Answer:
B.9.710-11M
Explanation:
plss tell me if im wrongSuppose of potassium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the potassium acetate is dissolved in it. Round your answer to significant digits.
The question is incomplete, the complete question is;
Suppose 0.377g of potassium acetate is dissolved in 250.mL of a 57.0mM aqueous solution of ammonium sulfate.
Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the potassium acetate is dissolved in it.
Round your answer to 3 significant digits.
Answer:
0.0152 M
Explanation:
The equation of the reaction is;
2CH3COOK(aq) + (NH4)2SO4(aq)------> K2SO4(aq) + 2CH3COONH4(aq)
Number of moles of potassium acetate = 0.377g/98.15 g/mol = 0.0038 moles
Number of moles of ammonium sulphate = 250/1000L × 57 × 10^-3 = 0.014 moles
2 moles of potassium acetate yields 2 moles of ammonium acetate
Hence;
0.0038 moles of potassium acetate yields 0.0038 moles of ammonium acetate
Also
1 mole of potassium sulphate yields 2 moles of ammonium acetate
0.014 of potassium sulphate yields 0.014 × 2/1 = 0.028 moles of ammonium acetate
So potassium acetate is the limiting reactant.
Since 0.0038 moles of ammonium acetate is produced, the final concentration of potassium acetate is = 0.0038 moles of ammonium acetate/0.25L = 0.0152 M
Hence final concentration of acetate ions =0.0152 M
11th grade chemistry question will mark brainliest
2.50 g of CO2 gas is confined in a rigid cylinder at a pressure
of 4.65 atm. If 0.42 g of gas is released from the cylinder,
what is the new pressure?
Answer:
3.88 atm
Explanation:
We'll begin by calculating the number of mole of CO₂ in each case. This can be obtained as follow:
For 2.50 g of CO₂:
Mass of CO₂ = 2.5 g
Molar mass of CO₂ = 12 + (2×16) = 44 g/mol
Mole of CO₂ =?
Mole = mass / molar mass
Mole of CO₂ = 2.5 / 44
Mole of CO₂ = 0.06 mole
For 0.42 g of CO₂:
Mass of CO₂ = 2.5 g
Molar mass of CO₂ = 44 g/mol
Mole of CO₂ =?
Mole = mass / molar mass
Mole of CO₂ = 0.42 / 44
Mole of CO₂ = 0.010 mole
Finally, we shall determine the new pressure. This can be obtained as follow:
Initial mole (n₁) = 0.06 mole
Initial pressure (P₁) = 4.65 atm
Final mole (n₂) = 0.06 – 0.010 = 0.05 mole
Final pressure (P₂) =?
NOTE: Temperature and volume is constant.
P₁ / n₁ = P₂ / n₂
4.65 / 0.06 = P₂ / 0.05
Cross multiply
0.06 × P₂ = 4.65 × 0.05
0.06 × P₂ = 0.2325
Divide both side by 0.06
P₂ = 0.2325 / 0.06
P₂ = 3.88 atm
Thus, the new pressure is 3.88 atm.
What were the advantages for Mendel in using pea plants for his breeding experiments?
Answer:
He chose pea plants because they were easy to grow, could be bred rapidly, and had several observable characteristics, like petal color and pea color.
Explanation:
Qual número atômico do ferro?
Answer:
The answer is Iron.
Explanation:
I hope this helps you out. Have a nice day!
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Which statement is true about the experiments? (5 points)
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
The same amount of heat is absorbed in both the experiments because the heat absorbed depends only on the final temperature.
The heat absorbed in Trial 2 is about 3,674 J greater than the heat absorbed in Trial 1.
The heat absorbed in Trial 2 is about 5,021 J greater than the heat absorbed in Trial 1.
Answer:
Explanation:
Using the formula below to calculate the heat absorbed in each trial:
Q = m × c × ∆T
Where;
Q = amount of heat absorbed (J)
m = mass of substance (g)
c = specific heat of water (4.184J/g°C)
∆T = change in temperature (°C)
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Q = 30 × 4.184 × (40 - 0)
Q = 30 × 4.184 × 40
Q = 5,020.8J
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Answer:
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
Explanation:
Explanation:
Using the formula below to calculate the heat absorbed in each trial:
Q = m × c × ∆T
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Q = 30 × 4.184 × (40 - 0)
Q = 30 × 4.184 × 40
Q = 5,020.8J
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Q=40*4.184*30
Q=5020.8J
when (S)-3-bromo-2,3-dimethylpentane is treated with sodium chloride with water as the solvent, the products formed are ___________. *Select all that apply
Answer:
(S)-3-chloro-2,3-dimethylpentane
Explanation:
When sodium chloride is treated with bromo dimethylpentane in presence of water, then chloro dimethylpentane is formed. This chemical reaction is stable reaction in water because of its solvency.
A reaction produces 3.0 mol of gas, which occupies 1.46 L. What is the volume of the product when 4.1 mol are produced at constant temperature and pressure?
Answer:
1.99 L
Explanation:
Given that,
A reaction produces 3.0 mol of gas, which occupies 1.46 L.
We need to find the volume of the product when 4.1 mol are produced at constant temperature and pressure.
We know that,
PV = nRT
i.e.
[tex]V\propto n\\\\\dfrac{V_1}{V_2}=\dfrac{n_1}{n_2}\\\\V_2=\dfrac{V_1n_2}{n_1}\\\\V_2=\dfrac{1.46\times 4.1}{3}\\\\V_2=1.99\ L[/tex]
So, the new volume is 1.99 L.
g 0.500 L of a solution with a concentration of 0.25 M is needed. To prepare this solution, a stock solution with a concentration of 1.25 M is prepared. What volume of the stock solution is needed to create the desired solution
Answer:
0.1 L
Explanation:
From the question given above, the following data were obtained:
Concentration of stock solution (C₁) = 1.25 M
Volume of diluted solution (V₂) = 0.5 L
Concentration of diluted solution (C₂) = 0.25 M
Volume of stock solution needed solution (V₁) =?
The volume of the stock solution needed can be obtained as follow:
C₁V₁ = C₂V₂
1.25 × V₁ = 0.25 × 0.5
1.25 × V₁ = 0.125
Divide both side by 1.25
V₁ = 0.125 / 1.25
V₁ = 0.1 L
Therefore, the volume of the stock solution needed is 0.1 L
The element antimony has two stable isotopes, antimony-121 with a mass of 120.90 amu and antimony-123 with a mass of 122.90 amu. From the atomic weight of Sb = 121.76 one can conclude that: ________
antimony-123 has the highest percent natural abundance
most antimony atoms have a mass of 121.76 amu
antimony-121 has the highest percent natural abundance
both isotopes have the same percent natural abundance
Answer:
antimony-121 has the highest percent natural abundance
Explanation:
percent natural abundance;
121.76 = 120.90 x + 122.90 (1 - x)
121.76 = 120.90 x + 122.90 - 122.90x
121.76 = -2x + 122.90
121.76 - 122.90 = -2x
x= 121.76 - 122.90/ -2
x= 0.57
Where x and 1 - x refers to the relative abundance of each of the isotopes
Percent natural abundance of antimony-121 = 57 %
Percent natural abundance of antimony-123 = (1 - 0.57) = 43%
Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.
We can clearly see that antimony-121 has the highest percent natural abundance.
Is the following sets of quantum numbers valid? Give suitable reason. n=3,l=2,m=3,s=+1/2
Answer:
The set of quantum numbers is not valid
Explanation:
There are four sets of quantum numbers;
1) Principal quantum number (n) which takes on integer values from n = 1,2,3 .......
2) Azimuthal quantum number (l) which takes on values 1, 2, ....(n - 1)
3) Magnetic quantum numbers (ml) which takes on values from (-l) to (+l)
4) spin quantum number (ms) which takes on values of ±1/2.
From the above, we can see that m can not have a value of 3 when l =2 because m has values between (-l) to (+l). Thus, the sets of quantum numbers is not valid.
What is the mass of 2.30x10^22 formula units of NaOH (molar mass =40.0g/mol)
Answer:
643(%=:(¥75 )(:7$"8"),"7$"()9_/"¥?:
Pablo, whose mass is 60 kg, is moving along at a velocity of 8 m/s. How much kinetic energy does he have?
Answer:
1920 Joules
Explanation:
* means multiply
^ means exponent
kinetic energy is in Joules or J
kinetic energy = 1/2 * mass * (velocity)^2
kinetic energy = 1/2 * 60 * (8)^2
kinetic energy = 1/2 * 60 * (8*8)
kinetic energy = 1/2 * 60 * (64)
kinetic energy = 1920 Joules or 1920 J
Answer:
KE = 1920 J
General Formulas and Concepts:
Energy
Kinetic Energy Formula: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]
m is mass (in kg)v is velocity (in m/s)Explanation:
Step 1: Define
[Given] m = 60 kg
[Given] v = 8 m/s
Step 2: Find KE
Substitute in variables [Kinetic Energy Formula]: [tex]\displaystyle KE = \frac{1}{2}(60 \ kg)(8 \ m/s)^2[/tex]Evaluate: [tex]\displaystyle KE = 1920 \ J[/tex]Topic: AP Physics 1 Algebra-Based
Unit: Energy
What effect does a high carbon level have on a deep ocean
Explanation:
High carbon concentration in the deep ocean means increased absorption of carbon to the atmosphere resulting to even greater and harmful amounts of carbon in the atmosphere. Therefore we need to keep a close eye of the deep ocean in the quest to monitor and pump out excess carbon from this part of marine life.
Explain why caffeine can be extracted from the tea leaves into hot water and how you extracted the aqueous solution and isolated the crude caffeine.
Answer:
The hot water dissolves the flavor and color components.
Explanation:
Caffeine can be extracted from the tea leaves into hot water because the hot water dissolves the flavor and color components away from the solid vegetable. This is an example of a solid-liquid extraction. We can extracted the aqueous solution and isolated the crude caffeine by converting the components of caffeine into their calcium salts which are insoluble in water. Then the caffeine can be extracted from the water by using methylene chloride.
Select the statement that correctly describes an endothermic process.
a. The enthalpy change for the process is negative.
b. Heat is lost by the system, but work is done on the system.
c. The enthalpy of the products is higher than that of the reactants.
d. The temperature change of the products is greater than that of the reactants.
e. Heat is released by the system, to the surroundings.
Answer:
C. The enthalpy of the products is higher than that of the reactants
Explanation:
An endothermic process is any process with an increase in the enthalpy of the system
The enthalpy of the products is higher than that of the reactants describes an endothermic process. Therefore, option (C) is correct.
What is the endothermic process?An endothermic process can be described as any thermodynamic process with an increase in the enthalpy H of the system. In such a process, a closed system commonly absorbs thermal energy from its surroundings.
An endothermic reaction leads to an increase in the temperature of the system and a decrease temperature of the surroundings. A physical process, such as the melting of ice cubes is an endothermic process.
Whether a process can take place spontaneously depends not only on the enthalpy change but also on the entropy change and absolute temperature.
An endothermic process commonly needs a favorable entropy increase (∆S > 0) in the system. In an endothermic process, the energy of the products is higher than that of the reactants.
Learn more about the endothermic process, here:
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A sealed vessel initially contains 100 g of chlorine gas and 90 g of hydrogen gas. The two gases undergo reaction to form HCl. Which of the following statements is true?
a. 100 g HCl is produced
b. 190 g HCl is produced
c. less than 90 g HCl is produced
d. between 100 and 190 g of HCl is produced
Answer:
d.
Explanation:
H2 + Cl2 = 2HCL
From the equation 2g hydrogen combine with 71g of chlorine.
So 35.5 g Cl2 combines with 1g of H2
There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g hydrogen.
This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.
The true statement is that d. between 100 and 190 g of HCl is produced.
How do find the mass of HCL?To find mass of HCL:
H2 + Cl2 = 2HCL
From the equation, 2g of hydrogen combines with 71g of chlorine.
So 35.5 g Cl2 combines with 1g of H2
There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g of hydrogen.
This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.
Hydrogen chloride may be formed by the direct combination of chlorine (Cl2) gas and hydrogen (H2) gas.
Learn more about Hydrogen chloride here: brainly.com/question/20323511
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What effect does hybridization have on chemical bond
20.1 cm3 of metal hydroxide (MOH) containing 4.8 g/dm3 react Completely with 23.0 cm3 of the HCL and Containing 3.8g/dm3. Identify the metal of M of MOH
Answer:
A metal M readily forms water-soluble sulphate MSO4, water-insoluble hydroxide M(OH)2 and oxide MO. The oxide and hydroxide are soluble in NaOH. The M is:
Methanoic acid is the simplest carboxylic acid molecule. It has one carbon atom. Draw the structural model for methanoic acid (using C and H).
ANSWER IS ABOVE
THE METHANOIC ACID
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
9.36 g
Explanation:
The equation of the reaction is;
C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)
Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles
1 mole of octane yields 9 moles of water
0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water
Number of moles of oxygen = 23g/32g/mol = 0.72 moles
12.5 moles of oxygen yields 9 moles of water
0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water
Hence oxygen is the limiting reactant;
Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g
Draw a Lewis structure for thiocyanic acid, HSCN, adding charges and lone electron pairs to the appropriate atoms.
Answer:
See explanation and image attached
Explanation:
Thiocyanic acid is made made up of hydrogen, sulphur, carbon and nitrogen atoms. Carbon is the central atom in the molecule.
The molecule has a total of sixteen valence electrons as shown in the image attached. There are no formal charges in the structure of the molecule as shown.
The molecule is linear in shape.
Design an experiment to demonstrate phototropism.
Answer:
Object: To demonstrate phototropism
equipments: A black box, window with light source, a well watered potted plant.
Experiment: A wellwatered potted plant is placed inside a darkened black box that is having a small window on one side. This is called as phototropic chamber. Window is remain closed for a day the plant shows normal growth.
Whereas if window is opened atwo days it will be found that shoot tip bends and grows towards light proving that it is positively phototropic.
A molecule with the formula AX 4 uses ________ to form its bonds. sp2 hybrid orbitals sp3d hybrid orbitals sp3 hybrid orbitals sp3d2 hybrid orbitals sp hybrid orbitals
Answer:
sp3 hybrid orbitals
Explanation:
The formula of a molecule gives us an idea of its structure and the nature of hybrid orbitals that are involved in the formation of the molecule.
AX4 corresponds to tetrahedral geometry. If a molecule is in tetrahedral geometry, it is most likely sp3 hybridized as usual.
Hence, a molecule with the formula AX 4 uses sp3 hybrid orbitals to form its bonds
Which of the following elements is the largest in size
a. O
b. He
c. K
d. H
Answer:
C. K
i took this class before
Answer: The largest element is K
Explanation: As K has the largest radius among O,He and H
2x^2+8x-7 solve the equation
Answer:
x = (-4 + √30) / 2, or x = (-4 – √30) / 2
Explanation:
x = (-b ± √b² – 4ac) / 2a →
y = ax² + bx + c
given
y = 2x² + 8x – 7 →
x = (-(8) ± √(8)² – 4(2)(-7)) / 2(2) →
x = (-8 ± √64 + 56) / 4 →
x = (-8 ± √120) / 4 →
x = (-8 ± 2√30) / 4 →
x = (-4 ± √30) / 2 →
x = (-4 + √30) / 2, or x = (-4 – √30) / 2
a 150 j of energy is added to a system that does 50 j of work is done.by how m uch will tyhe internal energy of system be raised?
The internal energy of system is raised by 3 times
Green plants use light from the Sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas form aqueous glucose (CH1206) and oxygen (O2) gas. Calculate the moles of glucose produced by the reaction of 2.40 mol of carbon dioxide. Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.
Answer:
.
Explanation:
The radius of a platinum atom is 139 pm. How many platinum atoms would have to be laid side by side to span a distance of 1.39 mm?
Answer:
5x10⁶ atoms of platium have to be laid side by side
Explanation:
1 atom of platium has a radius of 139pm = 139x10⁻¹²m. The distance that occupies 1 atom of platinum is 2 times its ratio:
139x10⁻¹²m*2 = 2.78x10⁻¹⁰m
Assuming there is no distance between to atoms of platinum, just its ratio. The amount of atoms necessary to occupy 1.39mm = 1.39x10⁻³m is:
1.39x10⁻³m / 2.78x10⁻¹⁰m =
5x10⁶ atoms of platium have to be laid side by sideA mixture is -
A. a pure substance made through chemical processes
B. a pure substance made through physical processes
C.
an impure substance made through chemical processes
D.
an impure substance made through physical processes
Answer:
C
Explanation:
an impure substance made through chemical process
Consider the balanced chemical equation below.
2 A ⟶ C + 4 D
How many moles of D would be produced if 6 moles of A were used?
Explanation:
[tex]here \: is \: your \: explanation : - \\ \\ given \: balanced \:equation \: = > \\ \\ 2 A=>C \: + \: 4D \\ \\ by \: this \: equation \: we \: get \: \\ \\ 2 \: moles \: of \: A \: produce \: 4 \: moles \: \\ \\ of \: D \\ \\ hence \: . \: 1 \: mole \: can \: produce \: = 4 \div 2 \\ \\ = > 2 \: moles \: \\ \\ so \: if \: 6 \: moles \: of \: A \: used \: then \: \\ \\ amount \: of \: D \: produced \: = (6 \times 2) \\ \\ = > 12 \: moles \: of \: D \\ \\ \mathcal\blue{ Hope \: it \: helps \: you \: (. ❛ ᴗ ❛.) }[/tex]