Answer:
0.3323
Explanation:
GIven that:
Density of the metal = 10.6 g/cm^3
atomic weight = 176.8 g/mol
atmic radius = 0.130 nm
values of a and c = 0.570 nm and 0.341 nm respectively
For us to determine the atomic packing factor, we need to first determine the volume of all spheres (Vs) and the volume of unit cell (Vc).
However, the number of atoms in the unit cell (n) can be computed as:
[tex]n = \dfrac{\rho * V_c *N_A}{A} \\ \\ n = \dfrac{(10.6) * (5.7)^2 (3.41)*(10^{-24}) *(6.022*10^{23})}{176.8}[/tex]
n = 4.0
Thus, the number of atoms in the unit cell is 4
∴
The atomic paking factor (APF) is calculated by using the formula:
[tex]\dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *R^3 }{a^2 *c} \\ \\ \\ \dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *(1.30*10^{-8})^3 }{(5.70*10^{-8})^2 *(3.41*10^{-8})}[/tex]
= 0.3323
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
Answer:
Solution given:
1 mole of KCl[tex]\rightarrow [/tex]22.4l
1 mole of KCl[tex]\rightarrow [/tex]74.55g
we have
0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g
74.55g of KCl[tex]\rightarrow [/tex]22.4l
10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
[tex]\:[/tex]
1 mole of KCl → 22.4l
1 mole of KCl → 74.55g
we have
0.14 mole of KCl → 74.55*0.14=10.347g
74.55g of KCl → 22.4l
10.347 g of KCl → 22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
How many protons does Tin have?
A. 50
B. 68
C. 118
Hello There!
Tin has 50 protons.Hope that helps you!
~Just a felicitous girlie
#HaveASplendidDay
[tex]SilentNature[/tex]
A sample of gas occupies 10.0 L at 240°C under a pressure of
80.0 kPa. At what temperature would the gas occupy 20.0 L if
we increased the pressure to 107 kPa?
Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
1090 degree Celsius
hope it helps
Given the following balanced reaction: 2Na(s) + F2(g) --> 2NaF(s)
a) How many moles of NaF will be made from 2.6 moles of F2?
b) How many moles of NaF will be made from 4.8 moles of Na?
Answer:
yes it is corrwect iyt is absolitle correct
Explanation:
What volume of 1.50 mol/L stock solution is needed to make 125 mL of 0.60 mol/L solution?
Chemistry 11 Solutions
978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85
Amount in moles, n, of the NaCl(s):
NaCl
2.5 g
m
n
M
58.44 g
2
4.2778 10 m l
ol
o
/m
u
Molar concentration, c, of the NaCl(aq):
–2 4.2778 × 10 mol
0.100
0.42778 mol/L
0.43 mol
L
/L
n
c
V
The molar concentration of the saline solution is 0.43 mol/L.
Check Your Solution
The units are correct and the answer correctly shows two significant digits. The
dilution of the original concentrated solution is correct and the change to mol/L
seems reasonable.
Section 8.4 Preparing Solutions in the Laboratory
Solutions for Practice Problems
Student Edition page 386
51. Practice Problem (page 386)
Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,
(NH4)2SO4(aq).
What volume of the stock solution do you need to use to prepare each of the
following solutions?
a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)
b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)
c. 250 mL of 0.300 mol/L NH4
+
(aq)
What Is Required?
You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution
needed to prepare each given dilute solution.
The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.
What is dilution?Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.
Given,
Initial volume = V₁
Initial molar concentration (M₁) = 1.50 mol/L
Final volume (V₂) = 125 mL = 0.125 L
Final molar concentration (M₂)= 0.60 mol/L
The dilution is calculated as:
M₁V₁ = M₂V₂
V₁ = M₂V₂ ÷ M₁
Substituting the values in the above formula as
V₁ = M₂V₂ ÷ M₁
V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L
V₁ = 0.05 L
= 50 mL
Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.
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What must happen to uranium before it can be used as a fuel source?
Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.
15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67
COO(s) + CO(8) = Co(s) + CO2(8) K = 490
Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.
The equilibrium constant for the reaction is K = 0.137
We obtain the equilibrium constant considering the following equilibria and their constants:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
COO(s) + CO(g) → Co(s) + CO₂(g) K₂ = 490
We write the first reaction in the forward direction because we need H₂(g) in the reactants side:
(1) COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):
(2) Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
From the addition of (1) and (2), we obtain:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
+
Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
-------------------------------------------------
H₂(g) + CO₂(g) → CO(g) + H₂O(g)
Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.
Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:
K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137
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g Identify the process in which the entropy increases. Group of answer choices a decrease in the number of moles of a gas during a chemical reaction the phase transition from a gas to a liquid the phase transition from a solid to a gas freezing water
Answer:
phase transition from a solid to a gas
Explanation:
Entropy refers to the degree of disorderliness in a system. The more disorderly a system is, the greater the entropy of the system.
Decrease in the number of moles of a gas decreases the entropy of the system. Similarly, the entropy of solids is less than that of liquids. The entropy of liquids is less than that of gases.
Therefore, a change of phase from solid to gas represents an increase in entropy of the system.
The density of toluene (C7H8) is 0.867 and that of thiophene (C4H4S) is 1.065 g/ml. A solution is made by dissolving 10.00g thiophene in 250.00ml of toluene. a)Calculate the molarity of the solution
b)Assuming the volume are addictive ,calculate the molarity of the solution
Answer:
Calcular la molaridad de una solución que se preparó disolviendo 14 g de KOH en suficiente
agua para obtener 250 mL de solución. (masa molar del KOH = 56 g/mol).
Resolución: de acuerdo a la definición de “molaridad” debemos calcular primero, el número de mol de soluto (KOH) que
se han disuelto en el volumen dado, es decir, “se transforma g de soluto a mol de soluto” por medio de la masa molar,
así:
56 g de KOH 14 g de KOH
----------------- = ------------------- X = 0,25 mol de KOH
1 mol X
Ahora, de acuerdo con la definición de molaridad, el número de mol debe estar contenido en 1000 mL (o 1 L) de
solución, que es el volumen estándar para esta unidad de concentración, lo que se determina con el siguiente planteamiento:
0,25 mol X
----------------------- = ------------------------- X = 1 mol de KOH
250 mL de solución 1000 mL de solución
Explanation:
HELP ASAS 15 POINTS
When using the process of evaporation to separate a mixture, what is left behind in the evaporating dish?
A. None of these.
B. The liquid evaporates and the solid is left in the dish.
C. The mixture does not separate, and the entire mixture evaporates.
D. The mixture does not separate, and the entire mixture remains in the dish.
Answer:
liquid will be evaporated while solid remains
Find the volume occupied by 128g of SO2.
The volume occupied by the given amount of sulfur dioxide will be 84 L.
please show working my dear citizen
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
17. According to Newton's first law, an object in motion will stay in what kind of
motion as long as the net forces acting on it are zero?
A. slowing motion in a straight line
B. accelerating motion in a straight line
C. constant motion in a straight line
D. constant motion in an unpredictable direction
Answer:
C
Explanation:
According to Newton's first law, an object in motion will stay in motion unless acted upon by an unbalanced force. The option which gives the same meaning is :
constant motion in a straight lineAccording to Newton's first law, an object in motion as long as the net forces acting on it are zero will stay in:
C. constant motion in a straight line
A constant motion in mechanics is a number that remains constant throughout the motion, effectively placing a restriction on the motion.
What is Newton's first law of motion?The foundation of classical mechanics is made up of the three laws of motion that Newton suggested. These three physical laws. In accordance with the three laws of motion, a body's connection to the forces operating on it and the motion that resulted from those forces are both described. As Galileo's law of inertia, Newton's First Law of Motion is also referred to. If a body is not forced by an outside force to change its condition, it will remain in its state of rest or uniform motion in a straight line. As a factor that has the potential to alter an object's condition, the law defines force and mentions it.In this way, Newton uses the inertia of rest, motion, and direction to describe his first law of motion.A body has an attribute called inertia that makes it resist changes to its state. A body's inertia of translational motion is measured by its mass.To learn more about Newton's first law of motion, refer to:
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What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+
Answer:
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Explanation:
We will balance the redox reaction through the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Br₂ ⇒ Br⁻
Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
2 e⁻ + Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻
Step 4: Make the number of electrons gained and lost equal
5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)
1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)
Step 5: Add both half-reactions
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Which of the following are examples of single replacement reactions? Select all that apply.
Answer:
Na2S(aq)+Cd(No3)2(aq)=CdS(s)+2NaNo3(aq)
Answer: it’s checkbox 2&3
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
The shape of a molecule is determined by:
A. All of these
B. The number of electron clouds around the atom.
C. The number of bonds.
D. Mutual repulsion between electrons.
A sample of gas is held at constant volume. If the number of moles of this sample of gas is doubled and the pressure of this sample of gas is halved, what happens to the absolute temperature of the gas?
Select one
a. The absolute temperature is doubled.
b. The absolute temperature is halved.
c. The absolute temperature is quadrupled.
d. The absolute temperature is quartered.
e. The absolute temperature stays the same.
Answer:
number of moles of gas increases the volume also increases.
Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ
How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Answer:
endet nach selam nw
4gh7
Hãy cho biết giá trị và ý nghĩa của số lượng tử n, l, m, ms khi mô tả trạng thái của electron trong nguyên tử?
which of the following measurements is equivalent to 5.461x10^-7m?
Answer:
B. 0.0000005461m
I used the method of moving the decimal.
What is Bose Einstein state of matter and their examples
Answer:
A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.
Examples - Superconductors and superfluids are the two examples of BEC.
Explanation:
There are three isotopes of carbon. They have mass number of 12, 13 and 14. The average atomic mass of carbon is 12.0107 amu. What does this say about the relative abundances of the three isotopes?
Answer:
lots more of the carbon 12 than the others
havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14
Explanation:
Hydrogengasand oxygengas react to form water vapor. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .
The question is incomplete. The complete question is :
Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]
= 11.0 mol
what characterizes a homogeneous mixture?
Answer:
a mixture that doesn't really show the ingredients or things put into the material or food.
At 445oC, Kc for the following reaction is 0.020. 2 HI(g) <--> H2 (g) + I2 (g) A mixture of H2, I2, and HI in a vessel at 445oC has the following concentrations: [HI] = 1.5 M, [H2] = 2.50 M and [I2] = 0.05 M. Which one of the following statements concerning the reaction quotient, Qc, is TRUE for the above system?
a. Qc = Kc; the system is at equilibrium.
b. Qc is less than Kc; more H2 and I2 will be produced.
c. Qc is less than Kc; more HI will be produced.
d. Qc is greater than Kc; more HI will be produced.
Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
Heating water makes most solids in it
soluble, and it makes gases
soluble.
Increasing the pressure on a gas above water makes the gas
soluble. Compounds with comparatively stronger ionic bonds are
soluble.
Answer:
1. more
2. less
3. more
4. less
Explanation:
It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of by-product formed. What is the by-product
Answer:
Biphenyl
Explanation:
The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.
The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.
Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.
Consider the reaction below. How much heat is absorbed if 5.00 moles of nitrogen react
with excess oxygen?
2 N2 (8) + O2(g) → 2 N20 (8) AHrxn- +163.2 kJ
Explanation:
The given chemical reaction is:
[tex]2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ[/tex]
When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.
When 5.00 mol of nitrogen requires how much energy?
[tex]5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ[/tex]
Hence, the answer is 408 kJ of heat energy is required.
A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices
0.0250 M
0.400 M
0.100 M
1.00 M
Answer:
Explanation:
1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all
Answer:
[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]
The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.
1. Moles of SoluteWe are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.
We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.
Na: 22.9897693 g/mol Cl: 35.45 g/molThe chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.
NaCl: 22.9897693 + 35.45 = 58.4397693 g/molThere are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.
[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]
We are converting 5.84 grams to moles, so we multiply by that value.
[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]
Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.
[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]
[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]
[tex]0.09993194823 \ mol \ NaCl[/tex]
2. MolarityWe can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]
[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]
3. Units and Significant FiguresThe original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.
[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]
1 mole per liter is 1 molar or M. We can convert the units.
[tex]molarity \approx 0.100 \ M \ NaCl[/tex]
The molarity of the solution is 0.100 M.
