Answer:
It is prepared small amounts of hydrogen cloride for uses in the lab.
It can be "generated in an HCl generator by dehydrating hydrochloric acid with either sulfuric acid or anhydrous calcium chloride."
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
If the starting material has no stereogenic centers, when carbonyl compounds are reduced with a reagent such as LiAlH4 or NaBH4 and a new stereogenic center is formed, what will the composition of the product mixture be?
A) Forms a racemic mixture of the two possible enantiomers.
B) Forms more of one enantiomer than another because of steric reasons around the carbonyl.
C) Forms more of one enantiomer than another depending on the temperature of the reaction.
D) Forms different products depending on the solvent used.
Answer:
A) Forms a racemic mixture of the two possible enantiomers
When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.
What is a chemical change?
Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.
There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.
There are three types of chemical changes:
1) inorganic changes
2)organic changes
3) biochemical changes
During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.
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t the centers of some stars, fluorine-19 can undergo a fusion reaction with a proton (a hydrogen-1 nucleus) to produce two different nuclei. One of the products is an alpha particle (a helium-4 nucleus). What is the other
Answer:
Explanation:
¹⁹F₉ + ¹H₁ = ⁴He₂ + ¹⁶X₈
The atomic weight of fluorine is 19 and atomic weight of hydrogen is 1 . Total becomes 20 . Helium , the product has atomic weight of 4 . Therefore atomic weight of product X must be 16 . Similarly we can calculate atomic number of product X . It is found to be 8 .
Hence atomic weight and number of product will be 16 and 8 respectively.
It is similar to oxygen atom
Hence the other product will be oxygen.
In order to dry wet clothes, we spread them on a clothline. This is because (i) spreading increases surface area (ii) clothes become brighter when spread
Answer:
this is because spreading it makes more sunlight hit the cloth which results in it drying faster
How many moles of gas occupy a volume of 101.3L?
Answer:
V= n VmV: gas volume , n : The number of moles of gas , Vm : molar volume*The molar volume of any gas at standard conditions of temperature and pressure is 22.4 L/mol
V= 101.3 L , n=? , Vm = 22.4 L/mol
V=n Vm101.3 = n × 22.4
n=101.3 / 22.4
n = 4.52 mol
I hope I helped you ^_^
Answer:
[tex]\boxed {\boxed {\sf 4.522 \ mol}}[/tex]
Explanation:
We are asked to find how many moles of gas occupy a volume of 101.3 liters.
1 mole of any gas at STP (standard temperature and pressure) has a volume of 22.4 liters. We can use this information to make a proportion.
[tex]\frac {1 \ mol}{22.4 \ L}[/tex]
We are converting 101.3 liters to moles, so we multiply the proportion by that value.
[tex]101.3 \ L *\frac {1 \ mol}{22.4 \ L}[/tex]
The units of liters (L) cancel.
[tex]101.3 *\frac {1 \ mol}{22.4}[/tex]
[tex]\frac {101.3}{22.4} \ mol[/tex]
Divide.
[tex]4.52232143 \ mol[/tex]
The original value of liters (101.3 L) has 4 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 3 in the ten-thousandths place to the right tells us to leave the 2 in the thousandths place.
[tex]4.522 \ mol[/tex]
101.3 liters of gas is equal to approximately 4.522 moles of gas.
4.
Ammonia gas occupies a volume of 450. mL at a pressure of 720 mm Hg. What volume in
liters will the gas occupy at standard atmospheric pressure?
Answer:
[tex]\boxed {\boxed {\sf 426 \ mL}}[/tex]
Explanation:
We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:
[tex]P_1V_1= P_2V_2[/tex]
The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.
[tex]450 \ mL * 720 \ mm \ Hg = P_2V_2[/tex]
The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.
[tex]450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2[/tex]
We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.
[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}[/tex]
[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2[/tex]
The units of millimeters of mercury (mm Hg) cancel.
[tex]\frac {450 \ mL * 720 }{760} = V_2[/tex]
[tex]\frac {324,000}{760} \ mL = V_2[/tex]
[tex]426.3157895 \ mL =V_2[/tex]
The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.
[tex]426 \ mL \approx V_2[/tex]
The volume at standard atmospheric pressure is approximately 426 milliliters.
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
a) potassium tetracyanonickelate(II)
b) sodium diamminedicarbonatoruthenate(III)
c) diamminedichloroplatinum(II)
Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
The empirical formula is OsO₄ :
Explanation:
Osmium oxide contains osmium and oxygen only.
Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:
Mass of compound = 2.89 g
Mass of Os = 2.16 g
Mass of O =?Mass of O = (Mass of compound) – (Mass of Os)
Mass of O = 2.89 – 2.16
Mass of O = 0.73 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Os = 2.16 g
Mass of O = 0.73 g
Empirical formula =..?Os = 2.16 g
O = 0.73 g
Divide by their molar mass of
Os = 2.16 / 190 = 0.011
O = 0.73 / 16 = 0.046
Divide by the smallest
Os = 0.011 / 0.011 = 1
O = 0.046 / 0.011 = 4
Empirical formula = OsO₄Learn more: https://brainly.com/question/23629778
How do I do this? What are the answers to the 5 questions shown?
Answer:
1. C₃H₆O₃
2. C₆H₁₂
3. C₆H₂₄O₆
4. C₆H₆
5. N₂O₄
Explanation:
1. Determination of the molecular formula.
Empirical formula => CH₂O
Mass of compound = 90 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂O]ₙ = 90
[12 + (2×1) + 16]n = 90
[12 + 2 + 16]n = 90
30n = 90
Divide both side by 30
n = 90/30
n = 3
Molecular formula = [CH₂O]ₙ
Molecular formula = [CH₂O]₃
Molecular formula = C₃H₆O₃
2. Determination of the molecular formula.
Empirical formula => CH₂
Mass of compound = 84 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂]ₙ = 84
[12 + (2×1)]n = 84
[12 + 2]n = 84
14n = 84
Divide both side by 14
n = 84/14
n = 6
Molecular formula = [CH₂]ₙ
Molecular formula = [CH₂]₆
Molecular formula = C₆H₁₂
3. Determination of the molecular formula.
Empirical formula => CH₄O
Mass of compound = 192 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₄O]ₙ = 192
[12 + (4×1) + 16]n = 192
[12 + 4 + 16]n = 192
32n = 192
Divide both side by 32
n = 192/32
n = 6
Molecular formula = [CH₄O]ₙ
Molecular formula = [CH₄O]₆
Molecular formula = C₆H₂₄O₆
4. Determination of the molecular formula.
Empirical formula => CH
Mass of compound = 78 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH]ₙ = 78
[12 + 1]n = 78
13n = 78
Divide both side by 13
n = 78/13
n = 6
Molecular formula = [CH]ₙ
Molecular formula = [CH]₆
Molecular formula = C₆H₆
5. Determination of the molecular formula.
Empirical formula => NO₂
Mass of compound = 92 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[NO₂]ₙ = 92
[14 + (2×16)]n = 92
[14 + 32]n = 92
46n = 92
Divide both side by 46
n = 92/46
n = 2
Molecular formula = [NO₂]ₙ
Molecular formula = [NO₂]₂
Molecular formula = N₂O₄
For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb
Answer:
11.12 → pH
Explanation:
This is a titration of a weak base and a strong acid.
In the first step we did not add any acid, so our solution is totally ammonia.
Equation of neutralization is:
NH₃ + HCl → NH₄Cl
Equilibrium for ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ Kb = 1.8×10⁻⁵
Initially we have 50 mL . 0.10M = 5 mmoles of ammonia
Our molar concentration is 0.1 M
X amount has reacted.
In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.
Expression for Kb is : x² / (0.1 - x) = 1.8×10⁻⁵
As Kb is so small, we can avoid the x to solve a quadratic equation.
1.8×10⁻⁵ = x² / 0.1
1.8×10⁻⁵ . 0.1 = x²
1.8×10⁻⁶ = x²
√1.8×10⁻⁶ = x → 1.34×10⁻³
That's the value for [OH⁻] so:
1×10⁻¹⁴ = [OH⁻] . [H⁺]
1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²
- log [H⁺] = pH
- log 7.45×10⁻¹² = 11.12 → pH
Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 10.1 mL of water (d= 1.00 g/mL) and 23.9 g of SO3? How much of the reagent in excess is left over?
Answer:
29.2 g of H₂SO₄ are produced
0.263 moles of water remain after the reaction goes complete.
Explanation:
We make the reaction in the first step:
Reactants are water and SO₃
H₂O + SO₃ → H₂SO₄
Let's determine moles of reactants:
23.9 g . 1 mol / 80.06g = 0.298 moles
We apply density, to determine mass of water:
D = m/ V so m = D . V
1 g/mL . 10.1 mL = 10.1 g
moles of water are: 10.1 g . 1 mol/ 18g = 0.561 moles
As ratio is 1:1, for 0.298 moles of SO₃ we need the same amount of water, and we have 0.561 moles. Then, water is the excess reagent and sulfur trioxide is the limiting.
0.561 - 0.298 = 0.263 moles of water that remain after the reaction goes complete.
As ratio is 1:1, again, 0.298 moles of SO₃ can produce 0.298 moles of acid.
We determine the mass: 0.298 mol . 98.06 g /mol = 29.2 g
Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.
Answer:
10.77%
Explanation:
Molar mass of Cu = mass deposited/number of moles of Cu
Molar mass of Cu = 0.4391 g/6.238x10^-3 moles
Molar mass of Cu = 70.391 g/mol
%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100
%error = 10.77%
What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationif an element has an atomic number of 9 what is the electronic structure of the same element
9 is the element Florine
Florine has 9 electrons as well as the 9 protons that determine its atomic number.
The ground state configuration is the lowest energy configuration.
what are the properety of covalent bond
Explanation:
1. boiling and melting point
2. electrical conductivity
3. Bond strength
4. bond length
A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.
how many moles of oxygen are present in 16 g of oxygen gas
Answer:
Mole = molecular weight / molecular mass
Mole = 16/16
Mole= 1
Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5
B)5,5
C)6,5
D)7,5
E)9,5
5. Calcule las concentraciones cuando se alcanza el equilibrio si partimos de unas concentraciones iniciales [A]=[B]=1M ; [C]=[D]=0M y una constante de equilibrio de 5.
Las concentraciones en el equilibrio para la reacción química presentada son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
Consideremos la siguiente reacción química genérica:
A + B ⇄ C + D
Para calcular las concentraciones en el equilibrio, debemos construir una Tabla ICE. Cada fila representa una instancia (Inicial, Cambio, Equilibrio) y la completamos con la concentración o cambio de concentración ("x" para concentraciones desconocidas). Como inicialmente no hay productos, la reacción se desplazará hacia la derecha para alcanzar el equilibrio.
A + B ⇄ C + D
I 1 1 0 0
C -x -x +x +x
E 1-x 1-x x x
La constante de equilibrio, Kc, es:
[tex]Kc = 5 = \frac{[C][D]}{[A][B]} = \frac{x^{2} }{(1-x)^{2} } \\\sqrt{5} = x/1-x\\x = 0.69[/tex]
Las concentraciones en el equilibrio son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
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What is the molarity of a solution that contains 0.75 mol Naci in 3.0 L of solution? Select one: O a. 4.0 M O b. 2.3 M O d. 3.8 M O d. 0.25 M Clear my choice
Answer:
[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]
The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.
moles of solute = 0.75 mol NaCl liters of solution = 3.0 LSubstitute these values into the formula.
[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]
[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]
Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.
[tex]molarity= 0.25 \ M \[/tex]
The molarity of the solution is 0.25 Molar and Choice D is correct.
Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 2.9 g of sulfuric acid is mixed with 3.53 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
1.07 g of water.
Explanation:
A reaction between an acid and a base makes water and a salt as product.
Our reaction is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Reactants are the acid and the base. Which is the limiting?
2.9 g . 1mol /98 g = 0.0296 moles of acid
3.53 g . 1mol / 40 g = 0.088 moles of base
2 moles of base react to 1 mol of acid
0.088 moles may react to (0.088 . 1)/2 = 0.044 moles of acid
And we only have 0.0296, sulfuric acid is the limiting
Ratio is 1:2. 1 mol of acid can produce 2 moles of water.
Our 0.0296 moles may produce (0.0296 . 2) /1 = 0.0592 moles of water.
We convert moles to mass:
0.0592 mol . 18g /mol = 1.07 g
Consider the following reaction:
Cr(NO3)3 (aq) + 2NaF (aq) --> 3NaNO3 (aq) + CrF3 (s)
If 21.0 grams of NaF are needed to precipitate all of the Cr+3 ions present in 0.125L of a solution of Cr(NO3)3, what is the molarity of the Cr(NO3)3 solution?
Your answer should be to 2 decimal places.
Answer:
2.01
Explanation:
First, let's convert grams to moles
(Na) 22.99 + (F) 18.998 = 41.988
Every mole of NaF is 41.988 grams
21/41.988 = 0.500143 moles of NaF
For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF
0.500143/2 = 0.250071
molarity = moles/liters
0.250071/0.125 = 2.0057 M
1. What happens to global temperature averages that start an ice age?
Answer:
Around 46° F (7.8° C)
Explanation:
"Scientists have predicted that the global average temperature during the ice age was around 46 degrees Farenheit (7.8 degrees Celsius.) However, the polar regions were far colder, around 25 degrees Fahrenheit (14 degrees Celsius) colder than the global average."
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For an atoms electrons, how many energy sublevels are present in the principal energy level n = 4?
A. 4
B. 9
C. 10
D. 16
E. 32
Answer:
by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?
Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
the density of oxygen 1.43 gm/liter at 0°c and pressure 1.0 atm. if a 20 liter cylinder is filled with oxygen at pressure of 25 atm and temperature of 27°c. what is the mass of oxygen in the cylinder
Answer:
640 g
Explanation:
Step 1: Given and required data
Volume of the cylinder (V): 20 LPressure of the oxygen (P): 25 atmTemperature (T): 27 °C (300 K)Ideal gas constant (R): 0.082 atm.L/mol.KStep 2: Calculate the moles of oxygen gas
We will use the ideal gas equation
P × V = n × R × T
n = P × V / R × T
n = 25 atm × 20 L / (0.082 atm.L/mol.K) × 300 K = 20 mol
Step 3: Calculate the mass corresponding to 20 moles of oxygen
The molar mass of oxygen is 32.00 g/mol.
20 mol × 32.00 g/mol = 640 g
A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide
Answer:
MoO2
Explanation:
The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.
To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.
The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:
Moles Mo2O3 -Molar mass: 239.9g/mol-
12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo
Mass Mo -95.95g/mol-:
0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo
Mass oxygen in the oxide:
13.197 - 9.895g = 3.302g Oxygen
Moles oxygen -Molar mass: 16g/mol-:
3.302g Oxygen * (1mol / 16g) = 0.206 moles O
Now, the ratio of moles O / moles Mo is:
0.206 moles O / 0.1031 moles Mo = 2
That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:
MoO2PLEASE HELP ME
a)
b)
c)
or d)?
Answer:
D / 15.0 g
Explanation:
3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions
.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle
Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.
Question Completion with Options:
O coarse...few...rapid
O fine...few...slow
O fine...multiple...rapid
O coarse...few...slow
O fine...multiple...slow
Answer:
The choice that best completes the sentence is:
O coarse...few...slow
Explanation:
Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow. This is because of the process that starts with recrystallization, recovery, and nucleation before growth can occur. While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.