Answer:
Q = 1.08x10⁻¹⁰
Yes, precipitate is formed.
Explanation:
The reaction of CoF₂ with NaOH is:
CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).
The solubility product of the precipitate produced, Co(OH)₂, is:
Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)
And Ksp is:
Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²
Molar concentration of both ions is:
[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = 3.79x10⁻⁴M
[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = 5.35x10⁻⁴M
Reaction quotient under these concentrations is:
Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²
Q = 1.08x10⁻¹⁰
As Q > Ksp, the equilibrium will shift to the left producing Co(OH)₂(s) the precipitate
A sample of krypton gas occupies a volume of 9.87 L at 51°C and 0.565 atm. If it is desired to decrease the volume of the gas sample to 8.05 L, while increasing its pressure to 0.678 atm, the temperature of the gas sample at the new volume and pressure must be___________ °C.
Answer:
44.1°C
Explanation:
Step 1:
Data obtained from the question include:
Initial volume (V1) = 9.87 L
Initial temperature (T1) = 51°C = 51°C + 273 = 324K
Initial pressure (P1) = 0.565 atm
Final volume (V2) = 8.05 L
Final pressure (P2) = 0.678 atm
Final temperature (T2) =..?
Step 2:
Determination of the new temperature of the gas.
The new temperature of the gas can be obtained by using the general gas equation as shown below:
P1V1 /T1 = P2V2 /T2
0.565 x 9.87/324 = 0.678 x 8.05/T2
Cross multiply
0.565 x 9.87 x T2 = 324 x 0.678 x 8.05
Divide both side by 0.565 x 9.87
T2 = (324 x 0.678 x 8.05)/(0.565 x 9.87)
T2 = 317.1K
Step 3:
Conversion of 317.1K to decree celsius.
This is illustrated below:
T (°C) = T(K) – 273
T (K) = 317.1K
T (°C) = 317.1 – 273
T (°C) = 44.1°C
Therefore, the new temperature of the gas is 44.1°C
The diagram below shows that the periodic table is divided into different blocks.
A periodic table is shown. The main table consists of seven rows; two additional rows are shown below. In each block, the first column is labeled and the remaining columns are empty. The s-block is shaded in yellow and comprises the first two columns, plus one cell at the far side of the table. The first column has seven rows with entries 1 s, 2 s, 3 s, 4 s, 5 s, 6 s, and 7 s. A lone cell labeled 1 s appears at the top far right corner, aligned with the 1 s cell in the first column. The d-block is shaded in blue and contains 10 columns and 3 or 4 rows. The first column is directly to the right of the s-block. The first entry in the first d-block column aligns with the 4 s block, and is labeled 3d; further entries in that column are 4 d, 5 d, and 6 d. The first three columns in the block are four entries long; the remaining columns are three entries long, losing the bottom entry. The p-block is shaded in orange, and has 6 columns and 5 rows. The top row aligns with the 2 s block; entrie
Elements that have complete valence electron shells are mostly found in the
s block.
d block.
p block.
Answer:
p block.
Explanation:
jus took the test
Answer:
c p block
Explanation:
What is the Lewis structure for *OPCl3 and AlCl6^3-? What are their electron/molecular geometry and Ideal Bond Angle ?
Answer:
Here's what I get
Explanation:
1. POCl₃
(a) Lewis structure
Set P as the central atom, with O and Cl atoms directly attached to it.
Electrons available = P + O + 3Cl = 5 + 6 + 3×7 = 11 + 21 = 32
Arrange these electrons to give every atom an octet. Put a double bond between P and O.
You get the structure shown below.
(b) Geometry
There are four bond pairs and no lone pairs about the P atom.
Electron pair geometry — tetrahedral
Molecular geometry — tetrahedral
(c) Ideal bond angles
Tetrahedral bond angle = 109.5°
2. AlCl₆³⁻
(a) Lewis structure
Set Al as the central atom, with the Cl atoms directly attached to it.
Electrons available = Al + 6Cl + 3(-) = 3 + 6×6 +3 = 6 + 36 = 42
Arrange these electrons to give every atom an octet. Assign formal charges.
You get the structure shown below.
(b) Geometry
There are six bond pairs and no lone pairs about the Al.
Electron pair geometry — octahedral
Molecular geometry — octahedral
(c) Ideal bond angles
Axial-equatorial = 90°
Equatorial-equatorial = 120°
Axial-axial = 180°
Describe the buffer capacity of the acetic acid buffer solution in relation to the addition of both concentrated and dilute acids and bases.
Answer:
The answer is in the explanation
Explanation:
Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.
That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.
When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:
CH₃COO⁻ + HX → CH₃COOH + X⁻
For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.
Now, if a base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:
CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.
In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.
The enthalpy change for the complete burning of one mole of a substance
is the enthalpy of _______
thermochemical equation
combustion
released
vaporization
fusion
absorbed
heat
Answer:
combustion
Explanation:
The enthalpy change for the complete burning of one mole of a substance
is the enthalpy of __combustion_____ .
How many moles of CO2 can be produced by the complete reaction of 1.0 g of lithium carbonate with excess hydrochloric acid (balanced chemical reaction is given below)? Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g) Question 1 options: 1.7 g 1.1 g 0.60 040 g
Answer:Mass of CO2 = 0.60g
Explanation:
Given the chemical rection
Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g
No of moles = mass / molar mass
molar mass Li2CO3 = Molecular mass calculation: 6.941 x 2 + 12.0107 + 15.9994 x 3 =
= 73.8909 g/mol
therefore Number of moles Li2CO3 = 1.0g / 73.89 g/mol
= 0.0135 moles Li2CO3
From our given Balanced equation, shows that
Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g
1 mole Li2CO3 produces 1 mole CO2
therefore 0.0135 mol Li2CO3 will produce 0.0135 moles of CO2
Also
No of moles = mass / molar mass
Mass = No of moles x molar mass
molar mass of CO2=12.0107 + 15.9994 x 2=44.0095 g/mol
Mass of CO2= 0.0135 X 44.0095 g/mol =0.594≈0.60g
Give the characteristic of a zero order reaction having only one reactant. a. The rate of the reaction is not proportional to the concentration of the reactant. b. The rate of the reaction is proportional to the square of the concentration of the reactant. c. The rate of the reaction is proportional to the square root of the concentration of the reactant. d. The rate of the reaction is proportional to the natural logarithm of t
Answer:
a. The rate of the reaction is not proportional to the concentration of the reactant.
Explanation:
The rate expression for a zero order reaction is given as;
A → Product
Rate = k[A]⁰
[A]⁰ = 1
Rate = K
GGoing through the options;
a) This is correct because in the final form of the rate expression, the rate is independent of the concentration.
b) This option is wrong
c) This option is also wrong
d) Like options b and c this is also wrong becaus ethere is no relationship between either the concentration or t.
Consider the reaction of aqueous potassium sulfate with aqueous g silver nitrate based on the solubility rule predict the product likely to be precipitate write a balanced molecular equation describing the reaction.
Answer:
K₂SO₄(aq) + 2AgNO₃ (aq) → 2KNO₃(aq) + Ag₂SO₄ (s) ↓
2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓
Explanation:
Our reactants are: K₂SO₄ and AgNO₃
By the solubility rules, we know that sulfates are insoluble when they react to Ag⁺, Pb²⁺, Ca²⁺, Ba²⁺, Sr²⁺, Hg⁺
We also determine, that salts from nitrate are all soluble.
The reaction is:
K₂SO₄(aq) + 2AgNO₃ (aq) → 2KNO₃(aq) + Ag₂SO₄ (s) ↓
2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓
In electrophilic aromatic substitution reactions, a chlorine substituent is ________. In electrophilic aromatic substitution reactions, a chlorine substituent is ________. an activator and a meta-director a deactivator and an ortho,para-director an activator and an ortho,para-director a deactivator and a meta-director none of the above
Answer:
d) is an activator and an ortho/para- director
Explanation:
We options for this question are:
a) is a deactivator and an ortho/para- director
b) is a deactivator and a meta-director
c) is an activator and a meta-director
d) is an activator and an ortho/para- director
e) is an activator and an ortho/meta/para director
f) none of the above
We have to remember that Cl is a halogen, therefore a lone pair of electrons will be involved in the resonance (see figure 1). So, we will have activation in the ortho and para positions. We have to take into account that the halogens have a high electronegativity. This means that inductively they are electron-withdrawing. So, we will have opposite phenomenons that will result in very weak activation
Enter an equation for the formation of C2H5OH(l) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)
Explanation:
Ethanol (C₂H₅OH) is an alcohol and it is formed by carbon (C), H (hydrogen) and O (oxygen) atoms. These elements in their standard states are:
C: C(s), it is solid, could be graphite, diamond, among others.
H: H₂(g), it is a diatomic gas.
O: O₂(g), it is a diatomic gas.
So, we can write the equation for the formation of C₂H₅OH from C(s), H₂ and O₂ as follows:
C(s) + H₂(g) + O₂(g) ⇒C₂H₅OH(l)
Finally, we have to balance the equation by adding the estequiometrical coefficients:
C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)
2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l)
Explanation:
Standard state of carbon: C(s)
Standard state of oxygen: [tex]O_{2} [/tex](g)
Standard State of Hydrogen: [tex]H_{2} [/tex](g)
Then balance the equation C2H5OH(l) to get 2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l).
the reaction below is at equilibrium. What would happen if more carbon were added ?
Answer:
B
Explanation:
If more carbon were added, the equilibrium position would shift to produce more products.
What is equilibrium?In a chemical reaction chemical equilibrium is defined as the state at which there is no further change in concentration of reactants and products.
If the concentration of one (or more) of the reactants or products is increased the equilibrium will shift to decrease the concentration.
Or if the temperature is decreased the equilibrium will shift to increase the temperature by favouring the exothermic reaction.
Hence, if more carbon were added, the equilibrium position would shift to produce more products.
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Which scenario describes an interaction between two of Earth's spheres?
Water flows from a stream to a lake.
Gravity moves rocks to another location.
Lions use energy to catch other animals for food.
Bears dig big holes in the ground to protect their young.
The correct answer is D. Bears dig big holes in the ground to protect their young
Explanation:
The Earth spheres include the biosphere (life in the Earth), the hydrosphere (water bodies), the geosphere (rocks and other elements that compose land and soil), and the atmosphere (gases that compose the air). In this context, there is an interaction between two spheres: the biosphere and the geosphere, when a bear digs holes in the ground because a living organism that is part of the biosphere is modifying the structure and shape of superficial soil, which is part of the geosphere.
Answer: D
Explanation:
A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitrogen is 75 K. What is the heat capacity of the sample in J/K
Answer:
heat capacity of the sample = 37.8 J/K
Explanation:
Step 1: Data given
Temperature of the sample = 275 K
The mass of liquid nitrogen = 2kg
temperature of liquid nitrogen = 70 K
The final temperature of the nitrogen is 75 K
Step 2: Calculate heat
Q = m*c*ΔT
⇒with m = the mass of liquid nitrogen = 2 kg = 2000 grams
⇒with c= the specific heat of the liquid nitrogen = 1.04 J/g*K
⇒with ΔT = the change of temperature of liquid nitrogen = T2 - T1 = 75 - 70 = 5K
Q = 2000 grams * 1.04 J/g*K * 5K
Q = 10400 J
Step 3: Calculate the heat capacity of the sample
heat capacity of the sample = 10400 J / 275 K
heat capacity of the sample = 37.8 J/K
Suppose that you add 27.6 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.69 oC compared to pure benzene. What is the molar mass of the unknown compound
Answer:
The molar mass of the unknown compound is 153.3 g/mol
Explanation:
Step 1: Data given
Mass of an unknown molecular compound = 27.6 grams
Mass of benzene = 0.250 kg
Kf of benzene = 5.12 °C/m
freezing point depression of 3.69 °C
Step 2: Calculate molality
ΔT = i*Kf*m
⇒with ΔT = reezing point depression of 3.69 °C
⇒with i = the van't Hoff factor of Benzene = 1
⇒with Kf = 5.12 °C/m
⇒ with m = molality = moles unknown compound / mass of benzene
3.69 = 1 * 5.12 * m
m = 0.72 molal
Step 3: Calculate moles of the unknown compound
molality = moles / mass benzene
0.72 molal = moles / 0.250 kg
Moles = 0.72 m * 0.250 kg
Moles = 0.18 moles
Step 4: Calculate molar mass of the unknown compound
molar mass = mass / moles
Molar mass = 27.6 grams / 0.18 moles
Molar mass = 153.3 g/mol
The molar mass of the unknown compound is 153.3 g/mol
Molar mass is the mass of the one mole of substance. The molar mass of the given unknown compound is 153.3 g/mol.
Molality of the compound can be calculated using
ΔT = i Kf m
Where,
ΔT = freezing point depression = 3.69 °C
i = Van't Hoff factor of Benzene = 1
Kf = constant of freezing = 5.12 °C/m
m = molality = ?
Put the values in the equation,
3.69 = 1 x 5.12 x m
m = 0.72 molal
Number of moles of the compound,
[tex]\bold {molality =\dfrac { moles} { mass\ benzene}}\\\\\bold {0.72\ molal = \dfrac {moles }{0.250\ kg}}\\\\\bold {Moles = 0.72\ m \times 0.250\ kg}\\\\\bold {Moles = 0.18}[/tex]
So, molar mass of the unknown compound,
[tex]\bold {Molar\ mass =\dfrac { mass}{ moles}}\\\\\bold {Molar\ mass = \dfrac {27.6\ grams }{0.18\ moles}}\\\\\bold {Molar\ mass = 153.3 g/mol}[/tex]
The molar mass of the given unknown compound is 153.3 g/mol.
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A solution of benzene in methanol has a transmittance of 93.0 % in a 1.00 cm cell at a wavelength of 254 nm. Only the benzene absorbs light at this wavelength, not the methanol. What will the solution's transmittance be if it is placed in a 10.00 cm long pathlength cell
Answer:
T = 48.39%
Explanation:
In this case we need to apply the Beer law which is the following:
A = CεL (1)
Where:
A: Absorbance of solution
C: Concentration of solution
ε: Molar Absortivity (Constant)
L: Length of the cell
Now according to the given data, we have transmittance of 93% or 0.93. We can calculate absorbance using the following expression:
A = -logT (2)
Applying this expression, let's calculate the Absorbance:
A = -log(0.93)
A = 0.03152
Now that we have the absorbance, let's calculate the concentration of the solution, using expression (1).
A = CεL
C = A / εL
Replacing:
C = 0.03152 / 1 *ε (3)
Now, we want to know the transmittance of the solution with a length of 10 cm. so:
A = CεL
Concentration and ε are constant, so:
A = (0.03152 / ε) * ε * 10
A = 0.3152
Now that we have the new absorbance, we can calculate the new transmittace:
T = 10^(-A)
T = 0.4839 ----> 48.39%
what’s the SI unit of time ?
Answer:
The answer is A
Explanation:
What is Key for the reaction 2503(9) = 2802(9) + O2(g)?
Answer:
Option C. Keq = [SO2]² [O2] /[SO3]²
Explanation:
The equilibrium constant keq for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, let us determine the equilibrium constant for the reaction given in the question.
This is illustrated below:
2SO3(g) <==> 2SO2(g) + O2(g)
Reactant => SO3
Product => SO2, O2
Keq = concentration of products /concentration of reactants
Keq = [SO2]² [O2] /[SO3]²
hno3 express your answer using four significant figures
Answer:
[tex]M=63.02g/mol[/tex]
Explanation:
Hello,
In this case, for the calculation of the molar mass of nitric acid, we should employ the following formula, knowing that there is one hydrogen atom, one nitrogen atom and three oxygen atoms:
[tex]M=m_H+m_N+3*m_O[/tex]
Now, we use the atomic mass of each atom to compute the total molar mass:
[tex]M=1.008g/mol+14.01g/mol+3*16.00g/mol\\\\M=63.02g/mol[/tex]
Best regards.
Please what's the missing minor products? And kindly explain in your own words how they were formed. Thank you!
Answer:
it's a two step elimination reaction
Explanation:
it follows a carbocationic pathway. When carbocation is stable, the equation is favourable, that is, double bond is formed by expelling hydrogen atom.
A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution were required? View Available Hint(s) A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution were required? 5.74 mL 0.315 mL 793 mL 315 mL
Answer:
315mL
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 0.135 M
Volume of stock solution needed (V1) =?
Molarity of diluted solution (M2) = 0.0851 M
Volume of diluted solution (V2) = 500mL
The volume of the stock solution needed can be obtain as follow:
M1V1 = M2V2
0.135 x V1 = 0.0851 x 500
Divide both side by 0.135
V1 = (0.0851 x 500) / 0.135
V1 = 315mL
Therefore, the volume of the stock solution needed is 315mL
El ejemplo más claro para definir una cadena de electricidad es una red de pesca, si fuera posible identificar una sola partícula representativa de esta fuerza , ¿ Cuál partícula se identificaría? *
Bosón W
Fotón
Glúon
Leptón
Answer:
Leptón
Explanation:
Lepton son partículas elementales de espín de medio entero (espín 1⁄2), y se sabe que no experimentan interacciones fuertes. Los leptones se clasifican en leptones cargados (los leptones similares a los electrones) y leptones neutros (conocidos como neutrinos). Un electrón es un leptón. Los leptones cargados pueden combinarse con otras partículas para formar átomos compuestos de partículas y positronio, mientras que los neutrinos rara vez interactúan con algo, lo que los hace raros de observar.
Dado que la electricidad es el resultado del movimiento o flujo de electrones, la única partícula representativa de esta fuerza es el leptón.
18. Investigue por qué en el bloque "d" se aplica la fórmula n-1 y en el bloque "f" n-2.
Answer:
La razón de esto es porque los elementos en el bloque d tienen sus electrones d en la capa (n-1) (penúltima capa) y puede usar electrones de la capa (n-1) (penúltima capa) para formar enlaces químicos.
Los elementos en el bloque f tienen sus electrones f en sus capas (n-2) y pueden llegar a usar electrones de la capa (n-2) (capa antes de la penúltima capa) para enlaces químicos.
Explanation:
Para explicar esto, comencemos con los elementos de bloque syp; Los elementos de bloque syp usan electrones que tienen el mismo número cuántico principal para enlaces químicos. Por ejemplo, el aluminio utiliza los electrones 3s más externos y el carbono los electrones 2s y 2p cuando forman enlaces químicos. A pesar de que el carbono (bloque p) utiliza electrones de dos conjuntos de orbitales (2s, 2p) para la unión covalente, su número cuántico principal es el mismo (2) ya que todavía provienen de la misma capa.
Pero, los metales de transición (elementos de bloque d) usan electrones del orbital "s" de la capa más externa y los orbitales "d" de la penúltima capa forman enlaces químicos. Por ejemplo, los elementos de la primera serie de transición como el manganeso, el cobre y el hierro pueden usar los electrones 4s más externos y los electrones 3d de la penúltima capa, con diferentes números cuánticos principales (4 y 3). Eso significa que los metales de transición pueden usar tanto el orbital ‘ns’ más externo como los orbitales d (n-1) de la penúltima caparazón.
Luego, para los metales de transición internos (elementos de bloque f), tienen sus electrones f en sus capas (n-2) y pueden llegar a usar electrones de la capa (n-2) (capa antes de la penúltima capa) para enlaces químicos.
¡¡¡Espero que esto ayude!!!
g The atomic mass of an element is equal to ________. The atomic mass of an element is equal to ________. its mass number one-twelfth of the mass of a carbon-12 atom a weighted average mass of all of the naturally occurring isotopes of the element its atomic number the average mass of all of the naturally occurring isotopes of the element
Answer:
Total numbe of protons and neutrons in a single atom of that element
Explanation:
Hello,
I'll answer the question by filling in the blank spaces
"The atomic mass of an element is equal to the total number of proton and neutron in a particular atom of the element. The atomic mass of an element is equal to the atomic weight. Its mass number one-twelfth of the mass of carbon-12 atom a weighted mass of all naturally occurring isotopes of the elements. Its atomic mass is the average mass of all the naturally occurring isotopes of the element."
The atomic mass of an element is the total number of protons and neutrons in a single atom of that element.
The atomic mass of an element is equal to a weighted average mass of all of the naturally occurring isotopes of the element. The correct answer is option 2.
Isotopes are elements with the same number of protons (atomic number) but differing numbers of neutrons (mass number).
Most elements exist in nature as a mixture of isotopes, each with a different mass number and abundance. The atomic mass of an element is computed by adding the masses of all isotopes, multiplying by their relative abundance, and dividing by the total abundance of all isotopes.
This gives a weighted average mass that corresponds to the normal mass of an element's atom in nature.
Therefore, the correct answer is option 2. to a weighted average mass of all of the naturally occurring isotopes of the element.
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The following reactions all have K < 1. 1) HCOO- (aq) + C6H5COOH (aq) HCOOH (aq) + C6H5COO- (aq) 2) C9H7O4- (aq) + C6H5COOH (aq) C6H5COO- (aq) + HC9H7O4 (aq) 3) HCOOH (aq) + C9H7O4- (aq) HC9H7O4 (aq) + HCOO- (aq) Arrange the substances based on their relative acid strength.
Answer:
Explanation:
C₉H₇O₄⁻ = weakest base
C₆H₅COO⁻ = strongest base
HCOO⁻ = intermediate base
HCOOH = not a Bronsted-Lowry base
HC₉H₇O₄ = not a Bronsted-Lowry base
C₆H₅COOH = not a Bronsted-Lowry base
A solution is prepared by mixing 5.00 mL of 0.100 M HCl and 2.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?
Answer:
0.129 M
Explanation:
0.100 M HCl = 0.100 mol/L solution HCl
5.00 mL = 0.00500 L solution HCl
0.100 mol/L HCl * 0.00500 L = 0.000500 mol HCl
HCl ------> H+ + Cl-
1 mol 1 mol
0.000500 mol 0.000500 mol
0.200 M NaCl = 0.200 mol/L solution NaCl
2.00 mL = 0.00200 L solution NaCl
0.200 mol/L NaCl*0.00200 L = 0.000400 mol NaCl
NaCl ------> Na+ + Cl-
1 mol 1 mol
0.000400 mol 0.000400 mol
Chloride ion altogether (0.000500 mol + 0.000400 mol) =0.000900 mol
Solution altogether (0.00500 L+0.00200 L) = 0.00700L
Molarity (Cl-)= solute/solution = 0.000900 mol/0.00700L = 0.129 mol/L=
= 0.129 M
am i correct if not correct me please
Answer:
D. Hund's rule
Explanation:
Not sure, but I would go with Hund's since it talks about filing electrons in each orbital before you can pair them up. The reason sulfur has lower ionization is because it has one set of paired electrons which makes the orbital unstable whereas phosphorus has 3 unpaired e's which means it is more stable. Thus it is easier to remove electron from sulfur hence lower ionization energy.
Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) bromide is dissolved in it.
Answer:
Molarity = 0.0428 M = 42.8 mM
Explanation:
Step 1: Data given
Mass of nickel(II) bromide = 1.87 grams
Molar mass of nickel(II) bromide = 218.53 g/mol
Volume = 200 mL = 0.200 L
Step 2: Calculate moles of nickel(II) bromide
Moles nickel (II) bromide = mass / molar mass
Moles nickel (II) bromide = 1.87 grams / 218.53 g/mol
Moles nickel (II) bromide = 0.00856 moles
Step 3: Calculate moles nickel (II) cation
For 1 mol NiBr2 we have 1 mol Ni^2+
For 0.00856 moles NiBr2 we have 0.00856 moles Ni^2+
Step 4: Calculate final molarity of Ni^2+
Molarity = moles / volume
Molarity = 0.00856 moles / 0.200 L
Molarity = 0.0428 M = 42.8 mM
Enter your answer in the provided box. On a cool, rainy day, the barometric pressure is 739 mmHg. Calculate the barometric pressure in centimeters of water (cmH2O) (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL).
Answer:
997.65cmH2O
Explanation:
Barometric pressure = 739 mmHg
density of Hg = 13.5 g/ml
density of water (H2O) = 1.00 g/ml
Calculate Barometric pressure in centimetres of water ( cmH20)
equate the barometric pressure of Hg and water
739 * 13.5 * 9.8 = x * 1 * 9.81
x ( barometric pressure of water in mmH2O ) = 739 *13.5 / 1 = 9976.5mmH2O
in cmH2O = 997.65cmH2O
Calculate the heat change in kilojoules for condensation of 195 g of steam at 100 ° C
Answer:
Q = 81.59kJ
Explanation:
Hello,
The heat of condensation is the energy required to to convert the steam into water.
Mass = 195g
Specific heat capacity of water = 4.184J/g°C
Initial temperature(T1) = 100°C
Final temperature(T2) = 0°C
Heat energy (Q) = ?
Heat energy (Q) = mc∇T
M = mass of the substance
C = specific heat capacity of the substance
∇T = T2 - T1 = change in temperature of the substance
Q = 195 × 4.184 × (0 - 100)
Q = -81588J
Q = -81.588kJ
The heat required for the condensation of 195g of steam is 81.59kJ
Carbon dioxide and water vapor are variable gases because _____.
Answer: their amounts vary throughout the atmosphere
Explanation:
There is very little that travels over the atmosphere
Vary=very little
Hope that helps