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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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.
A student asks why the ashes from a fire have a much lower mass than the wood that was burned.
Which is the correct answer to the student’s question?
Gases are released into the air.
Atoms in the wood are destroyed.
Heat causes the molecules to change color.
Water inside the wood solidifies.
(I am not in college)
calculate the mass of N2 gas which has a volume 0.227 at STP
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.
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.
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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.
Helpppp!!!!! Please help me
Answer:
10.)Weight and mass are different because only weight describes the force of gravity on an object.
11.)On a planet with more mass than earth your weight would be higher because greater gravitational pull.
On a planet with less mass than earth your weight would be less because of the lower gravitational pull.
Explanation:
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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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]
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plsss help!!!!! I'll give u brainlest and 10 points
Answer:
I would say it is true
Explanation:
Select the term that matches each definition:
a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion.
b) The mass of a salt in grams that will dissolve in 100 mL of water.
c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved.
d) The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation.
e) The maximum number of moles of a salt that will dissolve in 1 L of solution.
*** Answer options for all questions: ***
- Solubility
- Molar Solubility
- Solubility product constant
- Common ion effect
- Saturated Solution
Answer:
a) common ion effect
b) solubility
c) saturated solution
d) solubility product constant
e) molar solubility
Explanation:
When a substance, say BA2 is dissolved in a solution and another substance CA2 is dissolved in the same solution. The solubility of BA2 is decreased due to the addition of CA2. This is known as common ion effect.
The mass of a substance that will dissolve in a given Volume of solvent is known as it's solubility.
The molar solubility is the amount of moles of solvent that dissolves in 1 dm^3 of solvent.
A solution that contains just as much solute as it can normally hold at a given temperature is known as a saturated solution.
Lastly, the product of molar solubilites raised to the power of the molar coefficient is know as the solubility product constant.
The correct matches and their explanation are:
a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion: Option 4. common ion effect
It relates to the equilibrium effect that occurs due to the addition of common ions.b) The mass of salt in grams that will dissolve in 100 mL of water: Option 1. solubility
Solubility is the property of solute to form a solution with the solvent of another substance.c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved: Option 5. saturated solution
When the solution cannot accommodate any more addition of solute of a substance is called a saturated solution.d) The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation: Option 3. solubility product constant
It is an equilibrium constant for the solid solute dissolved in the solution.e) The maximum number of moles of a salt that will dissolve in 1 L of solution: Option 2. molar solubility
Before the saturation of a solution, the amount of solute it can dissolve is called molar solubility.To learn more about molar solubility and common ion effect follow the link:
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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.
2.
(3x – 4y = -10
(6x + 3y = –42
SOLUTION:
Answer:
Is like for solving for the solution for both equations??
To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?
Answer: 600 mL
Explanation:
Given that;
M₁ = 5.85 m
M₂ = 1.95 m
V₁ = 200 mL
V₂ = ?
Now from the dilution law;
M₁V₁ = M₂V₂
so we substitute
5.85 × 200 = 1.95 × V₂
1170 = 1.95V₂
V₂ = 1170 / 1.95
V₂ = 600 mL
Therefore final volume is 600 mL
which element Shows very similar chemical properties to barium?
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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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
Give Me An Atom With The Following Characteristics Lanthanide series
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].
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:
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
A stream of oxygen enters a compressor at 298 K and 1.00 atm at a rate of 127m3/h and is compressed to 358 K and 1000 atm. Estimate the volumetric flow rate of compressed O2 using the compressibility-factor equation of state.
Answer:
The value is [tex]V_2 = 0.246 \ m^3/h[/tex]
Explanation:
From the question we are told that
The temperature at which the gas enters the compressor is
[tex]T_i = 298 \ K[/tex]
The pressure at which the gas enters the compressor is
[tex]P_I = 1.0 \ atm[/tex]
The volumetric rate at which the gas enters the compressor is
[tex]V = 127 m^3/h[/tex]
The temperature to which the gas is compressed to is
[tex]T_f = 358 \ K[/tex]
The pressure to which the gas is compressed to is
[tex]P_f= 1000 \ atm[/tex]
Generally the volumetric flow rate of compressed oxygen is evaluated from the compressibility-factor equation of state as
[tex]V_2 = V_1 *\frac{z_2}{z_1} * \frac{T_2}{T_1} * \frac{P_1}{P_2}[/tex]
Here [tex]z_1[/tex] is the inflow compressibility factor with value [tex]z_1 = 1[/tex]
Here [tex]z_1[/tex] is the outflow compressibility factor with value [tex]z_2 = 1.61[/tex]
So
[tex]V_2 = 127*\frac{1.61}{1} * \frac{358}{298} * \frac{1}{1000}[/tex]
[tex]V_2 = 0.246 \ m^3/h[/tex]
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.
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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.
What is the density of a block of gold that occupies 1000 ml and has a mass of 3.5 kg? Show your work
Answer:
We are given:
mass of the block = 3500 grams
volume of the block = 1000 mL
Finding the density:
Density = mass of the object (in grams) / volume of the object (in mL)
Density = 3500 / 1000
Density = 3.5 g / mL
PLEASE PLEAS HELP Which of the following compounds is insoluble in water?
a) ZnSO4
b) K2SO4
c) Na2CO3
d) Ag2CO3
Answer:
Your answer is d
Explanation: silver carbonate Ag2CO3 is insoluble in water
13. (6C) An unknown chemical has the
following properties: it is a white crystal,
reacts with water, and has a high boiling
point. Which of these properties is
physical?
A White color only
B Reacts with water, high
boiling point
C Reacts with water only
D White color, high boiling
point
Answer: it is B
Explanation:
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