Answer:
C2H5P
is the right answer
Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr
Answer:
I do not speak Spanish.
Explanation:
Consider the following equation for the combustion of acetone (C3H6O), the main ingredient in nail polish remover.
C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(g) ΔHrxn = −1790kJ
If a bottle of nail polish remover contains 143 g of acetone, how much heat would be released by its complete combustion? Express your answer to three significant figures.
Molar mass of Acetone
C3H6O3(12)+6+1658g/molNow
1 mol releases -1790KJ heat .Moles of Acetone:-
143/58=2.5molAmount of heat:-
2.5(-1790)=-4475kJHow many atoms are in .45 moles of P4010
Answer:
5×6.02×1023
Explanation:
there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010
How much of a 24-gram sample of Radium-226 will remain unchanged at the end of three half-life periods?
Answer:
The right answer is "3 g".
Explanation:
Given:
Initial mass substance,
[tex]M_0=24 \ g[/tex]
By using the relation between half lives and amount of substances will be:
⇒ [tex]M=\frac{M_0}{2^n}[/tex]
[tex]=\frac{24}{2^3}[/tex]
[tex]=3 \ g[/tex]
Thus, the above is the correct answer.
The overall order of an elementary step directly corresponds to its molecularity.
a. True
b. False
Answer:
true
Explanation:
2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25
Answer:
5.25 moles of protons. Option e
Explanation:
Reaction between phosphoric acid and sodium hydroxide is neutralization.
We can also say, we have an acid base equilibrium right here:
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Initially we have 5.25 moles of base.
We have data from the acid, to state its moles:
M = mol/L, so mol = M . L
mol = 1.75 moles of acid
If we think in the acid we know:
H₃PO₄ → 3H⁺ + PO₄⁻³
We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)
If we have 1.75 moles of acid, we may have
(1.75 . 3) /1 = 5.25 moles of protons
These moles will be neutralized by the 5.25 moles of base
H₃O⁺ + OH⁻ ⇄ 2H₂O Kw
In a titration of a weak acid and a strong base, we have a basic pH
Chemical reactions can exhibit different rate constants at differing: Select the correct answer below: initial concentrations volumes of container temperatures none of the above
Explanation:
Chemical reactions can exhibit different rate constants at differing:
i)initial concentrations
ii)volumes of container
iii) temperatures
iv)none of the above.
The rate constant of a reaction is constant at a particular temperature.
It is not depending on the initial concentration of the reactants. It varies with temperature.
Thus, among the given options the correct answer is Temperatures.
Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
What is a rate constant?
The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances.
Chemical reactions proceed at vastly different speeds depending on the nature of the reacting substances, the type of chemical transformation, the temperature, etc.
For a given reaction, the speed of the reaction will vary with the temperature, the pressure, and the amounts of reactants present.
The rate constant goes on increasing as the temperature goes up, but the rate of increase falls off quite rapidly at higher temperatures.
On the other hand, the volume of the container, initial concentration does not affect the rate constant.
Therefore, Chemical reactions can exhibit different rate constant at different temperatures. Hence, the correct option is temperature.
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Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it reacts with exhaled CO2 to form Na2CO3 and O2. How many liters of respired air can react with 96.7 g of Na2O2 if each liter of respired air contains 0.0755 g of CO2
Answer:
725.15 L
Explanation:
The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:
Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂
From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.
Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:
Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol
moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂
Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂
Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:
1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂
In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:
1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L
Write the functional isomers of C2H6O?
Answer:
See explanation
Explanation:
Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.
The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).
Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.
-300g de acido comercial se disuelve en agua destilada contenidos en un cono, cuyo radio es de 0.005Km y 300cm de altura, si la densidades de 1.2g/m3 ¿Cuál es la concentración expresada en %m/m?
Answer:
[tex]\%m/m=76.1\%[/tex]
Explanation:
¡Hola!
En este caso, considerando la información dada, entendemos que se haría primero necesario calcular el volumen del cono en metros cúbicos, teniendo en cuenta que 0.005 km son 5 m y 300 cm son 3 m:
[tex]V=\frac{1}{3} \pi r^2h\\\\V=\frac{1}{3} \pi *(5m)^2(3m)=78.5m^3\\\\[/tex]
Ahora, convertimos esta cantidad a gramos por medio de la densidad para conocer la masa de la solución:
[tex]m_{sol}=78.5m^3*\frac{1.2g}{1m^3} =94.2g[/tex]
Finalmente, aplicamos la definición de %m/m para obtener:
[tex]\%m/m=\frac{300g}{300g+94.2g}*100\%\\\\ \%m/m=76.1\%[/tex]
¡Saludos!
A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?
Answer:
Likely [tex]\rm In[/tex] (indium.)
Explanation:
Number of atoms: [tex]N = 2.241 \times 10^{21}[/tex].
Dividing, [tex]N[/tex], the number of atoms by the Avogadro constant, [tex]N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}[/tex], would give the number of moles of atoms in this sample:
[tex]\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}[/tex].
The mass of that many atom is [tex]m = 0.4272\; \rm g[/tex]. Estimate the average mass of one mole of atoms in this sample:
[tex]\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
The average mass of one mole of atoms of an element ([tex]114.82\; \rm g \cdot mol^{-1}[/tex] in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass [tex]114.82[/tex]. Indium, [tex]\rm In[/tex], is the closest match.
The symbol of the element is In
StoichiometryFrom the question, we are to determine the identity of the element
First, we will determine the number of moles of sample present
Using the formula
[tex]Number \ of\ moles = \frac{Number\ of \ atoms }{Avogadro's\ constant}[/tex]
Number of moles of the sample = [tex]\frac{2.241\times 10^{21} }{6.022\times 10^{23} }[/tex]
Number of moles of the sample = 0.003721355 mole
Now, we will determine the Atomic mass of the sample
From the formula,
[tex]Atomic\ mass = \frac{Mass}{Number\ of\ moles}[/tex]
Therefore,
Atomic mass of the sample = [tex]\frac{0.4272}{0.003721355}[/tex]
Atomic mass of the sample = 114.8 amu
The element that has an atomic mass of 114.8 amu is Indium. The symbol of Indium is In.
Hence, the symbol of the element is In.
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An electron moved from shell n = 2 to shell n = 1. What most likely happened during the transition?
When electron goes from a higher shell to lower shell then it loses energy .
So, when an electron moved from shell n = 2 to shell n = 1 then a photon of energy is released.
Carbon monoxide and chlorine gas react to produce phosgene (COCl2) gas according to the following reaction at 100.0C CO(g) + Cl2 (g) <-> COCl2 (g) Kp = 1.49x10^8
In an equilibrium mixture of the three gases, PCO = PCl2 = 2.55 x 10^-4 what is the equilibrium partial pressure (in atm) of Phosgene?
Express your answer below in decimal form (you will not be able to use scientific notation) to THREE significant figures. If your answer is negative include the sign.
Answer:
0.000000000000000969 atm
Explanation:
Based on the reaction:
CO(g) + Cl2(g) ⇄ COCl2(g)
Where equilibrium constant, Kp, is defined as:
Kp = 1.49x10⁻⁸ = PCOCl2 / PCO * PCl2
As PCO = PCl2 = 2.55x10⁻⁴atm:
1.49x10⁻⁸ = PCOCl2 / PCO * PCl2
1.49x10⁻⁸ = PCOCl2 / 2.55x10⁻⁴atm * 2.55x10⁻⁴atm
9,69x10⁻¹⁶atm = PCOCl2
In decimal form:
0.000000000000000969 atmPLEASE HELP ASAP MOLES TO MOLECULES
Answer:
4.77mol is the correct answer
9. How can you separate sugar from a sugar solution contained in a glass without taste? Explain
Answer:
See explanation
Explanation:
Sugar is a polar crystalline substance. The sugar crystal is capable of dissolving in water since it is polar.
When sugar dissolves in water, a sugar solution is formed. If I want to separate the sugar from the water in the solution, I have to boil the solution to a very high temperature.
When I do that, the water in the sugar solution is driven off and the pure sugar crystal is left behind.
A solution contains only sucrose and water. If the mole fraction of sucrose is 0.0558, determine the molality of sucrose.
Answer:
The correct solution is "3.28 m".
Explanation:
According to the question,
Mol fraction of solvent,
= 0.0558
Molar mass of water,
= 18 g/mol
Mol of H₂O in 1000 g water,
= 55.55 mol
Now,
Let the mol of solute will be "x mol".
Total mol in solution will be "55.55 + x".
As we know,
⇒ The mol fraction of solvent = [tex]\frac{x}{55.55+x}[/tex]
[tex]0.0558=\frac{x}{55.55+x}[/tex]
[tex]x=0.0558[55.55+x][/tex]
[tex]x=3.09969+0.0558x[/tex]
[tex]x-0.0558x=3.09969[/tex]
[tex]x=\frac{3.09969}{0.9442}[/tex]
[tex]=3.38 \ m[/tex]
An unknown element, X, reacts with potassium to form the compound K2X. In other compounds this element also can accommodate up to 12 electrons rather than the usual octet. What element could X be
Answer:
Se
Explanation:
First of all, we must note that any element that we must choose is an element that is in group sixteen.
Elements of groups 16 have six electrons in their outermost shell which can be used for bond formation thereby yielding a total of twelve electrons on the valence shell.
However, this is only possible for the heavier members of the group 16 (from sulphur downwards) which are able to expand their octet.
Oxygen can not expand its octet hence it is not the answer.
A rock originally has 2200 atoms of 235U and no 207Pb. It now has 800 atoms of 235U. Assuming the only chemical process is the radioactive decay of uranium to lead, what is the age of the rock
Answer:
[tex]Age=1040.55 million\ years[/tex]
Explanation:
From the question we are told that:
Initial Rock atoms [tex]a_1=2200atoms of 235^U[/tex]
Final Rock atom [tex]a_2=800 of 235^U[/tex]
Age of half life 713 million years
Generally the equation for Age is mathematically given by
[tex]\frac{1}{2^n}=\frac{800}{2200}[/tex]
[tex]\frac{1}{2^n}=\frac{1}{2.750}[/tex]
[tex]n=1.46[/tex]
Therefore
[tex]Age=713*1.46[/tex]
[tex]Age=1040.55 million\ years[/tex]
3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?
4. Excess oxygen gas is added to 34.5 grams of aluminum and heated under pressure. How many grams of aluminum oxide are produced?
Please explain as well if possible!
Answer:
3) About 0.35 grams of hydrogen gas.
4) About 65.2 grams of aluminum oxide.
Explanation:
Question 3)
We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.
Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:
[tex]\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]
To balance it, we can simply add another sodium atom on the left. Hence:
[tex]\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}[/tex]
To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.
The molar mass of sodium is 22.990 g/mol. Hence:
[tex]\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}[/tex]
From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:
[tex]\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}[/tex]
And the molar mass of hydrogen gas is 2.016 g/mol. Hence:
[tex]\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]
Given the initial value and the above ratios, this yields:
[tex]\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}[/tex]
Cancel like units:
[tex]=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}[/tex]
Multiply. Hence:
[tex]=0.3463...\text{ g H$_2$}[/tex]
Since we should have two significant values:
[tex]=0.35\text{ g H$_2$}[/tex]
So, about 0.35 grams of hydrogen gas will be released.
Question 4)
Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:
[tex]\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}[/tex]
To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:
[tex]\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}[/tex]
To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.
The molar mass of aluminum is 26.982 g/mol. Thus:
[tex]\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}[/tex]
According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:
[tex]\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}[/tex]
And the molar mass of aluminum oxide is 101.961 g/mol. Hence: [tex]\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]
Using the given value and the above ratios, we acquire:
[tex]\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}[/tex]
Cancel like units:
[tex]\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}[/tex]
Multiply:
[tex]\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}[/tex]
Since the resulting value should have three significant figures:
[tex]\displaystyle = 65.2 \text{ g Al$_2$O$_3$}[/tex]
So, approximately 65.2 grams of aluminum oxide is produced.
Answer:
Solution given:
3.
[tex]2Na+H_2O→Na _2O+H_2[/tex]
2Na=2*23g.
2O=18g.
[tex]Na_2O[/tex]=62g
[tex]H_2[/tex]=2 g
we have
2*23g of Na produce 2g of [tex]H_2[/tex]
Now
7.9 g of Na produce 2*7.9/(2*23)
=0.34g of [tex]H_2[/tex]
:. 0.34g of [tex]H_2[/tex] is produced.4.
we have
[tex]3O_2+4Al→2Al_2O_3[/tex]
[tex]3O_2[/tex]=3*16g*2g
4Al=4*27g
[tex]2Al_2O[/tex]= 2*27*2g+2*16*3g
4*27g of Al produces 204g of [tex]Al_2O_3[/tex]
34.5g of Al produces 204g*34.5/(4*27)
=65.17g of [tex]Al_2O_3[/tex] is producedUsing the periodic table, choose the more reactive nonmetal.
Br or As
A beverage contains tartaric acid, H2C4H4O6, a substance obtained from grapes during wine making. If the beverage is 0.190 tartaric acid, what is the molal concentration? What is the mole fraction of tartaric acid and water? Calculate the mass percent of tartaric acid. The density of the solution is 1.016g/mL.
Answer:
1) [tex]molality = 0.19[/tex]
2) [tex]Mole\ fraction= 0.003486[/tex]
3)[tex]Mass\ percent = 2.8%[/tex]
Explanation:
Concentration of Tartaric acid=0.190mole /l
1)
Generally
[tex]Mass\ of\ tartaric\ acid = 150.087 *0.190[/tex]
[tex]Mass\ of\ tartaric\ acid = 28.5 g[/tex]
Since 1L of solution tartartic acid is
[tex]T_{1l}= density * volume[/tex]
[tex]T_{1l}= 1.016Kg / L X 1 L[/tex]
[tex]T_{1l}= 1016[/tex]
Therefore
[tex]Mass of solvent = 1016-28.5[/tex]
[tex]Mass of solvent = 987.5 g[/tex]
Generally the equation for molality is mathematically given by
[tex]molality = \frac{moles}{Kg}[/tex]
[tex]molality = \frac{0.190 * 1000}{987.5}[/tex]
[tex]molality = 0.19[/tex]
2.
Generally the equation for Moles of water is mathematically given by
[tex]Moles\ of\ water = \frac{mass}{mol wt}[/tex]
[tex]Moles\ of\ water = \frac{987.5}{18 }[/tex]
[tex]Moles\ of\ water= 54.86[/tex]
Therefore
[tex]Mole\ fraction = \frac{Moles\ of\ solute}{total\ moles}[/tex]
[tex]Mole\ fraction= \frac{0.190}{54.5}[/tex]
[tex]Mole\ fraction= 0.003486[/tex]
3
Generally the equation for Mass Percent is mathematically given by
[tex]Mass\ percent = \frac{mass\ of\ tartaric\ acid}{total mass}[/tex]
[tex]Mass\ percent = \frac{28.5* 100}{1016}[/tex]
[tex]Mass\ percent = 2.8%[/tex]
If 0.650 mL of benzaldehyde reacts with enough of the Grignard reagent, calculate the theoretical yield (in grams) of the alcohol product. Show calculation with units for full credit.
Answer:
Theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g
Explanation:
A Grignard reagent is any of the numerous organic derivatives of magnesium (Mg) which are commonly represented by the general formula RMgX (where R is a hydrocarbon radical e.g. methyl, ethyl, propyl, etc.; and X is a halogen atom, e.g. chlorine, bromine, or iodine)
A Grignard reaction converts an aldehyde to a secondary alcohol. In the grignard reaction involving benzaldehyde as in this experiment, the grignard reagent used is ethyl magnesium bromide, EtMgBr, and the resulting product is 1-Phenyl-1-propanol, a secondary alcohol. The reaction is shown in the figure attached below.
Mass of benzaldehyde in 0.650 mL = density × volume
Density of Benzaldehyde = 1.044 g/mL
Mass of benzaldehyde = 1.044 g/mL × 0.650 mL = 0.6786 g
Molar mass of benzaldehyde = 106 g/mol
Molar mass of 1-Phenyl-1-propanol = 136 g/mol
Mass of = mass of benzaldehyde × mole ratio of 1-Phenyl-1-propanol and benzaldehyde
Mass of 1-Phenyl-1-propanol = 0.6786 g × (136 g/mol)/(106 g/mol) = 0.871 g
Therefore, the theoretical yield of 1-Phenyl-1-propanol in grams = 0.871 g
How many grams of sodium nitrate (NaNO3) are needed to
prepare 100 grams of a 15.0 % by mass sodium nitrate
solution?
Answer:
15.0 g
Explanation:
15.0% =0.150
100.0 g × 0.150= 15.0g
Sodium nitrate is "an inorganic compound with the formula of NaNO₃.
What is an inorganic compound?Inorganic compound is "a chemical compound that lacks carbon–hydrogen bonds".
15% = 0.15
100.0 g × 0.15= 15g
Hence, 15g of Sodium nitrate are needed to prepare 100 gms of a 15% by mass sodium nitrate.
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Draw the structure of the neutral product formed in the reaction of dimethyl malonate and methyl vinyl ketone.
Answer:
Explanation:
The reaction between dimethyl malonate which is an active methylene group with an (∝, β-unsaturated carbonyl compound) i.e methyl vinyl ketone is known as a Micheal Addition reaction. The reaction mechanism starts with the base attack on the β-carbon to remove the acidic ∝-hydrogens and form a carbanion. The carbanion formed(enolate ion) attacks the methyl vinyl ketone(i.e. a nucleophilic attack at the β-carbon) to give a Micheal addition product, this is followed by the protonation to give the neutral product.
How many grams of Ag2CO3 will precipitate when excess K2CO3 solution is added to 56.0 mL of 0.671 M AgNO3 solution?
Answer:
The mass of silver carbonate precipitated is 5.18 grams.
Explanation:
Molarity of the silver nitrate solution = 0.671 M
Volume of the silver nitrate solution = 56.0 mL
[tex]1 mL = 0.001 L\\56.0 mL = 56.0\times 0.001 L=0.0560 L[/tex]
Moles of silver nitrate = n
[tex]Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}\\\\0.671 M=\frac{n}{0.0560 L}\\n=0.671 M\times 0.0560 L=0.0376 mol[/tex]
Moles of silver nitrate used = 0.0376 mol
[tex]K_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2KNO_3[/tex]
According to the reaction, 2 moles of silver nitrate gives 1 mole of silver carbonate, then 0.0376 moles of silver nitrate:
[tex]=\frac{1}{2}\times 0.0376 mol=0.0188 \text{mol of }Ag_2CO_3[/tex]
Moles of the silver carbonate formed = 0.0188 mol
Molar mass of silver carbonate = 275.7453 g/mol
Mass of silver carbonate :
[tex]=275.7453 g/mol\times 0.0188 mol=5.1840 g\approx 5.18 g[/tex]
The mass of silver carbonate precipitated is 5.18 grams.
Which of the following are examples of physical properties of ethanol? Select all that apply.
The boiling point is 78.37°C
It is a clear, colorless liquid
It is flammable
It is a liquid at room temperature
what are the major specials presents in a solution of a strong acid like HCl
Answer:
hydrogen ions (H⁺) and chloride ions (Cl⁻)
Explanation:
Hydrochloric acid (HCl) is a strong acid. That means that the compound dissociates completely into ions when is dissolved in water, as follows:
HCl → H⁺ + Cl⁻
The equilibrium is completely shifted to the right side (products). Thus, it is considered that the concentration of the non-dissociated compound (HCl) is negligible, and the major specials present in the solution are the hydrogen ions (H⁺) and chloride ions (Cl⁻).
How to solve this problem step by step
Answer:
[tex]V_2= 736mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law:
[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]
Thus, we solve for the final volume by solving for V2 as follows:
[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]
Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:
[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]
Regards!
C. A sample may contain any or all of the following ions Hg2 2, Ba 2, and Al 3. 1) No precipitate forms when an aqueous solution of NaCl was added to the sample solution. 2) No precipitate forms when an aqueous solution of Na2SO4 was added to the sample. 3) A precipitate forms when the sample solution was made basic with NaOH. Which ion or ions were present. Write the net ionic equation(s) for the the reaction (s).
Answer:
Al^3+
Explanation:
Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.
Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.
If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;
Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)
A 250-mg sample of carbon from wood underwent 15300 carbon-14 disintegrations in 36 hours. Estimate the time since the death of the sample.
Answer:
The correct answer is - 9935 years approximately.
Explanation:
Let z be the age in years to be found:
(15300 disintegrations) x (1.0 g / 0.250 g) / (1.84×10^4 disintegrations)
= 3.3260
half life of carbon = (1/2)^(z/5730 yr)
Solve for z
3.3260 = (1/2)^(z/5730)
Take the log of both sides:
log 3.3260 = (z/5730) log (1/2)
log 3.3260 / log (1/2) = z/5730
z = 5730 log 3.3260 / log (1/2)
= 1.73378816*5730
= 9935 years approximately.
