Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
Both chlorine and fluorine are represented by a green modeling piece that has 4 holes. Is using the same piece for two different atoms acceptable? Why or why not
Answer:
Yes, same piece can be used.
Explanation:
The same piece can be used for two different atoms are acceptable because both atoms has 7 electrons in their outermost valance shell. Both atoms belong to same group i. e. halogens so same piece can be used for both atoms. If the atoms belong to different groups and they have different number of electrons in their outermost shell so using same piece will be a problem so it is recommended to use different pieces for different atoms.
The use of the same modeling piece for chlorine and fluorine has been accepted as it has consisted of the same properties and belongs to the same group.
Chlorine and fluorine have been the elements of group 17. The elements are halogens with the presence of 7 valence electrons.
The elements have been belonging to the same group and have the same number of valence electrons thus resembling each other in the chemical properties.
Since both the elements are similar to each other, the use of the same piece for two different atoms has been acceptable.
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If the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?
Answer:
14.297 g
Explanation:
From the question;
1 mo of the compound requires 1320.0 kJ
From the molar mass;
1 ml of the compound weighs 30.55g
How many grams requires 617.30kJ?
1 ml = 1320
x mol = 617.30
x = 617.30 / 1320
x = 0.468 mol
But 1 mol = 30.55
0.468 mol = x
x = 14.297 g
What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?
Answer:
The rate would have doubled
Explanation:
When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration
Answer:
B) hydroxide concentration
Explanation:
Hello,
In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.
After that, due to the fact that the pH is related with the pOH as shown below:
pH=14-pOH
We can directly compute the pH.
Best regards.
The concentration of glucose, C6H12O6, in normal spinal fluid is 75 mg/100g. What is the molality of the solution
Answer:
4.16x10⁻³m
Explanation:
Molality is defined as the ratio between moles of a solute, in this case glucose, and kg of solvent.
As there are 100g of solvent, the kg are 0.1. Thus, we only need to calculate from the mass of glucose its moles to solve the molality of the solution.
Moles glucose:
There are 75mg = 0.075g of glucose. To conver mass to moles it is necessary molar mass.
Molar mass glucose:
6C = 12.01g/mol*6 = 72.06g/mol
12H = 12*1.008g/mol = 12.10g/mol
6O = 6*16g/mol = 96g/mol
72.06 + 12.10 + 96 = 180.16g/mol
Moles of 0.075g of glucose:
0.075g * (1 mol / 180.16g) =
4.16x10⁻⁴ moles of glucose
Molality of the solution:
4.16x10⁻⁴ moles of glucose / 0.1kg of solvent =
4.16x10⁻³mThe molarity of the solution is 4.16x10⁻³m
Calculation of the molarity:We know that the molarity refers to the ratio that arise between the moles of a solute.
Since there are 100 g of solvent so here the kg should be 0.1.
Likewise there is 75 mg so it should be 0.075g
Now the Molar mass glucose should be
6C = 12.01g/mol*6 = 72.06g/mol
12H = 12*1.008g/mol = 12.10g/mol
6O = 6*16g/mol = 96g/mol
So,
= 72.06 + 12.10 + 96
= 180.16g/mol
Now
Moles of 0.075g of glucose:
0.075g * (1 mol / 180.16g) =
4.16x10⁻⁴ moles of glucose
Now finally
Molality of the solution:
= 4.16x10⁻⁴ moles of glucose / 0.1kg of solvent
=4.16x10⁻³m
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The condition that a reaction takes place without outside help Choose... Solution in which no more solute can be dissolved in the solvent Choose... Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature Choose... The extent of randomness in a system Choose... Sum of the internal energy plus the product of the pressure and volume for a reaction
Answer:
Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature
Explanation:
The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).
∆G is given by;
∆G= ∆H -T∆S
Where;
∆H= change in enthalpy of the system
T= absolute temperature of the system
∆S= change in entropy
Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.
Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
For ethylenediamine, use (en) in the formula.
a) sodium hexachloroplatinate(IV)
b) dibromobis(ethylenediamine)cobalt(III) bromide
c) pentaamminechlorochromium(III) chloride
Answer:
sodium hexachloroplatinate(IV)- Na2[PtCl6]
dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br
pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2
Explanation:
The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.
The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.
For which one of the following reactions will the enthalpy change be approximately equal to the internal energy change?
A. H2 + I2 → 2HI
B. PCl5(g) → PCl3(g) + Cl2
C. 2H2O2 → 2H2O2 + O2
D. C(s) + O2(g) → CO2(g)
Answer: A. [tex]H_{2}_{(g)}+I_{2}_{(g)}=>2HI_{(g)}[/tex] and D.[tex]C_{(s)}+O_{2}_{(g)}=>CO_{2}_{(g)}[/tex]
Explanation: The relationship between internal energy change and enthalpy change during a chemical reaction occurs according to the following formula:
[tex]\Delta H=\Delta E+\Delta(PV)[/tex]
So, for changes in enthalpy and internal energy to be equal volume or pressure has to be constant, i.e., zero.
Change in the number of moles of gas during the reaction can make the difference between [tex]\Delta H[/tex] and [tex]\Delta E[/tex] be larger, so for them to be equal and pressure constant, number of moles must be the same in reagents and products.
Analysing each reaction above:
Reaction A has the same number of moles in reagents and products, so enthalpy change and internal energy change will be equal;
Reactions B and C don't have the same number of moles at both sides, so enthalpy and energy will be different.
Reaction D, although reagent side have 2 compounds, carbon is solid, so reaction have the same number of moles in both sides. Enthalpy and Energy will be equal.
If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?
Answer:
[tex]m=0.127g[/tex]
Explanation:
Hello,
In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:
[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]
In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:
[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]
Best regards.
How many grams of H2O will be formed when 32.0 g H2 is mixed with 73.0 g of O2 and allowed to react to form water
hope this helps u
pls mark as brainliest .-.
A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONaof their
(c) Ethanol is an alcoholic beverage which can be brewed
from cassava. Outline the process by which ethanol can
be prepared.
[3]
(d) Ethanol is used as a fuel.construct a balanced chemical
equation for its complete combustion.
[2]
[Total:10 Marks]
s?
Answer:
the chemical equation of ethanol as fuel is
C2H5OH(l)+3 02(g)------2CO2(g)+3H2O(g)
the preparation process of ethanol from cassava is
cassava flour---liquification---saccharification---cooling---fermentation---distillation---ethanol
The entropy of a substance above absolute zero will always be:
a. Negative
b. Positive
c. Neither Negative nor positive
Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ
Answer:
CO2 will diffuse more rapidly.
Explanation:
From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:
Rate (R) & 1/√Density (d)
R & 1/√d
But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.
Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:
Rate (R) & 1/√Molar mass (M)
R & 1/√M
From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.
Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.
This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71 g/mol
Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol
Summary
Gas >>>>>> Molar mass
Cl2 >>>>>> 71 g/mol
CO2 >>>>> 44 g/mol
From the illustration above, we can see that CO2 is lighter than Cl2.
Therefore, CO2 will diffuse more rapidly.
Answer: CO2
Explanation:
g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4
Answer:
0.0002 M
Explanation:
The molarity of the HCl required would be 0.0002 M.
First, let us consider the balanced equation of the reaction:
[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]
Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.
Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:
0.0041 x 2 = 0.0082 mole HCl
Volume of the HCl = 40.95 L
Molarity = mole/volume
Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M
The standard entropy of a substance refers to its entropy at:__________.
a. absolute zero and 1 bar
b. 0°C and 1 bar
c. 25 °C and 1 bar
d. 25 °C and 0 bar
Answer:
b. 0°C and 1 bar
Explanation:
Hello,
In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.
Best regards.
28.What is the correct IUPAC name for the following compound?A)12-crown-5B)12-crown-4C)4-crown-12D)12-crown-12E)Cyclododecane tetraether
Answer:
12-crown-4
Explanation:
We must recall that any structural moiety in organic chemistry having the R-O-R unit is an ether. If the oxygen form a ring in which they are sandwiched in between carbon atoms, the compound is known as a crown ether. The name emanates from the close resemblance of the compound to an actual crown.
If we want to name the crown ether, we first count the number of carbon atoms present and the number of oxygen atoms present. The correct name is now, total number of carbon + oxygen atoms -crown- number of oxygen atoms, in this case; 12-crown-4, hence the answer.
Answer:
12-crown-4
Explanation:
a sample of gas occupies a volume of 2.62 liters at 25 C and 1.00 atm. what will be the volume at 50 C and 2 atm
Answer:2.62 L
Explanation:
A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
What is ideal gas law ?The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.
An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.
By using ideal gas equation,
P₁ V ₁ ÷ T = P₂V₂ ÷ T
1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50
V₂ = 1 × 2.62 × 50 ÷ 25 × 2
V₂ = 2.62 liters.
Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
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Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
Answer:
The correct answer is 1332 KJ.
Explanation:
Based on the given information,
ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.
Now the balanced equation is:
C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)
ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2
ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)
ΔH°rxn = -1411.9 KJ
Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.
Now based on the balanced chemical reaction,
ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2
ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)
ΔS°rxn = -267.68 J/K or -0.26768 KJ/K
T = 25 °C or 298 K
Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get
ΔG = -1411.9 - 298.0 × (-0.2677)
ΔG = -1332 KJ.
Thus, the maximum work, which can obtained is 1332 kJ.
Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.
Answer:
whats the ph ofpoh=9.78
Explanation:
What is pH of a buffer made by combining 45.0mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate
Answer:
3.11
Explanation:
Any buffer system can be described with the reaction:
[tex]HA~->~H^+~+~A^-[/tex]
Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:
[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]
With all this in mind, we can write the reaction for our buffer system:
-) Nitrous acid: [tex]HNO_2[/tex]
-) Sodium nitrate: [tex]NaNO_2[/tex]
[tex]HNO_2~->~H^+~+~NO_2^-[/tex]
In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).
We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:
-) moles of [tex]HNO_2[/tex]:
[tex]mol=0.150~M*0.045~L=0.00657[/tex]
-) moles of [tex]NO_2^-[/tex]:
[tex]mol=0.175~M*0.020~L=0.0035[/tex]
The total volume would be:
0.020 L + 0.045 L = 0.065 L
With this in mind, we can calculate the molarity of each compound:
-) Concentration of [tex]HNO_2[/tex]
[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]
-) Concentration of [tex]NO_2^-[/tex]
[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]
The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:
[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]
The final pH value would be 3.11
I hope it helps!
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.
Nitrous acid is a weak acid and nitrate ion is its conjugate base.
What is a buffer?It is a solution used to resist abrupt changes in pH when acids or bases are added.
Step 1: Calculate the moles of each species.We do so by multiplying the molar concentration by the volume in liters.
HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol
NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol
Step 2: Calculate the total volume of the mixture.The total volume will be the sum of the volumes of each solution.
V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L
Step 3: Calculate the molar concentration of each species in the mixture.HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M
NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M
Step 4: Calculate the pH of the buffer.We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.
pH = pKa + log [NaNO₂]/[HNO₂]
pH = 3.16 + log 0.0538/0.104 = 2.87
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
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A 5-column table with 2 rows. Column 1 is labeled number of protons, with entries 20 and 9; column 2 is number of neutrons, with entries 20 and D; Column 3 is atomic number, with entries A and E; Column 4 is Mass Number, with entries B and 19, and Column 5 is Element (symbol) with entries C and F. Using the periodic table, complete the table to describe each atom. Type in your answers
Answer:
A ⇒ 20
B ⇒ 40
C ⇒ Ca
D ⇒ 10
E ⇒ 9
F ⇒ F
Explanation:
edge 2021
Answer:
the person above is correct
Explanation:
Given that π = n M R T, rearrange the equation to solve for V
Answer:
V= n/M
Explanation:
From;
π = nRT/V = MRT
Where;
n= number of moles
R= gas constant
T= absolute temperature
M= molar mass
V= volume of the solution
π= osmotic pressure
Thus;
nRT/V = MRT
nRT = VMRT
V= nRT/MRT
V= n/M
What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Answer:
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
Explanation:
When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.
Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05
The question is incomplete, the complete question is;
Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75
Answer:
A nucleus with a A nucleus with a neutron:proton ratio of 1.49
A nucleus with a mass of 187 and an atomic number of 75
Explanation:
The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1
Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.
Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.
Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.
Answer:
-A nucleus with a neutron:proton ratio of 1.49
-The nucleus of Sb-123
-A nucleus with a mass of 187 and an atomic number of 75
What volume (in mL) needs to be added to 69.6 mL of 0.0887 M MgF2 solution to make a 0.0224 M MgF2 solution
Answer:
The correct answer is 206 ml.
Explanation:
Based on the given information, the molarity or M₁ of MgF₂ solution is 0.0887 M, the molarity or M₂ of the final solution given is 0.0224 M. The initial volume of V₁ of the solution is 69.6 ml, for finding the final volume of V₂ of the solution, the formula to be used is,
M₁V₁ = M₂V₂
Now putting the values in the formula we get,
0.0887 × 69.6 = 0.0224 M × V₂
V₂ = 0.0887 × 69.6 / 0.0224
V₂ = 275.6 ml
Therefore, the volume in ml added to the initial volume of 69.6 ml to make the molarity of the solution 0.0224 will be,
= 275.6 ml - 69.6 ml = 206 ml
A chemical reaction that has the general formula of nA → (A)n is best classified as a ____ reaction. A. synthesis B. polymerization C. decomposition D. oxidation E. replacement
Answer:
B.
Explanation:
A chemical reaction that has the general formula of nA → (A)n is best classified as a polymerization reaction.
Answer:
B. Polymerization
Explanation:
I'm just smart
A simplified version of photosynthesis can be represented as carbon dioxide combining with water to form glucose and oxygen: 6CO2+6H20 C6H12O6+6O2 In this reaction, ________ is oxidized. 1.Carbon dioxide 2.Hydrogen 3.Carbon 4.Oxygen
Answer:
2, hydrogen
Explanation:
i think
Answer:
Answer is not hydrogen
Explanation:
did the test and got it wrong
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step in balancing a redox reaction under basic conditions is to: ______
Explanation:
After combining hydrogen ions and hydroxide ions combine to form water molecules, the next step is the last step when balancing a redox reaction under basic condition.
The last step is to cancel all common terms that arises as a result of the formation of the water molecule. Usually, there's need to balance the water molecules in the reactant and product side of the reaction.
g Which ONE of the following molecules and ions has trigonal planar molecular geometry? (NOTE: You must first determine what the Lewis structure of each substance is.) A) PCl3 B) HCN C) CO3 2– D) H3O+ E) NF3
Answer: C) [tex]CO^{2-}_{3}[/tex]
Explanation: The Valence Shell Electron Pair Repulsion Model (VSEPR Model) shows bonding and nonbonding electron pairs present in the valence, outermost, shell of an atom connecting to other atoms. It also gives the molecular geometric shape of a molecule.
To determine molecular geometry:
1) Draw Lewis Structure, i.e., a simplified representation of the valence shell electrons;
2) Count the number of electron pairs (count multiple bonds as 1 pair);
3) Arrange electron pairs to minimise repulsion;
4) Position the atoms to minimise the lone pair;
5) Name the molecular geometry from the atom position;
Trigonal planar molecular geometry is a model which molecule's shape is triangular and in one plane. Such molecule has three regions of electron density extending out from the central atom and the repulsion will be at minimum when angle between any two is 120°.
The Lewis structure of each molecule is shown in the attachment.
Analysing each one, it can be concluded that molecule with trigonal planar geometry is [tex]CO^{2-}_{3}[/tex]
