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)
Classify each of the reactions listed below as a single-displacement, double-displacement, synthesis,
decomposition, oxidation reduction or combustion reaction.
Reaction Type
: 2Na + Cl2 → 2NaCl
: C2H4 + 3O2 → 2CO2 + 2H2O
: 2Ag2O-> 4Ag + O2
: BaCl2 + Na2SO4->BaSO4 +2NaCl
: 2AI + Fe2O3-> 2Fe + Al2O3
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode where Co2 (aq) is reduced to Co (s) . Assume all aqueous solutions have a concentration of 1 mol/L.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Why does increasing the temperature of two reactants in solution make a
reaction proceed more quickly?
Answer:
-The particles of the two reactants will gain kinetic energy and collide with one another more frequently and forcefully, which makes the reaction take place more quickly
10g of a non-volatile and non-dissociating solute is dissolved in 200g of benzene.
The resulting solution boils At temperature of 81.20oC. Find the molar mass of solute.
Given that the BP of pure benzene is 80.10oC and Its elevation boiling point constant = 2.53 oC/m.
Answer: The molar mass of solute is 115 g/mol.
Explanation:
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]
OR
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Boiling point of pure solvent (benzene) = [tex]80.10^oC[/tex]
Boiling point of solution = [tex]81.20^oC[/tex]
i = Vant Hoff factor = 1 (for non-electrolytes)
[tex]K_b[/tex] = Boiling point elevation constant = [tex]2.53^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute = 10 g
[tex]M_{solute}[/tex] = Molar mass of solute = ? g/mol
[tex]w_{solvent}[/tex] = Mass of solvent = 200 g
Putting values in equation 1, we get:
[tex]81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol[/tex]
Hence, the molar mass of solute is 115 g/mol.
5. How many grams of tin metal can be produced from smelting (heating) of a 4.5 kilograms of tin (IV) oxide? (Note: Elemental tin and oxygen gas are the only products of this reaction).
Answer:
About 3500 grams of tin.
Explanation:
We want to determine amount of tin metal (in grams) that can be produced from smelting 4.5 kilograms of tin(IV) oxide.
First, write the chemical compound. Since our cation is tin(IV), it forms a 3+ charge. Oxygen has a 2- charge, so we will have two oxygen atoms. Hence, tin(IV) oxide is given by SnO₂.
By smelting it, we acquire elemental tin and oxygen gas. Hence:
[tex]\text{SnO$_2$}\rightarrow \text{Sn} + \text{O$_2$}[/tex]
(Note: oxygen is a diatomic element.)
The equation is balanced as well.
To convert from SnO₂ to only Sn, we can first convert from grams of SnO₂ to moles, use mole ratios to convert to moles of Sn, and then from there convert to grams.
Since Sn has a molar mass of 118.71 g/mol and oxygen has a molar mass of 15.999 g/mol, the molar mass of SnO₂ is:
[tex](118.71)+2(15.999) = 150.708\text{ g/mol}[/tex]
Therefore, given 4.5 kilograms of SnO₂, we can first convert this into grams using 1000 g / kg and then using the ratio:
[tex]\displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}[/tex]
We can convert this into moles.
Next, from the chemical equation, we can see that one mole of SnO₂ produces exactly one mole of Sn (and also one mole of O₂). So, our mole ratio is:
[tex]\displaystyle \frac{1\text{ mol Sn}}{1\text{ mol SnO$_2$}}[/tex]
With SnO₂ in the denominator to simplify units.
Finally, we can convert from moles Sn to grams Sn using its molar mass:
[tex]\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
With the initial value and above ratios, we acquire:
[tex]\displaystyle 4.5\text{ kg SnO$_2$}\cdot \frac{1000 \text{ g SnO$_2$}}{1\text{ kg SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol Sn}}{1 \text{ mol SnO$_2$}} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
Cancel like units:
[tex]=\displaystyle 4.5\cdot \frac{1000}{1}\cdot \displaystyle \frac{1}{150.708}\cdot \displaystyle \frac{1}{1} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1}[/tex]
Multiply. Hence:
[tex]\displaystyle = 3544.5696...\text{ g Sn}[/tex]
Since we should have two significant figures:
[tex]=3500 \text{ g Sn}[/tex]
So, about 3500 grams of tin is produced from smelting 4.5 kg of tin(IV) oxide.
Answer:
3546g
Explanation:
start w/ tin (IV) oxide n elemental tin and oxygen gas are the only products of this reaction
SnO2 -> Sn + O2
Sn molecular wt: 119
O2 molecular wt: 32
SnO2 molecular wt: 119+32 = 151
so Sn / SnO2 wt ratio = 119 / 151
4.5 kilograms of tin (IV) oxide will produce:
= 4.5 * 119 / 151
= 3.546 kg
or 3546 grams of tin metal
no need to involve moles ;)
define surface are tension of liquid
The hydrological cycle refers to the circulation of water within the earth's hydrosphere in different from I. e. the liquid, solid and the gaseous forms.
balance the following reaction using LCM method by showing each steps Pb (N3)2 + Cr(MnO4)2 Cr2O3 + MnO2 + Pb3O4+ NO
here's the answer to your question
Choose all the answers that apply. Silicon (Si) has 14 protons and an atomic mass of 28. Silicon has _____. three electron shells 14 electrons 14 neutrons two electron shells 28 electrons
Answer:
three electron shells
14 electrons
14 neutrons
Explanation:
Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.
Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.
The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.
1. Which of the following are covalent compounds?
Select all that apply:
Potassium Chloride: K CI
Octadecanol: C18H380
Dimethyl Sulfoxide: CHOS
Lithium Bromide: LiBr
Answer:
Octadecanol: C18H380
Dimethyl sulfoxide: CHOS
Explanation:
Covalent compounds are formed between non-metallic elements
If the temperature of a volume of dieal gas ncreases for 100 to 200, what happens to the average kinetic energy of the molecules?
Answer:
It increases but less than double
Explanation:
As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.
We know, the kinetic energy of an ideal gas is given by :
[tex]$V_{avg} = \sqrt{\frac{8R}{\pi M}}$[/tex]
where, R = gas constant
T = absolute temperature
M = molecular mass of the gas
From the above law, we get
[tex]$V_{avg} \propto \sqrt{T}$[/tex]
Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.
In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$[/tex]
[tex]$(V_{avg})_2 = 1.12\ (V_{avg})_1$[/tex]
Therefore, [tex]$(V_{avg})_2 > (V_{avg})_1$[/tex]
Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.
Thus, the answer is " It increases but less that double".
please help!
What is the definition of thermal chemistry?
a.The study of change that involves warm objects
b.The study of change that involves heat
c.The study of change that involves cool objects
d.The study of change that involves temperature
Where is CO2 concentration greatest in the body?
Answer:
I believe it might be blood exiting the lungs, not positive though.
Explanation:
i think it is blood exiting the lungs
For a particular catalyzed reaction, the change in enthalpy is 26kJmole and the activation energy is 67kJmole. Which can be the change in enthalpy and the activation energy for the uncatalyzed reaction
Answer:
26kJmole,84kJmole
Explanation:
A catalyst lowers the activation energy between reactants and products. Once this energy barrier is lowered, reactants are converted into products faster.
However, the enthalpy change of a reaction is constant both in the catalysed and uncatalysed reaction.
The activation energy of the uncatalysed reaction must be higher than the activation energy of the catalysed reaction. Hence the answer above.
A process will definitely be spontaneous if: Select the correct answer below: the entropy of the system increases in the process the entropy of the system decreases in the process the entropy of the universe increases in the process the entropy of the universe decreases in the process
Answer:
the entropy of the universe increases
Explanation:
The second law of thermodynamics can be stated in terms of entropy, in this statement of the law; ''a spontaneous process increases the entropy of the universe''.
Given that; ∆Suniverse = ∆Ssystem + ∆Ssurroundings, the definition of a spontaneous process is one in which ∆Suniverse >0.
Hence, a process is spontaneous when the entropy of the universe increases.
K always has the same value
at a given temperature
regardless of the amounts of
reactants or products that
are present initially.
Select one:
True
False
hope it will help you
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
8. Using the data from question 7 what is the molar concentration of KMnO4 ?
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Answer:
7. 0.1021 M
8. 1.167 M
10. Increase in volume of water would lower the rate of reaction
Explanation:
7. What is the molar concentration of H₂C₂O₄ ?
Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.
Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V
= 1.225 × 10⁻³ mol/12 × 10⁻³ L
= 0.1021 mol/L
= 0.1021 M
8. Using the data from question 7 what is the molar concentration of KMnO₄ ?
Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.
Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V
= 14 × 10⁻³ mol/12 × 10⁻³ L
= 1.167 mol/L
= 1.167 M
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.
What is the electron domain geometry around N in N2CL4
Answer:
trigonal bipyramidal.
Describe A Simple experiment that can be prepared in the laboratory to demonstrate the formation of Iron (III) Chloride from iron fillings
Answer:
Anhydrous iron(III) chloride may be prepared by treating iron with chlorine:[11]
{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}
Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.
Dissolving iron ore in hydrochloric acid
{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}
Oxidation of iron(II) chloride with chlorine
{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}
Oxidation of iron(II) chloride with oxygen
{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}
Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride. Hydrated iron(III) chloride can be converted to the anhydrous form by treatment with thionyl chloride.[12] Similarly, dehydration can be effected with trimethylsilyl chloride:[13]
{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}
Anhydrous iron(III) chloride may be prepared by treating iron with chlorine.
What is an iron filling?
Iron filings are small shavings of ferromagnetic material.
[tex]{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}[/tex]
Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.
Dissolving iron ore in hydrochloric acid.
Oxidation of iron(II) chloride with chlorine.
[tex]{\displaystyle {\ce {2FeCl_2_{(}aq){+~}Cl_2_{(}g)- > 2FeCl_3_{(}aq)}}}\\[/tex]
Oxidation of iron(II) chloride with oxygen.
Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride.
Hydrated iron(III) chloride can be converted to an anhydrous form by treatment with thionyl chloride. Similarly, dehydration can be affected by trimethylsilyl chloride.
[tex]{\displaystyle {\ce {FeCl_3.6H2O + 12 Me_3SiCl - > FeCl3 + 6 (Me_3Si)2O + 12 HCl}}}[/tex]
Learn more about the iron filings here:
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Name of this product
Answer:
Explanation:
ethyl 3-methylbenzoate
Why does the dehydration of an alcohol more often use concentrated sulfuric acid, H 2 S O 4 HX2SOX4, as the acid catalyst rather than dilute hydrochloric acid, H C l HCl
KAnswer:
See explanation
Explanation:
It is more common to use H2SO4 for dehydration reaction rather than HCl because HCl contains a good nucleophile,the chloride ion.
Owing to the presence of the chloride ion, a substitution reaction involving the chloride ion may also proceed also thereby affecting the elimination reaction.
Also, concentrated H2SO4 is a very good drying agent thus, as long as it is used, the alcohol substrate is completely dehydrated to yield the alkene.
Note that HCl is not a dehydrating agent.
A sample of hellium has a volume of 500 mL at STP. What will be its new volume be in mL if the temperature is increased to 325 K and its pressure is increased to 125 kPa?
Answer:
[tex]V_2=482.5mL[/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 due to the fact that we are dealing with variable volume, temperature and pressure:
[tex]\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}[/tex]
In such a way, we solve for the final volume, V2, considering that the initial volume, V1, is 500 mL, the initial temperature, T1, is 273 K (STP), the initial pressure, P1, is 1 atm (STP) and the final temperature, T2, is 325 K and the final pressure, P2, is 125 kPa (1.23 atm):
[tex]V_2=\frac{P_1V_1T_2}{T_1P_2} \\\\V_2=\frac{(1atm)(500mL)(325K)}{(273K)(1.23atm)} \\\\V_2=482.5mL[/tex]
Regards!
Based on the time and temperature data collected for the reaction of KMnO4 with either malonic acid or oxalic acid, one can conclude that generally as the temperature of a reaction is increased, the rate of the reaction increases. This is becaus:_________
a. the activation energy is lowered and the reactant molecules collide with greater energy
b. the activation energy is lowered and the reactant molecules collide more frequently
c. the activation energy is lowered, the reactant molecules collide more frequently and with greater energy per collision
d. the reactant molecules collide more frequently and with greater energy per collision
Answer:
The correct option is C ( the activation energy is lowered, the reactant molecules collide more frequently and with greater energy per collision).
Explanation:
The rate of a chemical reaction is defined as the quantity of products that are formed per unit time. Rates can be computed based on either how the reactants are used up or how products are formed.
There are factors that affects the rate of a chemical reaction and they include TEMPERATURE, catalyst, surface area of reacting substances and many among others.
TEMPERATURE increase in most chemical reactions increases the rate of the reaction. This is because molecules gain more energy at higher temperatures. This increases their kinetic energy resulting in more effective collision of the reactant molecules. The rate of reaction depends on the frequency of this effective collisions between the reacting particles.
Effective collision are those that result in reactions, which when they occur, the colliding particles become activated with increased kinetic energy. This kinetic energy must exceed a particular energy barrier for a particular reaction if the reaction must take place. This energy barrier that must be overcome is known as the ACTIVATION ENERGY.
As a result of the time and temperature data collected for the reaction of KMnO4 with either malonic acid or oxalic acid, the reactant molecules collide more frequently and with greater energy per collision.
It is very important to known that the more the reactant molecules is said to collide, the more they will tend to react with one another.
This also leads to a faster reaction rate in any system. So for one to produce a good collision, reactant particles is said to some minimum amount of energy. This energy, used to start the reaction process, is called the activation energy.
Learn more about activation energy from
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The rate law for a reaction can be derived from the: Select the correct answer below: stoichiometry of the overall reaction molecularity of the rate-determining step molecularity of the overall reaction none of the above
Answer:
molecularity of the rate-determining step
Explanation:
The rate determining step of a sequence of reactions is the slowest step in the sequence of non-elementary reactions.
The molecularity of the slowest step in the reaction mechanism gives us the rate law of reaction.
It is for this cause that the slowest step in the reaction sequence is called ''rate determining step'' since it determines the rate law of reaction.
What is a system called when neither energy nor matter is exchanged between the system and the surroundings?
Closed system
Free energy
Isolated system
Open system
Answer:
open system
Explanation:
Answer:
Isolated system
Explanation:
An isolated system is one that cannot exchange either matter or energy with its surroundings.
you ned to make an aqeous solution of .139M aluminum nitrate for an experiment in a lab, using 300. ml volumetric flask. how much solid aluminum nitrate
Answer:
8.88 grams of aluminum nitrate should be weighted.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to use the definition of molarity to calculate the moles of aluminum nitrate as follows:
[tex]M=\frac{n}{V}\\\\n=M*V\\\\n=0.139mol/L*0.300L=0.0417mol[/tex]
Now, since the molar mass of aluminum nitrate is 212.996 g/mol, we obtain the following mass:
[tex]m=0.0417mol*\frac{212.996 g}{1mol} \\\\m=8.88g[/tex]
Therefore, 8.88 grams of aluminum nitrate should be weighted.
Regards!
The SALT I agreement in 1972 brought about
O the creation of documents that officially ended the Cold War.
O the US recognition of China for the first time since 1949.
• the regulation of Chinese production or missiles that carried nuclear weapons.
• the regulation of Soviet production of missiles that carried nuclear weapons.
Answer:
the regulation of Soviet production of missiles that carried nuclear weapons
Explanation:
The aim of the SALT I agreement in 1972 was to stop or greatly reduce the arms race where world powers were stockpiling ballistic missiles and other nuclear arsenal.
Therefore, this agreement brought about the regulation of Soviet production of missiles that carried nuclear weapons
Answer:
D. the regulation of Soviet production of missiles that carried nuclear weapons.
Explanation:
Oxygen and hydrogen form the polyatomic hydroxide ion. What is its charge?
Answer:
negative
the chage on hydroxide ions os negative
Salts have varying solubilities in solvents such as water. Some salts dissolve well in water, whereas others are nearly insoluble. The solubility product Ksp is the equilibrium constant for the dissolution of a solid salt into water, and its magnitude is an indicator of the solubility of the salt.
Barium sulfate, BaSO4, is used in medical imaging of the gastrointestinal tract because it is opaque to X rays. A barium sulfate solution, sometimes called a cocktail, is ingested by the patient, whose stomach and intestines can then be visualized via X-ray imaging. If a patient ingests 280mL of a saturated barium sulfate solution, how much toxic Ba2+ ion has the patient consumed?
Answer:
Amount of toxic Ba²+ ingested by the patient = 0.404 mg
Explanation:
The solubility product constant, Ksp for barium sulfate is 1.1 x 10-¹⁰.
Considering the dissociation products of aqueous BaSO₄ solution:
BaSO₄ (aq) ⇄ Ba²+ (aq) + SO₄²- (aq)
Let the concentration of Ba²+ be x and that of SO₄²- be x as well since the mole ratio of the products is 1 : 1
Ksp = [Ba²+] × [SO₄²-]
Ksp = x × x = x²
x² = 1.1 x 10-¹⁰
x = 1.05 × 10-⁵ M
Therefore, the concentration of Ba²+ ions = 1.05 × 10-⁵ M
Volume of saturated barium sulfate ingested by the patient = 280 mL = 0.280 L
Number of moles of Ba²+ ions = molar concentration × volume
Number of moles of Ba²+ = 1.05 × 10-⁵ M × 0.280 L = 2.94 × 10-⁶ moles
Mass of Ba²+ ions = number of moles × molar mass
Molar mass of Ba²+ = 137.33 g/mol
Mass of Ba²+ = 2.94 × 10-⁶ moles × 137.33 g/mol = 4.04 × 10-⁴ g = 0.404 mg
Therefore, amount of toxic Ba²+ ingested by the patient = 0.404 mg
How many grams of calcium chloride are needed to produce 10.0 g of potassium chloride?
CaCl2(aq) + K2CO3(aq) → 2 KCl(aq) + CaCO3(aq)
Answer:
11.1g
Explanation:
since the equation is already balanced just drop down the elements you will work with and use the mole to mole ratios
Cacl2 : 2Kcl
1 : 2
since potassium chloride has alot of information find it's moles
number of moles=mass/molecular mass
=10g/74.5
=0.13g/mol
now use the mole to mole ratios to find the number of moles of calcium chloride
1 : 2
x:0.13
2x/2=0.13/2
x=0.067g/mol of cacl2
then you can calculate the mass of calcium chloride
m=n×mm
=0.067×111
=7.4g
I hope this helps
The mass of calcium chloride, CaCl₂ needed to produce 10 g of potassium chloride, KCl is 7.45 g
We'll begin by calculating the mass of CaCl₂ that reacted and the mass of KCl produced from the balanced equation.
CaCl₂ + K₂CO₃ —> 2KCl + CaCO₃
Molar mass of CaCl₂ = 40 + (35.5 × 2) = 111 g/mol
Mass of CaCl₂ = 1 × 111 = 111 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mass of KCl from the balanced equation = 2 × 74.5 = 149 g
From the balanced equation above,
149 g of KCl were produced by 111 g of CaCl₂.
Finally, we shall determine the mass of CaCl₂ needed to produce 10 g of KCl. This can be obtained as follow:
From the balanced equation above,
149 g of KCl were produced by 111 g of CaCl₂.
Therefore,
10 g of KCl will be produce by = (10 × 111) / 149 = 7.45 g of CaCl₂.
Thus, 7.45 g of CaCl₂ were obtained from the reaction.
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How many seconds in day?
Answer:
86400 seconds
Explanation:
Since 1 day equals 24 hours, and there are 60 minutes in 1 hour. Also, 60 seconds make a minute.
So, to find the number of seconds in a day, we need to find the number of minutes in a day since 1 day = 24 hours.
So, the number of minutes in a day = 1 day × 24 hours/day × 60 minutes/hour = 24 hours × 60 minutes/hour = 24 × 60 minutes = 1440 minutes.
So, we have that 1440 minutes = 1 day.
So, to find the number of seconds in 1 day, we find the number of seconds in 1440 minutes.
So, 1 day × 1440 minutes/day × 60 seconds/min = 1440 minutes/day × 60 seconds/minute = 1440 × 60 seconds = 86400 seconds.
So, there are 86400 seconds in 1 day.
