotope B (¹¹B):
Mass of B = 11
Abundance (B%) = 80.1%
Atomic mass of Boron =?
The atomic mass of boron can be obtained as illustrated below:
Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
= [(10 × 19.9)/100] + [(11 × 80.1)/100]
= 1.99 + 8.811
= 10.801 amu
Thus, the atomic mass of boron is 10.801 amu
The atomic mass of boron with natural abundance of 19.9% of 10 B and 80.1% of 11 B is 10.801 amu
Boron has 2 isotopes.
First isotopes
mass = 10
% abundance = 19.9%
Second Isotopes
mass = 11
% abundance = 80.1%
Therefore,
Atomic mass = (19.9% of 10) + (80.1% of 11)
Atomic mass = (19.9 / 100 × 10) + (80.1 / 100 × 11)
Atomic mass = 199 / 100 + 881.1 / 100
Atomic mass = 1.99 + 8.811
Atomic mass = 10.801
Atomic mass = 10.801 amu
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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.
Which of these would be classified as a chemical property that could be measured quantitatively?
Volume
Flammability
рн
Solubility
Answer:
рн
Explanation:
From the given choices, pH is the only chemical property that can be measured quantitatively.
A chemical property is one that tells us about what a substance can do as regards to whether or not the substance reacts with other substances.
Examples are flammability, rusting of iron, precipitation, decomposition of water e.t.c
pH is the degree of acidity or alkanility of a solution. It is usually determined quantitatively using a pH scaleThe scale is graduated from 1 to 14 1 to 7 is for acids7 is for neutral compounds7-14 is for alkalines.the half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, how many grams were in the original sample?
Answer:
768g
Explanation:
We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.
The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram were in the original sample.
What is half life?The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.
The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.
ln P = kt + C
P = amount of I-137 at time t
C = constant
k = 1/time
t = time
1st condition:
P = Po, t = 0 days
2nd condition: (half-life)
P = 0.5Po, t = 8.07 days
3rd condition:
P = 25 grams, t = 40.35 days
Po = 800 grams
mass of I-137 = 800 gram
Therefore, 800 gram were in the original sample.
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Look at the image HOH. What does this image represent?
Answer:
Formation of covalent bond structures. The image is essentially a Lewis dot structure.
The image of HOH has depicted the covalent bond formation in water. Thus, option B is correct.
The image has been the Lewis structure of the water molecule. It has been consisted of two hydrogen atoms bonded to oxygen molecules.
The oxygen has a presence of 4 dots that have been the representation of the valence electrons. Thus, oxygen has been consisted of 2 lone pairs.
The water molecule has the presence of shared electrons between hydrogen and oxygen. The bond formed by the sharing of electrons has been covalent.
Thus, the image of HOH has been the representation of the formation of covalent bonds in water. Thus, option B is correct.
For more information about Lewis structure, refer to the link:
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Y-Chart
Explain how an invasive species can influence each aspect of an ecosystem.
Answer:
Invasive species are an organism that causes ecological or economic harm in a new environment where it's not native.
Explanation:
An invasive species can harm both the natural resources in the ecosystem as well it threaten the human use of these resources and invasive species can be introduced to a new area via the ballast water of oceangoing ships, intentional and accidental releases of aquaculture species, aquarium specimens or bait, and etc.
Invasive species is capable of causing extinctions to native plants and animals, reducing biodiversity, competing with native organisms for limited resources, and altering habitats. This can also result a huge economic impacts and fundamental disruptions of coastal and the great lakes of the ecosystems.
I hope it helps you.
A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 116.9 CWhat is the specific heat (SH) of the substance?
Answer:
Approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].
Explanation:
The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)
Calculate the increase in the temperature of this sample:
[tex]\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C[/tex].
The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)
It took [tex]2911\; \rm J[/tex] of energy to raise the temperature of this sample by [tex]\Delta T = 92.6\; \rm ^\circ\! C[/tex]. Therefore, raising the temperature of this sample by [tex]1\; \rm ^\circ\! C[/tex] (unit temperature) would take only [tex]\displaystyle \frac{1}{92.6}[/tex] as much energy. That corresponds to approximately [tex]31.436\; \rm J[/tex] of energy.
On the other hand, the energy required to raise the temperature of this material by [tex]1\; \rm ^\circ\! C[/tex] is proportional to the mass of the sample (also assuming no phase change.)
It took approximately [tex]31.436\; \rm J[/tex] of energy to raise the temperature of [tex]57.07\; \rm g[/tex] of this material by [tex]1\; \rm ^\circ C[/tex]. Therefore, it would take only [tex]\displaystyle \frac{1}{57.07}[/tex] as much energy to raise the temperature of [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C\![/tex]. That corresponds to approximately [tex]0.551\; \rm J[/tex] of energy.
In other words, it takes approximately [tex]0.551\; \rm J[/tex] to raise [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C[/tex]. Therefore, by definition, the specific heat of this material would be approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].
The reaction between HCl and KOH results in an increase in temperature in the solution. Select the correct statement from the list below.
a) this is an endothermic reaction
b) this is a phase change reaction
c) this is a vaporization reaction
d) this is an exothermic reaction
Answer:
d) this is an exothermic reaction.
Explanation:
The reaction between HCl and KOH results in an increase in temperature in the solution. Select the correct statement from the list below.
a) this is an endothermic reaction . NO. This would cause a decrease in the temperature of the solution.
b) this is a phase change reaction . NO. All the species remain in the aqueous phase.
c) this is a vaporization reaction . NO. All the species remain in the aqueous phase.
d) this is an exothermic reaction. YES. The reaction releases heat, so it is exothermic.
Please help!!
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Explanation:
U need to draw the graph first and make a line at 17 pennies, where the line of 17 pennies and your graph meet is the mass of it(at y axis)
The radius of Pt atom in an fcc structure is 0.1386 nm. Pt has atomic mass of 195.09 g/mol. Calculate the density of this fcc structure. The Avogadro’s number NAis 6.022 x 1023 per mole
Answer:
21.51g/cm³
Explanation:
To answer this question we need to know that, in 1 unit FCC cell:
Edge length = √8 * R
Volume = 8√8 * R³
And there are 4 atoms per unit cell
The mass of the 4 atoms of the cell, in grams, is:
4 atom * (1mol / 6.022x10²³atom) * (195.09g / mol) = 1.2958x10⁻²¹g
Volume in cm³:
0.1386nm * (1x10⁻⁷cm / 1nm) = 1.386x10⁻⁸cm
Volume = 8√8 * (1.386x10⁻⁸cm)³
Volume = 6.02455x10⁻²³cm³
And density, the ratio of mass and volume, is:
1.2958x10⁻²¹g / 6.02455x10⁻²³cm³ =
21.51g/cm³
calculate the equilibrium concentration for the nonionized bases and all ions in a solution that is 0.25M
Answer:
The equilibrium concentration of [CH₃NH₂] = 0.23965 M.
The equilibrium concentration of CH₃NH₃⁺ and OH⁻ = 0.01035 M respectively.
Explanation:
The first step is to write out the dissociation reaction. Therefore, the equation showing the dissociation is given as below;
CH₃NH₂ + H₂0 <--------------------------------------------------------> CH₃NH₃⁺ + OH⁻.
Kindly note that ''<----------->'' arrow shows that the reaction is an equilibrium reaction.
Therefore, at the start of the reaction [that is time, t =0], we have that the concentration of CH₃NH₂ = 0.25M, thus, the concentration of CH₃NH₃⁺ and OH⁻ is zero respectively at this time, t =0.
At equilibrium, the concentration of CH₃NH₂ = 0.25M - x, thus, the concentration of CH₃NH₃⁺ and OH⁻ is x respectively.
Therefore, kb = 4.47 × 10-4 = [CH₃NH₃⁺ ][OH⁻]/[CH₃NH₂]. Hence, slotting in the values into this equilibrium equation showing the relationship between kb and concentration of the species involved, we have that;
kb = 4.47 × 10⁻⁴ = x² /0.25 - x.
x² + 4.47 × 10⁻⁴x - 1.1175 × 10⁻⁴ = 0.
Solving this quadratic equation gives us the value of x as 0.01035 M.
Thus, the concentration of [CH₃NH₂] = 0.25 M - 0.01035 M = 0.23965 M
The equilibrium concentration of [CH₃NH₂] = 0.23965 M.
The equilibrium concentration of CH₃NH₃⁺ and OH⁻ = 0.01035 M respectively.
calculate the mass of N2 gas which has a volume 0.227 at STP