Answer:
7.44%
Explanation:
[tex]\frac{6.44}{6.44+80.1} *100=7.44[/tex]
Write two sentences explaining how the illustration and student observations support the cell theory?
Answer:
The classical cell theory was proposed by Theodor Schwann in 1839. There are three parts to this theory.
Explanation:
In the image below if the frisbee represents lights, what does the fence represent?
Answer:
The atmosphere
Explanation:
It reflects and traps the light outside of the Earth's Atmosphere?
I think, hope i helped.
what is the iupac name of CH3-CH2CH(CH3)CH2-COOH
Answer:
Butanoic acid.
Explanation:
fireworks convert chemical energy into what kind of energy
Answer:
Kinetic energy
Explanation:
Fireworks convert chemical energy into kinetic energy to send them flying upward. Hope it is correct!
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Answer:
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Calculate Kb for a 0.1 M solution of cyanide ion, given that Ka for hydrocyanic acid is 6.2 x 10â10.a) 8.4 x 10â9b) 7.87 x 10â3c) 1.34 x 10â3d) 1.8 x 105e) 7.87 x 10â6f) 1.61 x 10â5g) 7.4 x 10â8
Answer:
Kb = 1.6 × 10⁻⁵
Explanation:
Step 1: Given data
Acid dissociation constant of hydrocyanic acid (Ka): 6.2 × 10⁻¹⁰
Concentration of cyanide ion (Cb): 0.1 M
Step 2: Calculate the basic dissociation constant (Kb) of cyanide ion
We have the Ka of HCN. We can calculate the Kb of its conjugate base using the following expression.
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Kb = 1.0 × 10⁻¹⁴/Ka
Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰
Kb = 1.6 × 10⁻⁵
The Kb for a 0.1 M solution of cyanide ion is :
- 1.6 × 10⁻⁵
Base dissociation constantGiven:
Acid dissociation constant of hydrocyanic acid (Ka)= 6.2 × 10⁻¹⁰
Concentration of cyanide ion (Cb)= 0.1 M
Base dissociation constant (kb)=?
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Kb = 1.0 × 10⁻¹⁴/Ka
Kb = 1.0 × 10⁻¹⁴/6.2 × 10⁻¹⁰
Kb = 1.6 × 10⁻⁵
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A buffer solution contains 0.0200 M acetic acid and 0.0200 M sodium acetate. What is the pH after 2.0 mmol of NaOH are added to 1.00 L of this buffer? pKa = 4.75 for acetic acid.
Answer:
The correct answer is 4.84
Explanation:
We use the Henderson Hasselbach's equation to calculate the pH of a buffer solution:
pH = pKa = log ( [conjugate base]/[weak acid])
In this case:
conjugate base: acetate ⇒ [conjugate base] = [acetate] = 0.0200 M
weak acid: acetic acid ⇒ [weak acid] = [acetic acid] = 0,0200 M
pH = pKa + log ([acetate]/[acetic acid])= pKa + log (0.0200 M/0.0200 M)
When a strong base- such as NaOH- is added, the acid reacts with OH⁻ to form the conjugate base. So, the conjugate base is increased while the acid is decreased. Thus, in 1 liter of solution we have:
acetate = (0,0200 mol/L x 1 L) + (2.0 mmol x 1 mol/1000 mmol) = 0.022 mol
acetic acid= (0,0200 mol/L x 1 L) - (2.0 mmol x 1 mol/1000 mmol) = 0.018 mol
Finally, we calculate the pH:
pH = 4.75 + log (0.022 mol /0.018 mol )= 4.84
Buffer has been the solution that resists the change in the pH of the solution with the addition of acid or base. The pH of buffer with NaOH has been 4.84.
What is pH?The pH has been the hydrogen ion concentration in the solution. The pH has been equivalent to pKa and can be given as:
[tex]\rm pka=log\dfrac{[base]}{[acid]}[/tex]
The base in the buffer has been the acetate, while acetic acid act as the acid.
The concentration of NaOH added to the solution has been given as:
[tex]\rm Molarity=\dfrac{moles}{volume}\\\\ NaOH=\dfrac{0.002\;mol}{1\;L}\\\\ NaOH=0.002\;M[/tex]
The addition of NaOH adds the concentration of base, and neutralization reduces the concentration of acid. Thus, the new pKa can be given as:
[tex]\rm pKa=log\dfrac{0.2+0.002}{0.2-0.002} \\pKa=log\dfrac{0.022}{0.018}[/tex]
The new pH of the solution has been the sum of the old and the new pKa value.
Thus, the pH of the new buffer has been given as:
[tex]\rm pH=4.75+log\dfrac{0.022}{0.018}\\ pH=4.84[/tex]
The pH of buffer with NaOH has been 4.84.
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At equilibrium, the value of [H ] in a 0.240M solution of an unknown acid is 0.00417M . Determine the degree of ionization and the Ka of this acid.
Answer:
[tex]ionization=1.74\%[/tex]
Explanation:
Hello!
In this case, since the degree of ionization of an acid is computed in terms of the concentration of hydrogen ions and the initial concentration of the acid:
[tex]ionization=\frac{[H^+]}{[HA]} *100\%[/tex]
Because the ionization reaction is represented by:
[tex]HA\rightleftharpoons H^++A^-[/tex]
Therefore, the degree or percent ionization turns out:
[tex]ionization=\frac{0.00417M}{0.240M} *100\%\\\\ionization=1.74\%[/tex]
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If Phosphorus, P, gained 3 electrons, the net ion charge would be
+5
-5
+3
-3
Answer:
-3
Explanation:
Phosphorus becomes phosphide with a -3 charge
Chem Muti Choice. Tell me the correct answer.
Answer:
my gues is red not a 100% but its in the 700s
Explanation:
How many moles of CO are produced when 1.8 moles C reacts?
Express your answer using two significat figures.
5C(s) + 2SO2(g) => CS2(l) +4CO(g)
Answer:
1.44mole of CO
Explanation:
The reaction equation is given as:
5C + 2SO₂ → CS₂ + 4CO
We check to see if the expression is balanced and it is so;
Now;
Given;
1.8mole of C reacted; how many moles of CO are produced;
From the balanced reaction equation:
5 mole of C is expected to produce 4 mole of CO
1.8 mole of C will then produce [tex]\frac{4 x 1.8}{5}[/tex] = 1.44mole of CO
Calculate the root mean square (rms) average speed of the atoms in a sample of krypton gas at 0.14 atm and -16 0C.
Answer:
8.52 m/s
Explanation:
Step 1: Given data
Molar mass of krypton (M): 83.80 g/molPressure of the sample (P): 0.14 atmTemperature of the sample (T): -16 °CStep 2: Convert "T" to the Kelvin scale
When working with gases, we need to consider the absolute temperature. We will convert from Celsius to Kelvin using the following expression.
K = °C + 273.15 = -16 + 273.15 = 257 K
Step 3: Calculate the root mean square speed of the gas
The root mean square speed measures the average speed of particles in a gas. We will calculate it using the following expression.
[tex]v_{rms} = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times 8.314 J/mol.K \times 257 K}{88.30 g/mol} } = 8.52 m/s[/tex]
How many neutrons does the isotope of hydrogen called
deuterium have?
a. four
b. three
C. one
d. two
Suppose the Cu ions are produced by the reaction of 0.94 g of copper turnings with excess nitric acid. How many moles of Cu are produced?
Answer:
0.0148 moles
Explanation:
Solid copper is oxidized with HNO₃:
Cu + 4HNO₃ → Cu²⁺ + 2 NO₃⁻ + 2H₂O + 2NO₂
Where 1 mole of Cu produce 1 mole of Cu²⁺ when nitric acid is in excess.
Moles of Cu that react (Molar mass Cu = 63.546g/mol):
0.94g * (1mol / 63.546g) = 0.0148 moles of Cu
And moles of Cu²⁺ produced are also:
0.0148 molesA solution of NaF is added dropwise to a solution that is 0.0173 M in Ba 2. When the concentration of F- exceeds___M BaF2 will precipiate. Neglect volume changes. For BaF2
Answer:
0.0099M = [F⁻]
Explanation:
For BaF2, Ksp = 1.7x10⁻⁶
When BaF₂ is in solution, the equilibrium between the solid and the dissociated ions occurs as follows:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Where Ksp = 1.7x10⁻⁶ is defined as:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
Where [] are equilibrium concentrations of each ion in solution.
That means you will add F⁻ until its concentration exceeds:
1.7x10⁻⁶ = [0.0173] [F⁻]²
9.827x10⁻⁵ = [F⁻]²
0.0099M = [F⁻]When more F⁻ is added, BaF₂ begins its precipitation.
The balanced equation for the combustion of Hydrogen is:
2 H2 + O2 → 2H2O
How many moles of H2O will be formed from the complete reaction of 0.750 moles of O2?
0.375 mol
1.50 mol
2.67 mol
0.750 mol
Answer:
1.5 moles H₂O
General Formulas and Concepts:
Chemistry - Stoichiometry
Using Dimensional AnalysisExplanation:
Step 1: Define
RxN: 2H₂ + O₂ → 2H₂O
Given: 0.750 moles O₂
Step 2: Stoich
[tex]0.750 \ mol \ O_2(\frac{2 \ mol \ H_2O}{1 \ mol \ O_2} )[/tex] = 1.5 moles H₂O
Step 3: Check
We are given 3 sig figs.
Since our answer is 2 sig figs, we don't need to round.
How many valence electrons are found in P3- ?
Answer:
5 valence elctrons
Explanation:
the outer most orbitals, 3s2 and 3p3 contains 5 electrons, thus valences electrons for P is 5
How many grams of CO2 are produced by the combustion of 344 g of a mixture that is 33.6% CH4 and 66.4% C3H8 by mass
Answer:
1,002.936 g
Explanation:
The combustion equation of each will be;
CH4 + 2O2 = CO2 + H2O
C3H8 + 5O2 = 3CO2 + 4H2O
We are told the mixture is 344g.
Thus;
For CH4 combustion;
Amount of CH4 = 0.336 × 344 = 115.584g
Molar mass of CH4 is 16 g/mol
Number of moles of CH4 is;
n = 115.584g/(16 g/mol)
n = 7.224 moles
n(CO2) = n(CH4) = 7.224 moles
For C3H8 combustion;
Amount of C3H8 = 0.664 × 344 = 228.416 g
Molar mass of C3H8 = 44 g/mol
Thus;
Number of moles of C3H8 = 228.416 g/(44 g/mol) = 5.19 mol
n(CO2) = 3n(C3H8) = 3 × 5.19 = 15.57 moles
Total moles of CO2 = 7.224 moles + 15.57 moles = 22.794 mol
Molar mass of CO2 = 44 g/mol
Thus amount of CO2 by mass = 22.794 mol × 44 g/mol = 1,002.936 g
Which lists the elements in order from least conductive to most conductive?
Answer:
Answer. Answer: Nitrogen (N), antimony (Sb), bismuth (Bi) is the order from least conductive to most conductive.
Answer:
nitrogen (N), antimony (Sb), bismuth (Bi)
What is the ratio of potassium atoms to oxygen atoms in potassium oxide (K2O)?
K20
K = 1+
0 = 2-
K/0 = 1/2
the answer is K:O = 1:2
Calculate the concentration (M) of arsenic acid (H3AsO4) in a solution if 25.00 mL of that solution required 35.21 mL of 0.1894 M KOH for neutralization.
Answer:
[tex]M_{acid}=0.08892M[/tex]
Explanation:
Hello!
In this case, since the reaction between arsenic acid and potassium hydroxide is:
[tex]H_3AsO_4+3KOH\rightarrow K_3AsO_4+3H_2O[/tex]
Thus, since the mole ratio between the acid and the base is 1:3, at the equivalence point we can write:
[tex]3n_{acid}=n_{base}[/tex]
That in terms of molarities and volumes is:
[tex]3M_{acid}V_{acid}=M_{base}V_{base}[/tex]
Thus, the concentration of the arsenic acid is:
[tex]M_{acid}=\frac{M_{base}V_{base}}{3V_{acid}}=\frac{0.1894M*35.21mL}{3*25.00mL} \\\\M_{acid}=0.08892M[/tex]
Best regards!
The concentration (M) of arsenic acid in a solution of 25.00 mL is 0.088M.
How we calculate the molarity?Molarity of any solution will be calculated as:
M = n/V, where
V = volume
n is the moles and it will be calculated as:
n = W/M, where
W = given mass
M = molar mass
Given chemical reaction is:
H₃AsO₄ + 3KOH → K₃AsO₄ + 3H₂O
Moles of 35.21 mL of 0.1894 M KOH will be calculated as:
n = (0.1894)(0.035) = 0.0066 moles
From the stoichiometry of the reaction it is clear that:
1 mole of H₃AsO₄ = react with 3 moles of KOH
0.0066 moles of KOH = react with 1/3×0.0066= 0.0022 moles of H₃AsO₄
Now we calculate the molarity of H₃AsO₄ in 25mL of solution as:
M = 0.0022/0.025 = 0.088 M
Hence, required molarity is 0.088 M.
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raw material for fusion
Answer:
Deuterium and Lithium.
Explanation:
I searched it up on google.
Select the best answer from the choices below. Light has many properties of waves, but sometimes it also behaves like:
Answer:
A photon
Explanation:
When electrons pass close to the standing wave of light, they hit the particles that are called photons
A sample of neon gas occupies 5.5 liters at 20°C and 650 mmHg. What is the expected volume of the gas at S.T.P.
Answer:
V₂ = 4.4 L
Explanation:
Given data:
Initial volume = 5.5 L
Initial pressure = 650 mmHg (650/760 =0.86 atm)
Initial temperature = 20 °C (20 +273 = 293 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂/T₁ P₂
V₂ = 0.86 atm× 5.5 L × 273 K / 293 k × 1 atm
V₂ = 1291.29 atm.K.L / 293 k.atm
V₂ = 4.4 L
Calculate the molarity of a solution that contains 183.51 grams of lead (II) bromide in 500.0 mL of the solution
Answer: 1M
Explanation:
Molarity = mols/L
moles of lead bromide: 183.51/ 367.0 = 0.5 mol
500 ml/ 1000 mL = .5L
.5 mol / .5 L = 1 mol/L = 1M
A student performs an experiment in which 5.0 g of potassium chloride, KCI, are dissolved into 100.0 mL of distilled water. The student records the
temperature both before and after dissolving the KCl into the water, as shown in the data table below. Use this data to answer the question that follows.
Initial Temperature, °C Final Temperature, °C
21.8°C
16.4°C
Which statement below is a correct conclusion based on these results ?
The temperature of the water decreases and so the reaction is exothermic
The temperature of the water increases and so the reaction is exothermic
The temperature of the water decreases and so the reaction is endothermic
The temperature of the water increases and so the reaction is endothermic
Answer:
The temperature of the water decreases and so the reaction is endothermic
Explanation:
This option is correct because when the 5.0g potassium chloride (KCl) was added to the 100.0ml distilled water, the KCl made the water absorb heat coming from it which made the water decrease in temperature thereby causing an endothermic reaction.
An endothermic reaction actually takes place when the temperature of a system which is isolated decreases. The surroundings of a non-isolated system actually gains heat. While in an exothermic reaction, the temperature of a system increases due to the release or evolution of heat. This heat is released into the surroundings.
An igneous intrusion show that a magma body emerged in a stratigraphic section. This magma body is _____ than the rocks around it.
A) older
B) the same age
C) newer
Answer:
Younger/newer
Explanation:
Stratagraphic means an intrusion or fault is younger than the layer it affects.
Calculate the following quantity: volume of 1.58 M calcium chloride that must be diluted with water
to prepare 994 mL of a 6.20 x 10^-2 M chloride ion solution.
Calculate the volume in milliliters of 1.57 M potassium hydroxide that contains 10.3 g of solute.
Answer:
[tex]V = 0.117 \ L[/tex]
General Formulas and Concepts:
Chem
Reading a Periodic TableWriting compounds and polyatomic ionsMolarity = moles of solute / liters of solutionExplanation:
Step 1: Define
1.57 M KOH (potassium hydroxide)
10.3 g KOH
Step 2: Define conversions
Molar Mass of K - 39.10 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H - 1.01 g/mol
Molar Mass of KOH - 39.10 + 16.00 + 1.01 = 56.11 g/mol
Step 3: Convert
[tex]10.3 \ g \ KOH(\frac{1 \ mol \ KOH}{56.11 \ g \ KOH} )[/tex] = 0.183568 mol KOH
Step 4: Solve for Volume
Substitute: [tex]1.57 \ M=\frac{0.183568 \ mol}{x \ L}[/tex]Move x: [tex]x1.57 \ M=0.183568 \ mol[/tex]Isolate x: [tex]x=\frac{0.183568 \ mol}{1.57 \ M}[/tex]Evaluate: [tex]x=0.116922 \ L[/tex]Step 5: Check
We are given 3 sig figs. Follow sig fig rules.
[tex]0.116922 \ L \approx 0.117 \ L[/tex]
Which of the following are fair conductors and are always solid at room temperature?
1.Metals
2.Nonmetals
3.Metalloids
4.Ions
Answer:
Metalloids and ions
Explanation:
Mercury is not solid at room temperature. (Metals)
Many nonmetals aren't good at conducting electricity. Oxygen for example. (Nonmetals)
Metalloids are always solid at room temp and are fair conductors. (Metalloids)
Ions are same as metalloids. (Ions)
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