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:
Vocabulary: dipole, dipole-dipole force, dipole-induced dipole force, electronegativity, intermolecular force, ionic bond, London dispersion force, molecule, nonpolar, nonpolar covalent bond, partial charges, polar, polar covalent bond, valence electron Prior Knowledge Questions (Do these BEFORE using the Gizmo.) 1. A big bully is having a tug-of-war with a small child. There is a ball attached to the middle of the rope. Toward whom will the ball move
Answer:
Towards the big bully
Explanation:
If a big bully and a small child are involved in a thug of war, it is clear that the bully is stronger than the child and he/she will pull the rope used in the thug of war with a greater force.
By so doing, the ball attached at the centre of the rope will naturally be drawn towards the stronger bully.
the Wavelength of a microwave is______than the wavelength of visible light
(Please help)
Answer:
The wavelength of a microwave is LONGER than the wavelength if visible light.
Is it possible for number of moles to be less than one?
Answer:
yes is very possible to be
PLEASE HELP!!!
Explain the various factors that impact the solubility of substances in water.
(At least 4 sentences) :)
The various factors that impact the solubility of substances in water are - nature of solute and solvent, temperature, pressure and pH.
The solubility of substances in water is influenced by several factors:
Nature of the solute and solvent: The chemical properties of both the solute and the solvent play a significant role. Substances with similar polarities and intermolecular forces tend to dissolve more readily in water. Polar solutes, such as salts and sugars, dissolve well in polar solvents like water, whereas nonpolar solutes, like oils and fats, have low solubility in water.Temperature: In general, an increase in temperature enhances the solubility of solid solutes in water, as it provides more energy for the solute particles to overcome intermolecular forces and mix with the solvent. However, the effect of temperature on solubility can vary depending on the specific solute. For some solutes, such as gases, solubility decreases with increasing temperature.Pressure (for gases): The solubility of gases in water is influenced by pressure. According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Therefore, an increase in pressure generally leads to an increase in gas solubility in water.pH: The pH of the solution can impact the solubility of certain substances. For example, the solubility of acidic or basic compounds may change with varying pH levels.Learn more about Solubility, here:
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c. rubidium sulfate
2. Write balanced molecular equation, complete ionic equation, and net ionic
equations for the mixing of the following solutions. Show states. If no reaction
occurs, show the ionic equation. (8 marks)
a.
NaNO3 +
Ag2SO4 → AgNO3 + Na2SO4
chemistry help!
This mysterious gas has a volume of 4.35 L and a pressure of 1.20 atm. If the pressure is changed to 0.95 atm, what will the new volume be? How do i set up this problem and solve it?
Refer to the attachment.
Hope this helps you...
What is the law of multiple proportions?
A. The proportion of elements to compounds is constant.
B. All elements are found in equal proportions in nature.
C. Different compounds may contain the same elements but may have different ratios of those elements.
D. All compounds contain the same elements in the same proportions.
Answer:
a. the proportion of elements to compounds is constant.
The Li2+ ion is very similar to the hydrogen atom, in that it has one electron and energy levels similar
to the hydrogen atom. However, the relation = − (
1
2 −
1
2) cannot be used for this ion but
rather the relation = −2+ (
1
2 −
1
2) where the constant 2+=1.96x10-17J.
Use this relation to determine the third ionization energy, which is energy required to remove the last
electron from a Li2+ ion in kJ/mol, if the ion starts off in the ground state (Li2+ → Li3+
+ e-
).
Og is the noble gas after Rn. To go from [Rn] to [Og], you must fill four subshells (s, p, d, and f) with a total of 32 electrons. Thus, the atomic numbers of 6th and 7th period elements of the same group differ by 32.
a. To go from [Og] to the next noble gas, however, you would theoretically fill five subshells (s, p, d, f, and g). How many electrons are needed to fill all five subshells?
b. Element 106 in the periodic table is Sg Determine the atomic number of the element just below Sg in the periodic table.
Answer:
See explanation
Explanation:
Since we have to fill five subshells in moving from Og to the next noble gas in the eight period, we have to know the maximum electrons contained in each of those subshells;
s= 2, p=6, d= 10, f= 14, g = 18
This means that we need a total of 50 electrons to fill all the five subshells.
Hence, the element just below Sg in the eight period will have an atomic number of 156.
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Question 1
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Initial Knowledge Check
Drafts
Sent
Write the symbol for every chemical element that has atomic number less than 14 and atomic mass greater than 23.2 u.
Answer:
8 oxygen. 9 flourine. 10. Neon. 5 Boron
En un recipiente cerrado y rígido se introdujo una mezcla gaseosa a cierta temperatura y las presiones parciales de cada gas son: p(F2) = 2,00 atm, p(BrF) = 1,50 atm y p(BrF3) = 0,0150 atm. A la temperatura que se preparó la mezcla tiene lugar la reacción representada por:
BrF3 (g) ⇌ BrF(g) + F2(g) Kp(T) = 64,0
Elegir la afirmación correcta.
Seleccione una:
Qp > Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) aumentan hasta alcanzar el equilibrio.
Qp < Kp, por lo tanto, la presión parcial de BrF3(g) disminuye hasta alcanzar el equilibrio.
Qp = Kp, por lo tanto, las presiones parciales de BrF3(g), BrF(g) y F2(g) no cambian.
Qp < Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) disminuyen hasta alcanzar el equilibrio.
Qp > Kp, por lo tanto, la presión parcial de BrF3(g) aumenta hasta alcanzar el equilibrio.
Answer:
Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
Explanation:
Paso 1: Escribir la ecuación balanceada
BrF₃ (g) ⇌ BrF(g) + F₂(g) Kp(T) = 64,0
Paso 2: Calcular el cociente de reacción (Qp)
Qp = pBrF × pF₂ / pBrF₃
Qp = 1,50 × 2,00 / 0,0150 = 200
Paso 3: Sacar una conclusión
Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
A calorimeter has been filled with 75 mL of water. The density of water at 25 °C is
0.998 g/mL. How many grams of water do you have at 25°C?
Answer:
isn't it 0.998
Explanation:
cause 0.998 is in the same grams / mole. I don't get
A calorimeter has been filled with 75 mL of water. The number of grams of water is 74.85 grams.
What is a calorimeter?A calorimeter is an instrument that is used to measure calorimetry. It is used to measure the heat of chemical reactions and physical and chemical changes.
Mass is the quantity of matter in a physical body. Volume is the space occupied by a three-dimensional object. Mass is the quantity of matter in a physical body. The product of the compound's molar mass and the substance's moles are defined as mass.
Given, that the volume of the water is 75 mL
The temperature of the water is 25 °C
The density of water is 0.998 g/mL
To calculate the mass, the volume is multiplied by density.
Mass = volume x density
putting the values in the formula
75 x 0.998 = 74.85 grams
Thus, the grams of water that have at 25°C is 74.85 grams.
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A chemist measures the amount of oxygen gas produced during an experiment. She finds that 276. g of oxygen gas is produced. Calculate the number of moles of oxygen gas produced.
Round your answer to 3 significant digits.
0.152 moles
Given that:The mass of the oxygen gas produced = 4.87 g
Also, The molar mass of oxygen gas, = = 32 g/mol
The formula for the calculation of moles is shown below:
moles = [tex]\frac{Mass-taken}{Molar-mass}[/tex] *
Thus,
moles = [tex]\frac{4.87g}{32g/mol }[/tex]
Both given values and the answer is in 3 significant digits.
brainlist pls?
*i put a dash between mass and taken as formulas do no spaces
The N H 2 group makes the molecule an amine. Like alcohols, amines can participate in hydrogen bonding. If the drawn molecule encounters water molecules, there will be (select all that apply): a hydrogen bond between the hydrogen of water and the carbon of the amine. a hydrogen bond between the hydrogen of water and the nitrogen of the amine. a hydrogen bond between the oxygen of water and a hydrogen from the -NH2 group. a hydrogen bond between the oxygen of water and a hydrogen from the -CH3 group. no hydrogen bonds.
Answer:
a hydrogen bond between the hydrogen of water and the nitrogen of the amine
a hydrogen bond between the oxygen of water and a hydrogen from the -NH2 group
Explanation:
A hydrogen bond is formed between molecules in which hydrogen is bonded to a highly electronegative element.
In amines, hydrogen is bonded to nitrogen while in water, hydrogen is bonded to oxygen. Both are highly electronegative elements hence hydrogen bonding is possible between amines and water.
This hydrogen bond may involve;
The hydrogen of water and the nitrogen of the amine
Or
The oxygen of water and a hydrogen from the -NH2 group
Please help
Why are different products formed in the combustion of butane ?
Explain please thanks :)
PLEASE ANSWER THESE:
Question 4: Biological Molecules (5 points)
A. Pepsin is an enzyme used by the human body to break down food in the stomach.
i. Explain, in terms of a chemical reaction, how pepsin helps a person. (3 points)
ii. What are the building blocks of the pepsin macromolecule? (2 points)
Answer: the answer is A
Explanation: add me hope i helped
1. What is the difference between accidental discoveries and scientific investigations?
Answer:
the discovery will always need to be tested more where if you're investigating you should already be in a more controlled environment
The difference between accidental discoveries and scientific investigations are as follows.
Accidental discoveriesThe discovery which is occur accidentally, means something discovered intensely.Examples of accidental discoveries like, microwave, teflon, penicillin, plastic etc.Scientific investigationA process of finding of answers of the question by using various research method.There are various scientific investigation, like fair testing, modeling, pattern, seeking, identification and classifying etcExample of scientific investigation are plant experiment.To learn more about accidental discoveries,
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Answer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.25 wt% carbon for each specimen. (b) Comparing the finished specimens at a depth of 0.20 mm, specimen A features the lowest carbon concentration. (c) Comparing the finished specimens as a whole, specimen D features the lowest overall amount of carbon.
Answer:
verdadero/a
falso/b
verdadero/c
Explanation:
The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the temperature of the gas rises (a) from 46 to 92 K and (b) from 35.4 to 69.0 oC.
Answer:
A. P₂ / P₁ = 2
B. P₂ / P₁ = 1.1
Explanation:
A. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 46 K
Final temperature (T₂) = 92 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/46 = P₂/92
Cross multiply
46 × P₂ = P₁ × 92
Divide both side by P₁
46 × P₂ / P₁ = 92
Divide both side by 46
P₂ / P₁ = 92 / 46
P₂ / P₁ = 2
B. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K
Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/308.4 = P₂/342
Cross multiply
308.4 × P₂ = P₁ × 342
Divide both side by P₁
308.4 × P₂ / P₁ = 342
Divide both side by 308.4
P₂ / P₁ = 342 / 308.4
P₂ / P₁ = 1.1
Consider the balanced chemical equation below.
3 A ⟶ C + 4 D
How many moles of D would be produced if 9 moles of A were used?
hope the picture above helps :)
1) Rank the following elements in order of decreasing atomic radius. Rank from largest to smallest radius. To rank items as equivalent, overlap them.
a. k
b. Li
c. Rb
d. Na
1. Largest
2. Smallest
2) Rank the following elements in order of decreasing atomic radius. Rank from largest to smallest radius. To rank items as equivalent, overlap them.
a.Mg
b. Al
c. Na
d. Si
Answer:
1) Ra
K
Na
Li
2) Na
Mg
Al and Si
We can use bond-line formulas to represent alkenes in much the same way that we use them to represent alkanes. Consider the following alkene: h5ch5e4 How many carbon atoms are sp2−hybridized in this alkene?
Answer:
2
Explanation:
The number of carbon atoms that are sp²-hybridized in this alkene is 2
Because all the single bonded carbon atoms in the alkene are sp²-hybridized
There are three(3) single formed via sp² orbitals and one ( 1 ) PI bond formed via Pure-P-orbital
attached below is the some part of the solution
What is the molarity of a solution if 325ml of the solution contains 46.8 grams of NaHCO3?
Answer:
1.714 M
Explanation:
We'll begin by calculating the number of mole in 46.8 g of NaHCO₃. This can be obtained as follow:
Mass of NaHCO₃ = 46.8 g
Molar mass of NaHCO₃ = 23 + 1 + 12 + (3×16)
= 23 + 1 + 12 + 48
= 84 g/mol
Mole of NaHCO₃ =?
Mole = mass / molar mass
Mole of NaHCO₃ = 46.8 / 84
Mole of NaHCO₃ = 0.557 mole
Next, we shall convert 325 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
325 mL = 325 mL × 1 L / 1000 mL
325 mL = 0.325 L
Thus, 325 mL is equivalent to 0.325 L.
Finally, we shall determine the molarity of the solution. This can be obtained as shown below:
Mole of NaHCO₃ = 0.557 mole
Volume = 0.325 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.557 / 0.325
Molarity = 1.714 M
Therefore the molarity of the solution is 1.714 M
Draw the Lewis structure for the polyatomic hydronium H3O cation. Be sure to include all resonance structures that
Answer:
Lewis structure of Hydronium ion is shown below :
Explanation:
Lewis structure : It is a representation of valence electrons on the atoms in a molecule
Here , Hydronium ion is given , which contains 1 atom of oxygen and 3 atoms of hydrogen .
Oxygen has a total of 6 valence electrons and hydrogen contains 1 valence electron .
Oxygen share its 3 valence electrons with 3 hydrogen atoms and left with 3 valence electrons. From these three valence electrons of oxygen atom two electrons will be shown as a pair of electrons on oxygen atom but a single electron can not be shown . So , to simplify this, one positive charge is shown overall .
Resonance structure will be same as the hybrid structure because all three atoms are same , that is hydrogen .
Which factors would increase the rate of a reaction?
1. Lowering the temperature
II. Increasing the concentration of readiants
JUL. Adding a catalyst
I and
Oland
O only
I only
Tony
Next
Adding a catalyst would increase the rate of a reaction
How many moles of carbon dioxide at STP will fit in a 50 liter container?
Answer:
n = 2.23 moles
Explanation:
Given the following data;
Standard temperature = 273 K
Standard pressure = 101.325 kPa
Volume = 50 liter
R = 8.314 J/mol·K
To find the number of moles, we would use the ideal gas law formula;
PV = nRT
Where;
P is the pressure.V is the volume.n is the number of moles of substance.R is the ideal gas constant.T is the temperature.Making n the subject of formula, we have;
[tex] n = \frac {PV}{RT} [/tex]
Substituting into the formula, we have;
[tex] n = \frac {101.325*50}{8.314*273} [/tex]
[tex] n = \frac {5066.25}{2269.722} [/tex]
n = 2.23 moles
Therefore, 2.23 moles of carbon dioxide at STP will fit in a 50 liter container.
Match the description with the type of precipitation being described.
1. Its formation requires very strong updrafts
2. Its formation requires falling through a layer of above freezing air
3. Precipitation from cumuliform clouds is typically of this nature
4. Precipitation from stratus clouds is typically of this nature
Options:
a. Hail
b. Drizzle
c. Shower
d. Freezing Rain
Answer:
1. Its formation requires very strong updrafts = a. Hail
2. Its formation requires falling through a layer of above-freezing air = d. Freezing Rain
3. Precipitation from cumuliform clouds is typically of this nature = c. Shower
4. Precipitation from stratus clouds is typically of this nature = Drizzle
Explanation:
Hail formation requires very strong updrafts, these updrafts are the upward moving air created in a thunderstorm. This period of noticeable thunderstorms creates hails.
Freezing rain requires the presence of warm air, it requires falling through a layer of above-freezing air to the colder air below to produce an ice coating on anything it drops on.
Showers are produced by cumuliform clouds which look like cotton balls. Since cumuliform clouds precipitate too, these clouds can have fluctuating rain in a day in the form of showers.
Drizzle which raises low visibility is considered a type of liquid precipitation since it also falls from a cloud. Drizzle which is obviously smaller in diameter when compared to that of raindrops, however, is common with stratus clouds.
An acetic acid buffer containing 0.50 M acetic acid (CH3COOH) and 0.50 M sodium acetate (CH3COONa) has a pH of 4.74. What will the pH be after 0.0020 mol of HCl has been added to 100.0 mL of the buffer
Answer:
pH = 4.71
Explanation:
We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:
pH = pKa + log [CH3COONa] / [CH3COOH]
Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.
As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:
pH = pKa + log 1
4.74 = pKa
To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl
Moles CH3COONa: Initial CH3COONa - Moles HCl.
Moles CH3COOH:
0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =
0.052 moles CH3COOH
Moles CH3COONa:
0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =
0.048 moles CH3COONa
Using H-H equation:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
pH = 4.71The pH be after 0.0020 mol of HCl has been added to 100.0 mL of the buffer is 4.71.
What is buffer solution?Buffer solution is a mixture of weak acid and its conjugate base or vice versa.
We can calculate the pH of buffer solution by using Henderson - Hasselbalch Equation:
pH = pKa + log[CH₃COONa] / [CH₃COOH]
Initially concentration of CH₃COONa is equal to the concentration of CH₃COOH and equal becomes:
4.74 = pKa + log(1)
pKa = 4.74
Given moles of added HCl moles = 0.002 mole
Given molarity of each CH₃COOH & CH₃COONa = 0.50M
Given Volume = 100mL
We can calculate the moles by using the formula:
n = M × V
Moles of CH₃COOH & CH₃COONa = 0.100 × 0.50 = 0.050 moles
Moles of CH₃COOH = 0.050moles CH₃COOH + 0.0020moles HCl =
0.052moles CH₃COOH
Moles of CH₃COONa = 0.050moles CH3COONa - 0.0020moles HCl =
0.048moles CH₃COONa
Now, resultant pH will be:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
pH = 4.71
Hence, pH of resultant solution is 4.71.
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A solution of the ionic salt NaF
would have _____
pH.
A. An acidic
B. A basic
C. A neutral
Answer:
The answer is basic
Explanation:
Choose letter b or basic. You welcome.
The strong acid and strong base has high rate constant of dissociation. The rate constant for weak acid and base for the dissociation is low, they do not easily dissociate in water. Therefore option B is the correct option.
What are acid and base?Acid is a solution which releases H⁺ hydrogen ion when dissolved in water. Base releases hydroxide ion OH⁻ ion when dissolved in water.
pH is a measurement of amount of hydronium ion H₃O⁺ in a given sample. Strength of acidic nature is directly proportional to the concentration of hydronium ion.
On subtracting pH from 14, we get pOH which measures the concentration of hydroxide ion in a given solution. Temperature affect the pH. At room temperature pH scale is between 0 to 14. 7 is the pH of neutral solution. A solution of the ionic salt NaF is basic in nature.
Therefore, option B is the correct option.
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Acetic acid and water react to from hydronium cation and acetate anion, like this: HCH3CO2 (aq) + H2O (I) → H3O+(aq) + Ch3CO2-(aq)Imagine 226. mmol of CH3CO2- are added to a flask containing a mixture of HCH3CO2, H2O, H3O + and CH3CO2- at equilibrium, and then answer the following questions. 1. What is the rate of the forward reaction before any HCH3CO2 has been removed from the flask? a. Zero.b. Greater than zero, but less than the rate of the reverse reaction. c. Greater than zero, and equal to the rate of the reverse reaction. d. Greater than zero, and greater than the rate of the reverse reaction. 2. What is the rate of the forward reaction just after the HCH3CO2 has been removed from the flask?a. Zero.b. Greater than zero, but less than the rate of the reverse reaction. c. Greater than zero, and equal to the rate of the reverse reaction. d. Greater than zero, and greater than the rate of the reverse reaction.3. What is the rate of the forward reaction when the system has again reached equilibrium?a. Zero.b. Greater than zero, but less than the rate of the reverse reaction. c. Greater than zero, and equal to the rate of the reverse reaction. d. Greater than zero, and greater than the rate of the reverse reaction.
Answer:
1) Greater than zero, and equal to the rate of the reverse reaction
2) Greater than zero, but less than the rate of the reverse reaction
3) Greater than zero, and equal to the rate of the reverse reaction
Explanation:
A reaction system is said to be in equilibrium when the rate of forward reaction is equal to the rate of reverse reaction.
Before we remove HCH3CO2 from the system, the system was in equilibrium. Recall that when a system is in equilibrium, the rate of forward reaction is equal to the rate of reverse reaction. The rate of reaction is greater than zero because products are being formed as the reactants interact with each other.
When HCH3CO2 is removed from the system, the equilibrium position shifts towards the left hand side hence the rate of reverse reaction is greater than the rate of forward reaction.
When the system attains equilibrium again, the rates of forward and reverse reaction become equal.