Answer:
353.10g/l
Explanation:
1000ml make 1 l
725ml are equivalent to 0.725l
Thus 625/0.725 makes a solution of
353.10g/l
If you mixed 265g of sugar to get a final volume of 725 ml, The concentration of the solutions will be 353.10g/l.
What is concentration?
Concentration is the presence of the amount of substance in a specific place ion specific amount. Concentration is also called molarity in chemistry.
The solution is a mixture of two or more substances. The solution is made when the ions of two substances mix with each other and show no difference.
Volume is the concentration or total space occupied by an object.
Here, the mass of the sugar is given 265g
The volume of the solution is 725 ml
The concentration of the solution is to be calculated
1000 ml equals to 1 l
So, 725ml are equivalent to 0.725 l
Hence, 625 / 0.725 = 353.10g/l
Thus, the concentration of the solutions is 353.10g/l.
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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:
MoO2SiO2 has a melting point of 1700°C, whereas CO2 has a melting point of -78°C. What is the reason for this huge difference between two elements in the same group?
The difference is because SiO2 is solid while CO2 is gaseous
Silicon and carbon are two elements that belong to the same group (Group 4) of the periodic table. They, however, both form oxides. Silicon bonds to two oxygen atoms to form silicon dioxide (SiO2) while carbon also bonds to two oxygen atoms to form CO2. According to this question, SiO2 has a melting point of 1700°C, whereas CO2 has a melting point of -78°C. The huge difference in the melting point values can be attributed to the state at which both molecules exist. SiO2 is a solid, hence, will require great heat to break the strong intermolecular forces that hold it's molecules. On the other hand, CO2 is a gas.The difference is the melting point values despite belonging to the same group is their STATE OF MATTER.
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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
Propane is a major component of natural gas used as fuel in homes. Write a balanced equation for the complete oxidation reaction that occurs when propane () burns in air.
Answer:
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
Explanation:
The chemical formula of propane is C₃H₈. When it burns in air, it reacts with gaseous oxygen (O₂), so carbon dioxide (CO₂) and water (H₂O) are formed. The combustion reaction is the following:
C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g)
Finally, we have to balance the equation. For this, we write a coefficient 3 in CO₂, then a coefficient 4 in H₂O, and a coefficient 5 in O₂. We obtain the balanced equation:
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
Please help answering 46)
Answer:
CO2 is a trigonal planar.
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
A sample of a compound is analyzed and found to contain 0.420 g nitrogen, 0.480g oxygen, 0.540 g carbon and 0.135 g hydrogen. What is the empirical formula of the compound? a. C2H5NO b. CH3NO c. C3H9N2O2 d. C4HN3O4 e. C4H13N3O3
Answer:
c. C3H9N2O2
Explanation:
The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:
Moles N -Molar mass: 14.01g/mol-
0.420g N * (1mol/14.01g) = 0.0300 moles N
Moles O -Molar mass: 16g/mol-
0.480g O * (1mol/16g) = 0.0300 moles O
Moles C -Molar mass: 12.01g/mol-
0.540g C * (1mol/12.01g) = 0.0450 moles C
Moles H -Molar mass: 1.0g/mol-
0.135g H * (1mol/1g) = 0.135moles H
Dividing in the moles of N (Lower number of moles) the ratio of atoms is:
N = 0.0300 moles N / 0.0300 moles N = 1
O = 0.0300 moles O / 0.0300 moles N = 1
C = 0.0450 moles C / 0.0300 moles N = 1.5
H = 0.135 moles H / 0.0300 moles N = 4.5
As the empirical formula requires whole numbers, multiplying each ratio twice:
N = 2, O = 2, C = 3 and H = 9
And the empirical formula is:
c. C3H9N2O2
A gas at 273K temperature has a pressure of 590 MM Hg. What will be the pressure if you change the temperature to 273K? 
Explanation:
here's the answer to your question
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
What is the mass of a piece of iron if its density is 1.98 g/mL and its volume is 2.45 mL?
0.80 g
4.858
1.248
5.998
2.71 g
Answer:
4.858 g
Explanation:
Start with the formula
density = [tex]\frac{mass}{volume}[/tex]
density = 1.98 g/mL
volume = 2.45 mL
mass = ??
rearrange the formula to solve for mass
(density) x (volume) = mass
Add in the substitutes and solve for mass
1.98 g/mL x 2.45 mL = 4.858 g
For each of the scenarios, determine if the ionic strength of the solution would increase, decrease, or not change.
a. If a solution of HNO3 were added to a solution of KOH , the ionic strength of the KOH solution would:_________
1. Increase
2. Decrease
3. Not change
b. If a dilute solution of KOH were added to a solution of CaCl2 (Ca(OH)2 (s) is formed), the ionic strength would:
1. Increase
2. Decrease
3. Not change
Answer:
Increase
Decrease
Explanation:While in solution, ionic substances produce ions. The ions in solution determine the conductivity of the solution.
The ionic strength of a solution shows the concentration of ions in a given solution. The more the number of ions in the solution, the greater the ionic strength of the solution and vice versa.
When HNO3 is added to a solution of KOH, the number of ions in the solution increases and so does the ionic strength of the solution.
When KOH is added to a solution of CaCl2 then Ca(OH)2 is formed. The formation of a solid precipitate decreases the concentration of ions in solution as well as the ionic strength of the solution.
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.
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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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
Find the oxidation number of:
A. sulfur in S032-
B. nickel in NiO2
c. iron in Fe(OH)2
Answer:
A. 4+.
B. 4+
C. 2+.
Explanation:
Hey there!
In this case, according to the given substances, it turns out possible for us to find the oxidation number of each element by applying the concept of charge balance in all of them as shown below:
A. sulfur in S032- : overall charge is 2- and the oxidation number of oxygen is 2-, thus:
[tex]x-6=-2\\\\x=6-2\\\\x=4+[/tex]
B. nickel in NiO2 : overall charge is 0 and the oxidation number of oxygen is 2-, thus:
[tex]x-4=0\\\\x=4+[/tex]
C. iron in Fe(OH)2: overall charge is 0 and the oxidation state of the OH ion is 1-, thus:
[tex]x-2=0\\\\x=2+[/tex]
Regards!
The standard enthalpies of combustion of fumaric acid and maleic acid (to form carbon dioxide and water) are - 1336.0 kJ moJ-1 and - 1359.2 kJ moJ-1, respectively. Calculate the enthalpy of the following isomerization process:
maleic acid ----> fumaric acid
Answer:
Explanation:
maleic acid ⇒ fumaric acid
ΔHreaction = ΔHproduct - ΔHreactant
ΔHproduct = -1336.0 kJ mol⁻¹
ΔHreactant = - 1359.2 kJ mol⁻¹.
ΔHreaction = -1336.0 kJ mol⁻¹ - ( - 1359.2 kJ mol⁻¹.)
= 1359.2 kJ mol⁻¹ -1336.0 kJ mol⁻¹
= 23.2 kJ mol⁻¹ .
Enthalpy of isomerization from maleic to fumaric acid is 23.2 kJ per mol.
Calculate the temperature in k of 3.05 moles of gas occupying 3.70 L at 4.12 atm
Answer:
[tex]{ \bf{PV = nRT}} \\ { \tt{(4.12 \times 3700) = 3.05 \times 0.083 \times T }} \\ { \tt{15244 = 0.25315 \: T}} \\ { \tt{T = 6.02 \times {10}^{4} \: kelvin }}[/tex]
The temperature of the given gas is 60.95 K when it is occupying 3.70 L at 4.12 atm.
What is the ideal gas equation?The ideal gas law can be described as a general equation of the state of an ideal gas. This equation gives the relationship between the volume and pressure of one-mole of gas equal to the multiplication of the universal gas constant and temperature.
The mathematical relationship can be shown for the ideal gas equation as:
PV = nRT
Where P is the pressure of the gas, n is the moles, V is the volume of the gas, and R is the gas constant.
Given, the volume of gas, V = 3.70 L
The pressure of the given gas, P = 4.12 atm
The value of the gas constant, R =0.082 atmL/K mol
The number of moles of the given gas, n = 3.05 mol
Substitute the values V, R, P, and n in the ideal gas equation, and we get:
T = PV/nR
T = 4.12 × 3.70/(0.082 × 3.05)
T = 60.95 K
Therefore, the temperature of the given gas is 60.95 K.
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The second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings. What is the thermodynamic driving force for dissolving a solid in a liquid if it is an endothermic process (which reduces the entropy of the surroundings)
Answer:
See explanation
Explanation:
Truly, the second law of thermodynamics requires that spontaneous processes generate entropy, either in the system or in the surroundings.
When a solid is dissolved in a liquid, the solid dissociates into ions. These ions increases the number of particles and hence the entropy of the system thereby making the process spontaneous.
Hence, the dissolution of a substance via an endothermic process is spontaneous because of increase in the number of particles which in turn increases the entropy of the system.
here is the question
Answer:
1. Nitrate ions, NaNO3 - Sodium nitrate.
2. Sulphide ions, K2S - Potassium sulphide.
3. Sulphate ions, CaSO4 - Calcium sulphate.
4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.
5. Carbonate ions, CaCO3 - Calcium carbonate.
6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.
7. Phosphite ions, PH3 - Hydrogen phosphite.
8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).
9. Ethanoate ions, CH3COONa - Sodium ethanoate.
10. Methanoate ions, HCOONa - Sodium methanoate.
11. Fluoride ions, HF - Hydrogen fluoride.
12. Chloride ions, KCl - Potassium chloride.
13. Bromide ions, HBr - Hydrogen bromide.
14. Iodide ions, NaI - Sodium iodide.
15. Phosphate ions, K3PO3 - potassium phosphate.
tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
How much energy is released when 31.0 g of water freezes? The heat of fusion for water is 6.02 kJ/mol.
Express your answer in kilojoules to three significant figures.
Answer:
The molar heat of fusion for a given substance basically tells you how much heat is required to melt one mole of that substance at its melting point from two angles.
Explanation:
:)
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
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
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 recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization
This question is incomplete, the complete question is;
In recrystallization from boiling water of benzoic acid contaminated with acetanilide, you begin with an impure sample of 5 grams. If the % composition of the acetanilide impurity in the sample is 6.3 %, what is the minimum amount in mL of solvent (water) required for the recrystallization
Compound Solubility in water at 25°C Solubility in water at 100°C
Benzoic Acid 0.34 g/100mL 5.6 g/100mL
Acetanilide 0.53 g/100mL 5.5 g/100mL
Answer:
The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL
Explanation:
Given the data in the question;
mass of sample = 5 g
percentage composition of the acetanilide impurity = 6.3%
mass of the acetanilide in impure sample will be;
⇒ 6.3% × 5 g = 0.315 g
Mass of benzoic acid in impure sample;
⇒ 5 g - 0.315 g = 4.685 g
now, solubility in water at 100°C for benzoic acid = 5.6 g/100mL
hence 4.685 g of benzoic acid is soluble in x mL
x = [ 100 mL × 4.685 g ] / 5.6 g
x = 83.66 ≈ 84 mL
Also, solubility in water at 100°C for acetanilide = 5.5 g/100mL
hence 0.315 g of benzoic acid is soluble in x mL
x = [ 100 mL × 0.315 g ] / 5.5 g
x = 5.727 ≈ 6 mL
So, the minimum amount in mL of solvent (water) required for the recrystallization will be;
⇒ 85 mL + 6 mL = 91 mL
The minimum amount in mL of solvent (water) required for the recrystallization is 91 mL
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
The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. Decide, in each case, whether or not an insoluble precipitate is formed.
a. K2S and NH4Cl
b. CaCl2 and NH4CO3
c. Li2S and MnBr2
d. Ba(NO3)2 and Ag2SO4
e. RbCO3 and NaCl
Answer:
a) [tex]K_{2} S[/tex] and [tex]NH_{4} Cl[/tex] :
There are no insoluble precipitate forms.
b) [tex]Ca Cl_{2}[/tex] and [tex](NH_{4} )_{2} Co_{3}[/tex] :
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c) [tex]Li_{2}S[/tex] and [tex]MnBr_{2}[/tex] :
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d) [tex]Ba(No_{3} )_{2}[/tex] and [tex]Ag_{2} So_{4}[/tex] :
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e) [tex]Rb_{2}Co_{3}[/tex] and [tex]NaCl[/tex]:
There are no insoluble precipitates forms.
Explanation:
a)
Solubility rule suggests:- [tex]K_{2} S[/tex] ⇒ soluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
KCl ⇒ soluble, [tex](NH_{4})_{2} S[/tex] ⇒ soluble.
There are no insoluble precipitate forms.
b)
Solubility rule suggests:- [tex]Ca Cl_{2}[/tex] ⇒ soluble, [tex](NH_{4} )_{2} Co_{3}[/tex] ⇒ soluble.
[tex]CaCo_{3}[/tex] ⇒ insoluble, [tex]NH_{4} Cl[/tex] ⇒ soluble.
There are the insoluble precipitates of [tex]CaCo_{3}[/tex] forms.
c)
Solubility rule suggests:- [tex]Li_{2}S[/tex] ⇒ soluble, [tex]MnBr_{2}[/tex] ⇒ soluble.
[tex]LiBr[/tex] ⇒ soluble, [tex]MnS[/tex] ⇒ insoluble.
There are the insoluble precipitates of [tex]MnS[/tex] forms.
d)
Solubility rule suggests:- [tex]Ba(No_{3} )_{2}[/tex] ⇒ soluble, [tex]Ag_{2} So_{4}[/tex] ⇒insoluble.
As [tex]Ag_{2} So_{4}[/tex] is insoluble, therefore no precipitate forms.
e)
Solubility rule suggests:- [tex]Rb_{2}Co_{3}[/tex] ⇒ soluble, [tex]NaCl[/tex] ⇒ soluble.
[tex]RbCl[/tex] ⇒ soluble, [tex]Na_{2} Co_{3}[/tex] ⇒ soluble.
There are no insoluble precipitates forms.
The numerical value of 0.001 is written with the prefix:
Answer:
Milli, m.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to answer to this question by bearing to mind the attached file whereas the most common prefixes and their factors are shown both in standard and scientific notation.
In such a way, we will be able to infer that the prefix related to the numerical value of 0.001 is milli, m, for example 1 mm which is 0.001 m.
Regards!
Identify each of the following properties as more typical of organic or inorganic compound
a. contains Li and F
b. is a gas at room temperature
c. contains covalent bonds
d. produces ion in water
this is the difference between organic and inorganic if this doesn't help you can research more on it
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.
