Answer:
Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.
A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.
Explanation:
1 mole HClO = 74.44g
0.0941g = [tex]\frac{0.0941}{74.44}[/tex] = 0.00126 moles
Concentration = no. of moles/volume in L
Hence, Concentration of HClO = 0.00126/ 0.250L
= 0.005M.
C1V1 =C2V2
0.005 × 250 mL = 0.2 × V2
Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.
A pentavalent cation atom has 20 and 15 neutrons as protons. Find the electron quantity and mass number respectively. (40 pts.) a) 20 and 15 b) 15 and 20 c) 15 and 35 d) 35 and 15 e) 10 and 20
Answer:
C.
Explanation:
Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.
Why Graphite is Diamagnetic?
Answer: Through years of studying and research ;
Graphite has shown that in weak and quantizing magnetic fields it is material is a highly anisotropie diamagnetic substance whose non-oscillating part of the magnetic suscepti- bility weakly depends on magnetic field.
Explanation:
Diamagnetism is a (very) weak form of magnetism which is caused (induced) by a change in the orbital motion of electrons mostly due to an applied magnetic field.
Explain why crude oil is a fossil fuel
Answer:
fossil fuels are exhaustible, available in limited quantity, takes a long time to replenish & is found under the earth.
Since crude oil satisfies all these conditions, it is a fossil fuel.
need this asap , help please
Answer:
Path A-B-D involves a catalyst and is slower than A-C-D
Explanation:
The diagram above illustrates both the catalyzed path and the uncatalyzed path of a chemical reaction.
The catalysed path is the path expressed with broken lines and the uncatalyzed path is the path expressed with thick small line as shown in the diagram above.
The catalyzed path has a higher activation energy than the uncatalyzed path.
Therefore, the catalyzed path will be slower that the uncatalyzed path because, the catalyzed path will require a higher energy to overcome the activation energy in order for the reaction to proceed to product.
On the other hand, the uncatalyzed path has a lower activation energy and a lesser amount of energy is needed to overcome it in order for the reaction to proceed to product.
Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe: PbO(s) NaCl(aq) H2O(l) --> Pb(OH)Cl(s) NaOH(aq) How many grams of PbO and how many grams of NaCl would be required to produce 10.0 g of Pb(OH)Cl
Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
[tex]= \frac{Mass}{Molar\ mass}[/tex]
[tex]= \frac{10.0 g}{259.65g / mol}[/tex]
= 0.0385 mol
Mass of PbO needed is
[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.
A student wants to prepare a salt starting with H2SO4. Select all of the compound types that can react with H2SO4 to form a salt.
1. salt
2. acid
3. acid salt
4. basic oxide
5. base
6. metal
7. acidic oxide
Answer:
4 and 6 would work for this
(-)-Cholesterol has a specific rotation of -32o. A mixture of ( )- and (-)-cholesterol was analyzed by polarimetry, and the observed rotation was 14o. What is the percent composition of the ( ) isomer in this mixture
Answer:
(+)-cholesterol = 71.88%
(-)-cholesterol = 28.12%
Explanation:
Asuming 1 gram of sample is dissolved in 1mL of water and the sample cell was 1dm long.
Enantiomeric excess is defined as the amount of pure enantiomer in a sample. The formula is:
ee = [α]mixture / [α]pure enantiomer.
Replacing:
ee = 14° / 32°×100 = 43.75%
As the sample is 14°, There is an excess of (+)-cholesterol and 56.25% is a 1:1 mixture of enantiomers.
That means percent composition of enantiomers is:
(+)-cholesterol = 43.75% + 56.25%/2 = 71.88%(-)-cholesterol = 56.25%/2 = 28.12%Identify which of the following are carbohydrates.
Check oh that apply
Answer:
3d and 4th
Explanation:
Carbohydrates general formula (CH2O)n.
Not 1st one because it has NH2-group.
2nd - C3H6O4, also it is acid.
3d - C3H6O3
4th - C6H12O6
Among the given compounds only compound present in 3rd option is a carbohydrate.
What are carbohydrates?Carbohydrates are the organic molecules which are made up of oxygen atom, carbon atom and hydrogen atom and general molecular formula of carbohydrates is CₙH₂ₙOₙ, where n is the number of atoms.
In the carbohydrate molecule, alcoholic group and aldehyde groups may present.
First given compound is not a carbohydrates as in that NH₃ group is present.Second given compound is also not a carbohydrate as in that carboxylic (-COOH) group is present.Third compound is a carbohydrate as in that compound same number of carbon and oxygen atoms are present and number of hydrogen atom is double to C & O atom.Fourth compound is also not a carbohydrate, as in that molecule number of carbon and oxygen atom is same but number of hydrogen is not double with respect to C & O.Hence compound C is carbohydrate.
To know more about carbohydrates, visit the below link:
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What is the half-life for the first order decay of 14C according to the reaction, 146C — 147N +e- ?
The rate constant for the decay is 1.21 x10-4 year-1
Answer:
5727 years or 5730 (rounded to match 3 sig figs) whichever one your teacher prefers
Explanation:
First Order decay has a half life formula of Half Life = Ln (2) / k = 0.693/K
Half-life = 0.693/k = 0.693/1.21 x10-4 = 5727 years or 5730 (rounded to match 3 sig figs)
This should be correct because if you google the half-life of 14 C it is ~ 5700 years
g Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.94 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 23.82 mL. What volume of base was required for the titration
Answer:
21.88mL is the volume of base required for the titration.
Explanation:
For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.
If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.
In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:
23.82mL - 1.94mL
21.88mL is the volume of base required for the titration.Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [
Answer: pH=2.38
Explanation:
To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.
HCOOH ⇄ H⁺ + HCOO⁻
I 1.0M 0 0
C -x +x +x
E 1.0-x x x
For the steps below, refer to the ICE chart above.
1. Since we were given the initial of HCOOH, we can fill this into the chart.
2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.
3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.
4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].
5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.
[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]
6. Once we plug this into the quadratic equation, we get x=0.00415.
7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).
-log(0.00415)=2.38
Our pH for the weak acid solution is 2.38.
Each unknown mixture contains 5 metal constituents. Select the 5 metal ions that you have identified as being present in your mixture. Please double check your selections before you hit the submit button. a. Ca b. Co c. Cr d. Fe e. K f. Mn g. Zn
Explanation:
A metal ion is a type of atom compound that has an electric charge.
Such atoms willingly lose electrons in order to build positive ions called cations. The selected Ions are :
[tex]1. Mn^2^+\\\ 2. Ca^2^+\\\ 3. Co^2^+\\\ 4. Fe^2^-\\\ 5. K^+[/tex]
How many moles of carbon atoms are there in 0.500 mol of C2H6?
The number of moles of carbon atoms in 0.500 mol of ethane (C₂H₆) is equal to one mole.
What is a mole?A mole can be defined as a scientific unit that is utilized to calculate the quantities such as atoms, molecules, ions, or other particular particles. The mass of one mole of a given chemical element is atomic mass and that of 1 mole of a chemical compound is molar mass.
The number of entities found in one mole is equal to 6.023 × 10 ²³ which is known as Avogadro’s constant.
Given, the number of moles of C₂H₆ = 0.500 mole
One molecule of ethane has carbons = 2
One mole of ethane has moles of carbons = 2 moles
0.500 mol of ethane has moles of carbon atoms = 0.500×2 = 1 mol
Therefore, one mole of carbon atoms is present in 0.500 mol of ethane C₂H₆.
Learn more about the mole, here:
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Which element has the largest atomic radius
Answer:
Francium
Explanation:
The atomic radius increases from top to bottom in a group, and decreases from left to right across a period.
so francium (Fr) is the largest atom or has highest radii.
Hope this helps & please mark as brainiest!
Answer:
Francium has the largest atomic radius.
The general trend for atomic radii is increasing from top to bottom and decreasing from left to right so the one with the largest atomic radius will be in the bottom left of the periodic table.
How many oxygen molecules are needed to make 10 carbon dioxide molecules according to the following balanced chemical equation? 2 CO + O2 → 2 CO2
five oxygen molecules
step by step explanation.
according to the equation,one molecule of oxygen is enough to react with two carbon molecules thus 10 carbon molecules need 5oxygen molecules
Question 11: How does the energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital compare to the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital?
Answer:
Explanation:
The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital
What is the specific heat of a 85.01 g piece of an unknown metal that exhibits a 45.2°C temperature change upon absorbing 1870 J of heat?
Answer:
[tex]0.48~\frac{J}{g~^{\circ}C}[/tex]
Explanation:
In this question, we have to remember the relationship between Q (heat) and the specific heat (Cp) the change in temperature (ΔT), and the mass (m).
[tex]Q=m*Cp*ΔT[/tex]
The next step is to identify what values we have:
[tex]Q~=~1870~J[/tex]
[tex]m~=~85.01~g[/tex]
[tex]ΔT~=~45.2~^{\circ}C[/tex]
[tex]Cp~=~X[/tex]
Now, we can plug the values and solve for "Cp":
[tex]1870~J=~85.01~g~*Cp*45.2~^{\circ}C[/tex]
[tex]Cp=\frac{1870~J}{85.01~g~*45.2~^{\circ}C}[/tex]
[tex]Cp=0.48~\frac{J}{g~^{\circ}C}[/tex]
The unknow metal it has a specific value of [tex]0.48~\frac{J}{g~^{\circ}C}[/tex]
I hope it helps!
which factor is most responsible for the fact that water is a liquid rather than a gas at room temperature
The answer is Hydrogyn bonding. It keeps the water molocules bonded together and in a liquid state, without it it'd be in a gashious state.
Answer:Hydrogen bonds keep them together in room temperature, hope this helps!
Explanation:
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Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. When 10.1 g of Mg reacts with 10.5 g of O2, 11.9 g of MgO are collected.
a) Write the balanced equation for this reaction, including all states of matter.
b) What is the limiting reactant in this reaction?
c) What is the theoretical yield for this reaction?
d) What is the percent yield for this reaction?
Answer:
A. 2Mg(s) + O2(g) —> 2MgO(s)
B. Mg is the limiting reactant.
C. Theoretical yield of MgO is 16.83g.
D. The percentage yield is 70.7%
Explanation:
A. The balanced equation for the reaction. This is given below:
2Mg(s) + O2(g) —> 2MgO(s)
B. Determination of the limiting reactant.
First, we shall determine the mass of Mg and O2 that reacted and the mass of MgO produced from the balanced equation. This is illustrated below:
Molar mass of Mg = 24g/mol
Mass of Mg from the balanced equation = 2 x 24 = 48g.
Molar mass of O2 = 16x2 = 32g/mol.
Mass of O2 from the balanced equation = 1 x 32 = 32g
Molar mass of MgO = 24 + 16 = 40g/mol
Mass of MgO from the balanced equation = 2 x 40 = 80g
Summary:
From the balanced equation above,
48g of Mg reacted with 32g of O2 to produce 80g of MgO.
Now, we can obtain the limiting reactant as follow:
From the balanced equation above,
48g of Mg reacted with 32g of O2.
Therefore, 10.1g of Mg will react with = (10.1 x 32)/48 = 6.73g of O2.
From the calculations made above, we can see that only 6.73g out of 10.5g of O2 given is needed to react completely with 10.1g of Mg.
Therefore, Mg is the limiting reactant and O2 is the excess reactant.
C. Determination of the theoretical yield of MgO.
The limiting reactant is used in this case as it will produce the maximum yield of the reaction since all of I is used up in the reaction.
The theoretical yield can be obtain as illustrated below:
From the balanced equation above,
48g of Mg reacted to produce 80g of MgO.
Therefore, 10.1g of Mg will react to produce = (10.1 x 80)/48 = 16.83g of MgO.
Therefore, the theoretical yield of MgO is 16.83g.
D. Determination of the percentage yield.
This is illustrated below:
Actual yield of MgO = 11.9g
Theoretical yield of MgO = 16.83g
Percentage yield =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 11.9/16.83 x 100
Percentage yield = 70.7%
when dissolved in water, an acid or a base breaks down into a. a proton and an electron b. two negative ions c. a positive and a negative ion d. a positive ion and a proton
Answer:
C. A positive and a negative ion
Explanation:
Acids and bases are made up of charged particles known as ions. The ions present in acids are oppositely charged and are held together by strong electrostatic forces. When acids or bases are dissolved in water, the electrostatic forces holding their individual molecules together are weakened and these ions are free to move apart in a process known as dissociation. Dissociation occurs because of the attraction between the positive and negative ions in the acid and bases and the negative and positive polarity of water.
For example, when an acid like hydrochloric acid is dissolved in water it dissociates into positive and negative ions as follows:
HCl(aq) -----> H+ + Cl-
When a base like sodium hydroxide is dissolved in water, it dissociates into positive and negative ions as follows:
NaOH(aq) ----> Na+ + OH-
Answer:
yeah C is correct
Explanation:
When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.430 g of a particular hydrocarbon was burned in air, 0.446 g of CO, 0.700 g of CO2, and 0.430 g of H2O were formed.
Required:
a. What is the empirical formula of the compound?
b. How many grams of O2 were used in the reaction?
c. How many grams would have been required for complete combustion?
Answer:
(a) The empirical formula of the compound is
m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).
(b) The grams of O2 that were used in the reaction is 1.146 g
(c) The amount of O2 that would have been required for complete combustion is 1.401 g.
Explanation:
a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)
(b) Using law of conservation of mass from above
m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)
m(O2) = 0.446 + 0.700 + 0.430 - 0.430
m(O2) = 1.146 g
The grams of O2 that were used in the reaction is 1.146 g
(c) for complete combustion, we need to oxidized CO to CO2
Then, 2CO +O2 = 2CO2
m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}
m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g
Note; Molar mass of O2 = 32, CO = 28
m(total)(O2) = m(O2) + m(add)(O2)
m(total)(O2) = 1.146 + 0.255 = 1.401 g
The amount of that grams would have been required for complete combustion is 1.401 g.
Note (add) and (total) were used subscript to "m"
Description (with words) of water just above melting temperature. What intermolecular forces do you expect to find in water in liquid state
Answer:
intermolecular dipole-dipole hydrogen bonds
Explanation:
Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.
Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.
Please solve quickly
Answer:
Explanation:
mass of one virus = 9.0 x 10⁻¹² mg
mass of one mole = 6.02 x 10²³ x mass of one virus
= 6.02 x 10²³ x 9.0 x 10⁻¹²
= 54.18 x 10¹¹ mg
= 54 x 10⁸ g .
= 54 x 10⁵ kg .
b )
let n be no of moles of virus that will be equal to weight of oil tanker
n x 54 x 10⁵ = 3 x 10⁷
n = 5.5555
rounding off to 2 significant figure
5.6 moles Ans .
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and placed in 112.0 mL of water at 20.5 ∘C . What is the final temperature of the water when thermal equilibrium is reached?
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
Explain with examples following characteristics of chemical reactions: a. Change of colour b. Evolution of gas c. Change of smell d. Change of state
Answer:
Explanation:
a. change of colour:
A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products. The products have different molecular structures than the reactants. Different atoms and molecules radiate different colours of light. Hence, there usually is a change in colour during a chemical reaction.
Eg: copper reactions with the elements
b. Evolution of gas:
A gas evolution reaction is a chemical reaction in which one of the end products is a gas such as oxygen or carbon dioxide.
Eg: ammonium hydroxide breaks down to water and ammonia gas.
c. Change of smell :
Production of an Odor Some chemical changes produce new smells. ... The formation of gas bubbles is another indicator that a chemical change may have occured.
Eg: The chemical change that occurs when an egg is rotting produces the smell of sulfur.
d. Change of state:
A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.
Eg: candle wax (solid) melts initially to produce molten wax (liquid)
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When a strip of magnesium metal is placed in an aqueous solution of copper(II) nitrate, elemental copper coats the surface of the magnesium strip and aqueous magnesium nitrate forms.
1. As a reactant, what is the charge on copper?
2. As a product, what is the charge on copper?
Answer:
1. 2+ ([tex]Cu^{2+}[/tex]).
2. 0 ([tex]Cu^0[/tex]).
Explanation:
Hello,
In this case, the described chemical reaction is a redox reaction in fact, since the oxidation states of both magnesium and copper change as shown due to the displacement:
[tex]Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)[/tex]
Therefore:
1. Since copper is the cation in the copper (II) nitrate, the (II) means that its charge is 2+ ([tex]Cu^{2+}[/tex]).
2. Since copper is alone, it means no electrons are being neither shared not given, its charge is 0 ([tex]Cu^0[/tex]).
Best regards.
what is the molarity of a solution that contains 49.8 grams of nai and is dissolved in enough water to make 1.50 liters
Answer: The molarity of solution is 0.221 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in L
moles of [tex]NaI[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{49.8g}{149.89g/mol}=0.332mol[/tex]
Now put all the given values in the formula of molality, we get
[tex]Molarity=\frac{0.332mol}{1.50L}[/tex]
[tex]Molarity=0.221mol/L[/tex]
Therefore, the molarity of solution is 0.221 M
What volume of 6.00 M hydrochloric acid is needed to prepare 500 mL of 0.100 M solution?
Answer:
8.33mL or .0083L
Explanation:
Use m1 * V1 = m2 * V2
6.00M(x) = 0.100M(500mL)
solve for x
x= (.1 * 500) / 6
x=8.333 mL
What is the mass of 3.45 moles
NO2?
(N = 14.01 g/mol, O = 16.00 g/mol)
Answer:
158.7 g
Its the right answer
Write the empirical formula
Answer:
[tex]1) NH_{4}IO_{3}\\2) Pb(IO_{3})_{4} \\3) NH_{4}(C_{2}H_{3}O_{2})\\4) Pb(C_{2}H_{3}O_{2})_{4}[/tex]
Explanation:
[tex]1) NH_{4}^{+}IO_{3}^{-} ---> NH_{4}IO_{3}\\2) Pb^{4+}(IO_{3}^{-})_{4} --->Pb(IO_{3})_{4} \\3) NH_{4}^{+}(C_{2}H_{3}O_{2})^{-} ---> NH_{4}(C_{2}H_{3}O_{2})\\4) Pb^{4+}(C_{2}H_{3}O_{2})^{-} _{4} --->Pb(C_{2}H_{3}O_{2})_{4}[/tex]