Answer:
1.15 moles of excess reactant will remain after precipitation is complete.
Explanation:
The balanced reaction is:
2 AgNO₃ (aq) + Na₂SO₄ (aq) → Ag₂SO₄ (s) + 2 NaNO₃ (aq)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
AgNO₃: 2 molesNa₂SO₄: 1 moleAg₂SO₄: 1 moleNaNO₃: 2 molesThen you can apply the following rule of three:: if by stoichiometry 2 moles of AgNO₃ reacts with 1 mole of Na₂SO₄, 3.80 moles of AgNO₃ reacts with how much moles of Na₂SO₄?
[tex]amount of moles of Na_{2}SO_{4} =\frac{1mole of Na_{2}SO_{4} * 3.80 moles of AgNO_{3} }{2 mols of AgNO_{3} }[/tex]
amount of moles of Na₂SO₄= 1.9 moles
But 3.05 moles of Na₂SO₄ are available. Since you have more moles than you need to react with 3.80 moles of AgNO₃, Na₂SO₄ will be the excess reagent.
To calculate the amount of excess reagent that will remain, you must make the difference between the amount you initially have and the amount that reacts:
3.05 moles - 1.9 moles= 1.15 moles
1.15 moles of excess reactant will remain after precipitation is complete.
Gaseous ethane (CH,CH,) will react with gaseous oxygen (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). Suppose 4.21 g of
ethane is mixed with 31. 9 of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has
the correct number of significant digits.
Answer: The mass of [tex]CO_2[/tex] produced is 12.32 g
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For ethane:Given mass of ethane = 4.21 g
Molar mass of ethane = 30 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ethane}=\frac{4.21g}{30g/mol}=0.140mol[/tex]
For oxygen gas:Given mass of oxygen gas = 31.9 g
Molar mass of oxygen gas= 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{31.9g}{32g/mol}=0.997mol[/tex]
The chemical equation for the combustion of ethane follows:
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
By stoichiometry of the reaction:
If 2 moles of ethane reacts with 7 moles of oxygen gas
So, 0.140 moles of ethane will react with = [tex]\frac{7}{2}\times 0.140=0.49mol[/tex] of oxygen gas
As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, ethane is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 2 moles of ethane produces 4 moles of [tex]CO_2[/tex]
So, 0.140 moles of ethane will produce = [tex]\frac{4}{2}\times 0.140=0.28mol[/tex] of [tex]CO_2[/tex]
We know, molar mass of [tex]CO_2[/tex] = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }CO_2=(0.28mol\times 44g/mol)=12.32g[/tex]
Hence, the mass of [tex]CO_2[/tex] produced is 12.32 g
Balance the equation by ion electron method
Answer:i believe you are to decompose the formula (i think)
Which of the choices below has more heat being transferred as thermal energy from one place to another?
A. A bowl of ice water
B. A pot of boiling water
Answer:
B
Explanation:
So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.
propose a synthetic route for the synthesis of a named alkanal starting with ethyl formate and grignard reagent.
Answer:
See explanation and image attached
Explanation:
A Grignard reagent is an alkyl magnesium halide. If it reacts with ethyl formate, an intermediate is formed as shown.
This intermediate can undergo water hydrolysis to form a diol, ethanol and MgBrOH.
Oxidation of the diol obtained now yields the corresponding alkanal which in this case is ethanal.
The scheme of the reaction is shown in the image attached to this answer.
g When aqueous solutions of and are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M is mixed with an excess of an aqueous solution of .
The question is incomplete, the complete question is:
When aqueous solutions of NaCl and [tex]Pb(NO_3)_2[/tex] are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M NaCl is mixed with an excess of an aqueous solution of
Answer: The mass of lead chloride produced is 1.96 g
Explanation:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{ \text{Volume of solution (mL)}}[/tex] .....(1)
Given values:
Molarity of NaCl = 0.1000 M
Volume of the solution = 140.7 mL
Putting values in equation 1, we get:
[tex]0.1000=\frac{\text{Moles of NaCl}\times 1000}{140.7}\\\\\text{Moles of NaCl}=\frac{0.1000\times 140.7}{1000}=0.01407mol[/tex]
The chemical equation for the reaction of NaCl and lead nitrate follows:
[tex]Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)[/tex]
By the stoichiometry of the reaction:
If 2 moles of NaCl produces 1 mole of lead chloride
So, 0.01407 moles of NaCl will produce = [tex]\frac{1}{2}\times 0.01407=0.007035mol[/tex] of lead chloride
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)
Molar mass of lead chloride = 278.1 g/mol
Plugging values in equation 2:
[tex]\text{Mass of lead chloride}=(0.007035mol\times 278.1g/mol)=1.96g[/tex]
Hence, the mass of lead chloride produced is 1.96 g
which of the following illustrates a reversible change a cooking corn be rusting c frying egg and the boiling water
Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.
CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________
Answer:
Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.
CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________
Explanation:
Given alcohol is propanol.
When it reacts with TsCl, the hydrogen in -OH group is replaced with tosyl group.
Pyridine is a weak base and it neutralizes the HCl (acid) formed during the reaction.
The reaction is shown below:
20ml of water is mixed with 40gm of fine powder. Calculate the concentration of the solution obtained.
Answer:
[tex]\%m=66.7\%[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.
Next, we apply the following equation to obtain the required concentration:
[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]
Regards!
Which statement best describes the formula equation Cl2(g) + 2KBr(aq) Right arrow. 2KCl(aq)+ Br2(l)?
Carbon iodide reacts with potassium bromide to form potassium carbon iodide and bromine.
Bromine gas reacts with a solution of potassium chloride to form potassium bromide and chlorine gas.
Potassium bromine gas reacts with liquid chlorine to form potassium chloride in solution and bromine gas.
Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
Answer:
Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
Answer:
DDDDD
Explanation:
Calculate [OH - ] given [H 3 O + ] = 5.69x10 -5 M.
Answer:
7
Explanation:
The rate equation for a reaction between substances C and D is:
rate = k[C]^2 [D]^2
The initial rate is found to be 7.5 x 10^-3mol dm^-3s^-1when the initial concentration of
C is of 0.25 mol dm^-3and the initial concentration of D is 0.50 mol dm^-3.
Calculate the value of the rate constant, k, at this temperature and deduce its units.
*Important Asap please *
Answer:
[tex]rate = k[C] {}^{2} [D] {}^{2} \\ 7.5 \times {10}^{ - 3} = k {(0.25)}^{2} {(0.50)}^{2} \\ k = \frac{7.5 \times {10}^{ - 3} }{ {(0.25)}^{2} {(0.50)}^{2} } \\ k = 0.48 \: {mol}^{ - 3} {dm}^{9} {s}^{ - 1} [/tex]
NCl3 + 3H20 - NH3 + 3HCIO
How many grams of ammonia can be produced from 1.33 grams of nitrogen trichloride?
Answer:
0.189 g
Explanation:
Step 1: Write the balanced equation
NCl₃ + 3 H₂O ⇒ NH₃ + 3 HCIO
Step 2: Calculate the moles corresponding to 1.33 g of NCl₃
The molar mass of NCl₃ is 120.36 g/mol.
1.33 g × 1 mol/120.36 g = 0.0111 mol
Step 3: Calculate the moles of NH₃ produced from 0.0111 moles of NCl₃
The molar ratio of NCl₃ to NH₃ is 1:1. The moles of NH₃ produced are 1/1 × 0.0111 mol = 0.0111 mol.
Step 4: Calculate the mass corresponding to 0.0111 moles of NH₃
The molar mass of NH₃ is 17.03 g/mol.
0.0111 mol × 17.03 g/mol = 0.189 g
Helppp
What do you need to know in order to find the mass of 3.00 moles of carbon?
Answer:
36g
Explanation:
you need to know the equation mass=moles*mr (in this case mr of carbon which is 12)
so 3*12=36g
hope this helps :)
a polluted lake is 0.300 μg (micrograms) per liter of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 15.0 square miles and an average depth of 27.0 feet?
Answer:
95.9 kg
Explanation:
First we convert 15.0 mi² to m²:
15.0 mi² * ([tex]\frac{1609.34 m}{1mi}[/tex])² = 3.88x10⁷ m²Then we convert 27.0 ft to m:
27.0 ft * [tex]\frac{0.3048m}{1ft}[/tex] = 8.23 mNow we calculate the total volume of the lake:
3.88x10⁷ m² * 8.23 m = 3.20x10⁸ m³Converting 3.20x10⁸ m³ to L:
3.20x10⁸ m³ * [tex]\frac{1000L}{1m^3}[/tex] = 3.20x10¹¹ LNow we calculate the total mass of mercury in the lake, using the given concentration:
0.300 μg / L * 3.20x10¹¹ L = 9.59x10¹⁰ μgFinally we convert μg to kg:
9.59x10¹⁰ μg * [tex]\frac{1kg}{1x10^9ug}[/tex] = 95.9 kgPhosphine, PH3, a reactive and poisonous compound, reacts with oxygen as follows: 4PH3(g) 8O2(g) - P4O10(s) 6H2O(g) If you need to make 6.5 moles of P4O10, how many moles of PH3 is required for the reaction
Answer: 26 moles of [tex]PH_3[/tex] are required for the reaction.
Explanation:
We are given:
Moles of [tex]P_4O_{10}[/tex] = 6.5 moles
The given chemical reaction follows:
[tex]4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)[/tex]
By the stoichiometry of the reaction:
If 1 mole of [tex]P_4O_{10}[/tex] is produced by 4 moles of [tex]PH_3[/tex]
So, 6.5 moles of [tex]P_4O_{10}[/tex] will be produced by = [tex]\frac{4}{1}\times 6.5=26mol[/tex] of [tex]PH_3[/tex]
Hence, 26 moles of [tex]PH_3[/tex] are required for the reaction.
Please help me, it’s my last try
Answer:
Group 1A: alkali metals, or lithium family.
Group 2A: alkaline earth metals, or beryllium family.
Group 7A: the manganese family.
Group 8A: the iron family.
Explanation:
Answer:
1A: Alkali Metals
2A: Alkaline Earth Metals
7A: Halogens
8A: Noble Gases
HCIO4 is identified as what acid
When 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48°C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat (in GJ) produced by combustion of a ton of coal.
Answer:
32.0 kJ
General Formulas and Concepts:
Thermochemistry
Specific Heat Formula: q = mcΔT
q is heat (in J) m is mass (in g) c is specific heat (in J/g °C) ΔT is change in temperature (in °C)Explanation:
Step 1: Define
Identify variables
[Given] m = 1.00 g
[Given] ΔT = 1.48 °C
[Given] c = 21.6 kJ/g °C
[Solve] q
Step 2: Find Heat
Substitute in variables [Specific Heat Formula]: q = (1.00 g)(21.6 kJ/g °C)(1.48 °C)Multiply [Cancel out units]: q = (21.6 kJ/°C)(1.48 °C)Multiply [Cancel out units]: q = 31.968 kJStep 3: Check
Follow sig fig rules and round. We are given 3 sig figs.
31.968 kJ ≈ 32.0 kJ
A pressure cooker contains 5.68 L of air at a temperature of 390 4K if the absolute pressure of the air in the pressure cooker is 205 Pa how many moles of air are in the cooker
Answer:
3.59x10⁻⁴ mol
Explanation:
Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:
P = 205 PaV = 5.68 Ln = ?R = 8314.46 Pa·L·mol⁻¹·K⁻¹T = 390.4 KWe input the data given by the problem:
205 Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 KAnd solve for n:
n = 3.59x10⁻⁴ molDiscuss the four impure forms of carbon
Various structures, or allotropes, of carbon, are precious stone, graphite, and fullerenes. In jewel, every carbon iota is attached to four other carbon iotas, shaping an unbending construction that makes precious stones hard.
Which of the following ionization energies indicates an atom is most likely to gain electrons and form an anion or not form an ion at all?
Group of answer choices
578 kJ/mol
9460 kJ/mol
496 kJ/mol
786 kJ/mol
Answer:
Explanation:
578kj/mol
Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kinetics and the rate of degradation was measured to be 0.12 day−1 (at a particular sun- light intensity and temperature). Given this, how long does it take for the concentration of hypochlorous acid to reach nondetectable levels (0.05 mg · L−1) if the initial concentration were 3.65 mg · L−1?
Answer:
35.75 days
Explanation:
From the given information:
For first-order kinetics, the rate law can be expressed as:
[tex]\mathsf{In \dfrac{C}{C_o} = -kt}[/tex]
Given that:
the rate degradation constant = 0.12 / day
current concentration C = 0.05 mg/L
initial concentration C₀ = 3.65 mg/L
[tex]\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}[/tex]
㏑(0.01369863014) = -(0.12) t
-4.29 = -(0.12)
t = -4.29/-0.12
t = 35.75 days
A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.950 M NaOH. The acid required 27.4 mL of base to reach the equivalence point. After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?
Solution :
The equation is :
[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]
The number of the moles of HA os 0.00285, and the volume is 25 mL.
15 mL of the 0.0950 M NaOH is added.
The total volume of a solution is V = 25 mL + 15 mL = 40 mL
The pH of the solution is 6.50
Calculating the [tex]K_a[/tex] of HA
[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
Let s calculate the concentration of HA and NaOH
[tex]$[HA] = \frac{^nH_A}{V}$[/tex]
[tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]
= 0.07125 M
[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]
[tex]$=\frac{0.001425 mol}{0.04L}$[/tex]
= 0.0356 M
[tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]
Initial conc. (M) 0.07125 M 0.0356 M 0 M
Change in conc. (M) -0.0356 M -0.0356 M + 0.0356 M
Equilibrium conc. (M) 0.03565 M 0 M 0.0356 M
Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M
0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]
Now for [tex][H^+][/tex]
[tex]$[H^+] = 10^{-pH}$[/tex]
[tex]$=10^{-6.5}$[/tex]
[tex]$=3.16 \times 10^{-7}$[/tex]
Calculating the value of [tex]K_a[/tex],
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
[tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]
[tex]$=3.16\times 10^{-7}$[/tex]
Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].
calculate the maximum theoretical percent recovery from the recrystallization of 1.00g of benzoic acid
Answer:
The maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%
Note: The question is incomplete. A similar but complete question is given below:
The solubility of benzoic acid in water is 6.80g per 100mL at 100 degrees C and 0.34 g per 100mL at 25 degrees C.
Calculate the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water, assuming the solution is filtered at 25 degrees C.
Explanation:
Solubility of benzoic acid in water at 100 degrees C = 6.80g per 100mL
Solubility of benzoic acid in water at 25 degrees C = 0.34 g per 100mL
Mass of benzoic acid to be theoretically recovered from 100 mL of water = 6.80 g - 0.34 g = 6.46 g
At 25 degrees;
0.34 g of benzoic acid is present in 100 mL of water
x g of benzoic acid will be present in 15 mL of water
x = 0.34 × 15 / 100 = 0.051 g
Mass of benzoic acid to be theoretically recovered from 25 mL of water = 1.00 g - 0.051 g = 0.949 g
Maximum theoretical percent recovery = (mass recovered / original mass dissolved) x 100%
Maximum theoretical percent recovery = (0.949 / 1.00) × 100% = 94.9 %
Therefore, the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%
which type of chemical bond would be formed between two elements having electron configuration of 1s2 2s2 2p6 3s2 and 1s2 2s2 2p4
Which rock type is most likely to compsoed of just one mineral?
rock salt
conglomerate
basalt
rhyolite
maybe is answer is rhyolite
Answer:
the answer is rhyolite.
Explanation:
i'm pretty sure it is my guy
QUESTIONS :
1.
Many of the flavours and smells of fruits are esters. A learner prepared an ester with a sme
Ilke banana in the school laboratory using pentanol and ethanoic acid. She set up the
apparatus as shown in the diagram below.
PAPER TOWEL DIPPED
-WATER BATH
IN COLD WATER
PEITANOL ETHANOIC
ACID+ 4 DROPS OF
SULPHURIC ACID
1.1 Which property of sulphuric acid makes it suitable to use as a catalyst for the
preparation of esters?
1.2 Why do we heat the test tube in a water bath and not directly over a flame?
1.3 With reference to the characteristic smells of esters, name TWO examples where
esters are used in different industries.
1.4 State ONE function of the wet paper towel in the opening of the test tube.
1.5 Write down the IUPAC name of an ester fomed.
Answer:
See explanation
Explanation:
Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.
Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.
The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.
Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.
The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.
The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.
Which of the following could not be a resonance structure of CH3NO2?
a)
H
H-C-NO
H
b)
H .0:
H-C-N
H
c)
H:03
H-C-NC2
H:06
d)
H
H-C=N
H :9-H
e) Both c and d
Answer:
the answer is b.CH3NO2 I guess I'm correct
Someone please help me with this
Answer:
I think A should be the answer because oxygen is the chemical change of carbon.
The compound sodium hydrogen sulfate is a strong electrolyte. Write the reaction when solid sodium hydrogen sulfate is put into water:
Answer:
NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)
Explanation:
Sodium hydrogen sulfate is a strong electrolyte, that is, when dissolved in water it completely dissociates into the cation sodium and the anion hydrogen sulfate. The corresponding chemical equation is:
NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)
