Answers
Formal charge = (Valence electrons) - (electrons shared in bonds) - (non-bonded electrons)
H: Formal charge = 1 - 1 - 0 = 0
Cl: Formal charge = 7 - 2 - 4 = +1
O: Formal charge = 6 - 1 - 6 = -1
This structure is not the actual way these atoms arrange. The atoms will actually arrange in the order of H-O-Cl to form hypochlorous acid with a single bond between the three atoms and formal charges of 0 on each atom.
The Lewis structure of HClO consists of bonds from Cl to each of H and O, with Cl having 5 pairs of non-bonding electrons and O having 2 pairs.
Explanation:To create the Lewis structure for HCLO, start by counting valence electrons. Hydrogen (H) has 1, Oxygen (O) has 6, and Chlorine (Cl) has 7. This totals to 14. Next, choose the atom with the least electronegativity (Cl in this case) to be the central atom, and draw bonds to other elements. A bond to each H and O from Cl uses up 4 electrons, leaving 10. Now fill the octets of the H and O atoms.
The Lewis structure should then look like this:
H - Cl - O
With 5 pairs of non-bonding electrons around Cl and two pairs of non bonding electrons around O.
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Related Questions
which of the following solar phenomena is thought to cause short-term climate changes
Answers
A 0.210 mol sample of pcl5(g) is injected into an empty 2.45 l reaction vessel held at 250 °c. calculate the concentrations of pcl5(g) and pcl3(g) at equilibrium.
Answers
PCl5(g) ↔ PCl3(g) + Cl2(g)
initial C (0.21/2.45) 0 0
change -X +X +X
equilibruimC (0.21/2.45)-X X X
So by substitution in Ka formula: when we have Ka at 250°C = 1.8 (must be given- missing in your question)
Ka = [PCl3][Cl2]/[PCl5]
1.8 = (X)(X) / (0.21/2.45)-X
1.8*0.086- 1.8X = X^2
X^2 + 1.8X - 0.1548 = 0 by solving this equation
X= 0.082 mol
∴[PCl5] = (0.21/2.45) - X
= 0.0857 - 0.082 = 0.0037 mol
∴[PCl3] = X = 0.082 mol
A certain reaction has an activation energy of 69.0 kj/mol and a frequency factor of a1 = 3.40×1012 m−1s−1 . what is the rate constant, k, of this reaction at 22.0 ∘c ? express your answer with the appropriate units. indicate the multiplication of units explicitly either with a multiplication dot (asterisk) or a dash.
Answers
What is the concentration of a phosphoric acid solution of a 25.00 mL sample if the acid requires 42.24 mL of 0.135 M NaOH for neutralization?
Please explain your steps.
Answers
do this by:
finding the moles of NaOH which will be 5.702 E -3 m
next find the moles of H3PO4 which will be 1.90 E -3 m
calulcate 25 ml sample molarity = 0.07603 m, just put 0.0760
Answer: The volume of 0.10 M NaOH required to neutralize 30 ml of 0.10 M HCl is, 30 ml.
Explanation:
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of NaOH solution = 0.135 M
[tex]V_1[/tex] = volume of NaOH solution = 42.24 ml
[tex]M_2[/tex] = molarity of [tex]H_3PO_4[/tex] solution = ?M
[tex]V_2[/tex] = volume of [tex]H_3PO_4[/tex] solution = 25 ml
[tex]n_1[/tex] = valency of [tex]NaOH[/tex] = 1
[tex]n_2[/tex] = valency of [tex]H_3PO_4[/tex] = 3
[tex]1\times (0.135M)\times 42.24=3\times M_2\times 25[/tex]
[tex]M_2=0.076M[/tex]
Therefore, the concentration of 0.076 M of phosphoric acid of a 25 ml is required to neutralize 42.24 ml of 0.135 M NaOH.
Determine how many grams of co2 are produced by burning 4.37 g of c4h10.
Answers
the balanced chemical reaction for combustion of C₄H₁₀ (butane) is as follows;
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
the stoichiometry of C₄H₁₀ to CO₂ is 2;4, simplified ratio is 1:2
this means that for every 1 mole of butane used up,4 moles of CO₂ are formed
molar mass of butane - (12 g/mol *4) + (1 g/mol * 10) = 58 g/mol
58 g of butane - 1 mol
Therefore 4.37 g of butane - 1/58 g/mol * 4.37g = 0.075 mol
1 mol of butane forms --> 4 mol of CO₂
Therefore 0.075 mol of butane forms = 4 x 0.075 mol = 0.3 mol of CO₂
molar mass of CO₂ = 44 g/mol
mass of CO₂ formed = 0.3 mol * 44 g/mol = 13.2 g of CO₂ is formed
To determine how many grams of CO2 are produced by burning 4.37 g of C4H10, we can use stoichiometry. The balanced equation shows that 1 mole of C4H10 produces 4 moles of CO2. By converting the mass of C4H10 to moles and then using the stoichiometric ratio, we can calculate the mass of CO2 produced.
Explanation:To determine how many grams of CO2 are produced by burning 4.37 g of C4H10, we can use stoichiometry. The balanced equation shows that 1 mole of C4H10 produces 4 moles of CO2. First, we need to convert the mass of C4H10 to moles by dividing it by the molar mass of C4H10. Then, we can use the stoichiometric ratio to calculate the number of moles of CO2 produced. Finally, we can convert the moles of CO2 to grams by multiplying it by the molar mass of CO2
Given:
Mass of C4H10 = 4.37 gMolar mass of C4H10 = 58.12 g/molMolar mass of CO2 = 44.01 g/molUsing the equation:
Mass of C4H10 (g) → Moles of C4H10 → Moles of CO2 → Mass of CO2 (g)
we can calculate:
4.37 g C4H10 * (1 mol C4H10 / 58.12 g C4H10) * (4 mol CO2 / 1 mol C4H10) * (44.01 g CO2 / 1 mol CO2) = 17.80 g CO2
Therefore, burning 4.37 g of C4H10 produces approximately 17.80 g of CO2.
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The shape of a molecule is tetrahedral. How many lone pairs are most likely on the central atom
Answers
The answer is zero....
This solution contains equal concentrations of both hf(aq) and naf(aq). write the proton transfer equilibrium (net ionic) reaction that includes f– as a reactant. include states of matter.
Answers
The proton transfer equilibrium reaction involving F¯ as a reactant in a solution with equal concentrations of HF and NaF is expressed as HF(aq) + H2O(l) ⇌ H3O+(aq) + F¯(aq). The net ionic equation illustrates the dissociation of hydrofluoric acid and the effect of the common ion from NaF on this equilibrium.
The proton transfer equilibrium reaction that includes F(aq) as a reactant in a solution containing equal concentrations of HF(aq) and NaF(aq) can be written as follows:
HF(aq) + H2O(l) ⇌ H3O+(aq) + F¯(aq)
This shows the dissociation of hydrofluoric acid (HF) into fluoride ions (F¯) and hydronium ions (H3O+). When NaF is dissolved in water, it completely dissociates into Na+ and F¯ ions. The increased concentration of the common ion F¯ from NaF will shift the equilibrium to the left (common ion effect), reducing the ionization of HF and consequently reducing the concentration of H3O+.
The net ionic equation focuses on the species that change during the reaction. In this case, the sodium ion (Na+) is a spectator ion and does not participate in the equilibrium process, so it is omitted from the net ionic equation.
Potassium-40 can decay into either calcium-40 or argon-40. All three of these atoms have essentially the same weight. Ninety percent of the potassium-40 will decay into calcium-40, and only ten percent will decay into argon-40. When argon-40 is produced by the radioactive decay of potassium-40 inside a rock, the argon-40 produced by the decay is a gas and is trapped inside the rock. The amount argon-40 trapped in a rock can be measured by grinding up the rock and capturing the liberated argon-40 gas.
Suppose the amount of potassium-40 inside a rock is measured to be 0.81 milligrams, and the amount of argon-40 gas trapped in the rock is measured to be 0.377 milligrams.
1. How much of the potassium-40 that was originally present inside the rock has undergone radioactive decay to produce argon-40?
Answers
3.77 mg of K-40 decayed into Ar-40.
Data:
1) K-40, Ca-40, Ar-40: all three have the same atomic mass
2) 90% of the potassium-40 will decay into calcium-40
3) 10% of the potassium-40 will decay into argon-40.
4) K-40 inside the rock = 0.81 mg
5) Ar-40 trapped = 0.377 mg
Soltuion:
1) 0.377 mg of Ar-40 is the 10% of the mass of the K-40 that decayed
=> x * 10% = 0.377 mg => x * 0.1 = 0.377mg
=> x = 0.377 mg / 0.1 = 3.77 mg
That means that 3.77 mg of K-40 decayed into Ar-40. And this is the answer to the question.
Additionaly, you can analyze the content of all K-40 and Ca-40, to understand better the case.
2) The mass of the K-40 that decayed into Ca-40 is 9 times (ratio 9:1) the amount that decayed into Ar-40 =>
mass of K-40 that decayed into Ca-40 = 9 * 0.377 = 3.393 mg
3) Total amount of K-40 that decayed = amount that decayed into Ar-40 + amount that decayed into Ca-40 = 0.377mg + 3.393mg = 3.77 mg
4) Original amount of K-40 = amount of K-40 that decayed + amount of K-40 present in the rock = 3.77mg + 0.81 mg = 4.58 mg
5) amount of K-40 that decayed into Ar-40 as percent
% = [3.77 mg / 4.58mg] * 100 = 82.31 %.
For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or increasing the volume of the reaction vessel. complete the sentences to determine if the reaction is endothermic or exothermic and to determine which side of the equation has more molecules.
Answers
First, we can model the chemical reaction in the following way:
nA(g) ⇄ mB(g)
Now, we can start answering the two questions:
1) Determine if the reaction is endothermic or exothermic:
The reaction is endothermic. We know that because the fraction of products is increased when the temperature is increased. According to Le Chatelier's principle, the system will move in the opposing direction of the changes to establish a new equilibrium. We can model an endothermic reaction in the following way, knowing that it needs heat to be completed
nA(g) + Δ ⇄ mB(g)
So, adding heat in the left side of the reaction will displace the equilibrium in the opposing direction (the right side) increasing the fraction of products.
2) Determine which side of the equation has more molecules:
The right side of the equation has more molecules than the left side. According to Le Chatelier's Principle, in a gas phase reaction, increasing the volume results in a decrease in the internal pressure of the system, so the reaction shifts to the side with more molecules to restore the pressure of the system. As the fraction of products increases with the increase in volume, the right side will have more molecules. We can then express this reaction in the following way:
nA(g) ⇄ mB(g) n<m
Have a nice day!
Final answer:
The reaction is endothermic because increasing the temperature increases the fraction of products, as energy acts as a reactant. Moreover, since the product fraction increases with volume, the products' side has more molecules, showing that the reaction has a larger number of gaseous products.
Explanation:
For a gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or increasing the volume of the reaction vessel. To determine if the reaction is endothermic or exothermic, we can use the behavior of the system in response to these changes. When temperature is increased for an endothermic reaction, energy can be considered as a reactant, therefore, the equilibrium shifts towards the products. Conversely, for an exothermic reaction, which releases energy, an increase in temperature would shift the equilibrium towards reactants. Since increasing the temperature increases the fraction of products, the reaction is endothermic.
Regarding the change in volume, an increase in volume results in a decrease in pressure and favors the side of the reaction with more gas molecules. If the fraction of products increases with increased volume, this indicates that the side with the products has more gas molecules. Therefore, the reaction equation has more molecules on the products' side.
What happens when naoh is added to a buffer composed of ch3cooh and ch3coo−?
Answers
When NaOH is added to a buffer composed of CH₃COOH and CH₃COO⁻, the following reactions happen:
-First, the NaOH is neutralized by CH₃COOH:
NaOH + CH₃COOH → H₂O + CH₃COONa
-Second, the CH₃COONa dissociates in its ions:
CH₃COONa → CH₃COO⁻ + Na⁺
-Finally, CH₃COO⁻ (a weak base) reacts with water to form OH⁻ ions and regenerate CH₃COOH
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻
By this sequence of reactions, the buffer can mitigate the effect of the strong base added.
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pH of buffer solution containing [tex]{\mathbf{C}}{{\mathbf{H}}_{\mathbf{3}}}{\mathbf{COOH}}[/tex] and [tex]{\mathbf{C}}{{\mathbf{H}}_{\mathbf{3}}}{\mathbf{COOH}}[/tex] does not change on addition of NaOH to the buffer.
Further explanation:
Buffer solution:
The aqueous solution that consists of weak acid and its conjugate base is known as buffer solution. Such solutions resist any change in their pH on addition of strong base or acid.
pH is used to determine acidity or basicity of solutions. Solutions with pH less than 7 are acidic in nature, those with pH 7 are neutral and those with pH more than 7 are basic.
Classification of buffers:
Acidic buffer:
Solutions of weak acid and its conjugate base with pH less than 7 are acidic. Mixture of acetic acid and sodium acetate is an example of acidic buffer.
Basic buffer:
Solutions of weak base and its conjugate acid with pH more than 7 are basic in nature. Mixture of ammonium chloride and ammonium hydroxide is an example of basic or alkaline buffer.
[tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}[/tex] is a weak acid while NaOH is a strong base so these react with each other to form respective salt and water. Reaction between these two occurs as follows:
[tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + {\text{NaOH}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}} + {{\text{H}}_{\text{2}}}{\text{O}}[/tex]
The salt formed by reaction of [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}[/tex] with NaOHis then dissociated to form its ions as follows:
[tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {\text{N}}{{\text{a}}^ + }[/tex]
Ionic identity [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }[/tex] reacts with water to form uncharged [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}[/tex] again. Reaction for this is as follows:
[tex]{\text{CHCO}}{{\text{O}}^ - } + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + {\text{O}}{{\text{H}}^ - }[/tex]
By going through above series of reactions, effect of addition of NaOH is neutralized by buffer containing [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}[/tex] and [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }[/tex]. Hence pH of buffer solution does not undergo any change in it.
Learn more:
The mass of ethylene glycol: https://brainly.com/question/4053884 Basis of investigation for the scientists: https://brainly.com/question/158048Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Buffer solutions
Keywords: buffer solution, pH, CH3COOH, NaOH, CH3COO-, weak acid, strong base, acidic buffer, basic buffer, less than 7, more than 7, H2O, OH-, CH3COONa, reaction, conjugate base.
How many grams of caf2 would be needed to produce 8.41×10-1 moles of f2?
Answers
Explanation:
1) ratio
Since 1 mole of CaF2 contains 1 mol of F2, the ratio is:
1 mol CaF2 : 1 mol F2
2) So, to produce 8.41 * 10^ -1` mol of F2 you need the same number of moles of CaF2.
3) use the formula:
mass in grams = molar mass * number of moles
molar mass of CaF2 = 40.1 g/mol + 2 * 19.0 g/mol = 78.1 g/mol
mass in grams = 78.1 g/mol * 8.41 * 10^ -1 mol = 65.7 grams
Final answer:
To produce 8.41×10⁻¹ moles of F2, 65.7 grams of CaF₂ are needed, calculated using stoichiometry and the molar mass of CaF₂ (78.08 g/mol).
Explanation:
The question asks how many grams of CaF₂ are needed to produce 8.41×10⁻¹ moles of F₂. Based on stoichiometry, the reaction involves CaF₂ decomposing to produce Ca and F₂. Since each formula unit of CaF₂ contains two fluorine atoms, it will produce 1 mole of F₂ for every mole of CaF₂ decomposed. Therefore, to produce 8.41×10⁻¹ moles of F₂, we also need 8.41×10⁻¹ moles of CaF₂. Using the formula mass of CaF₂ (78.08 g/mol), we can calculate the mass of CaF₂ needed.
Mass of CaF₂ = moles of CaF₂ × molar mass of CaF₂
= 8.41×10⁻¹ moles × 78.08 g/mol
= 65.7 g of CaF2
which of the following substances has a molar mass of 208g
Answers
Answer is: barium chloride (BaCl₂).
a) M(P₂O₅) = 2 · Ar(P) + 5 · Ar(O) · g/mol.
M(P₂O₅) = 2 · 31 + 5 · 16 · g/mol = 142 g/mol.
b) M(BaCl₂) = Ar(Ba) + 2 · Ar(Cl) · g/mol.
M(BaCl₂) = 137,3 + 2 · 35,45 · g/mol = 208,2 g/mol.
c) M(AlCl₃) = Ar(Al) + 3 · Ar(Cl).
M(AlCl₃) = 27 + 3 · 35,45 · g/mol = 133,35 g/mol.
d) M(MgCl₂) = Ar(Mg) + 2 · Ar(Cl) · g/mol.
M(MgCl₂) = 24,3 + 2 · 35,45 · g/mol = 95,2 g/mol.
a) P2O5
b) BaCl2
c) AlCl3
d) MgCl2
Answer:BaCl2
Explanation:
I will find the molar mass of each substance
The molar mass is the sum of the masses of each atom in the unit formula.
1) P2O5
Molar mass = 2 * atomic mass of P + 5 * atomic mass of O = 2 * 31 g/mol + 5 * 16 g/mol = 62g/mol + 80g/mol = 140 g/mol.
∴ This is not the answer.
2) BaCl2
Molar mass = atomic mass of Ba + 2 * atomic mass of Cl = 137.3 g/mol + 2 * 35.5 g/mol = 208.3 g/mol ≈ 208 g/mol
∴ This is the answer
3) AlCl3
Molar mas = atomic mass of Al + 3 * atomic mass of Cl = 27 g/mol + 3 * 35.5 g/mol = 133.5 g/mol
∴ This is not the answer
4) MgCl2
molar mass = atomic mass of Mg + 2*atomic mass of Cl = 24.3 g/mol + 2*35.5 g/mol = 95.3 g/mol
∴ this is not the answer
Write the name of the following compound MgSO4 • H20
Answers
Write a chemical equation that shows the release of hydroxide ions as each of these bases dissolves in water.
a. koh(s), potassium hydroxide
b. ba(oh)2(s), barium hydroxide
Answers
Ba(OH)2 ------> Ba+2. 2OH-
The chemical equations show the release of hydroxide ions for the given bases:
[tex]KOH \longrightarrow K^+ + OH^-[/tex]
[tex]Ba(OH)_2 \longrightarrow Ba^{2+} +2 OH^-[/tex]
What are acids and bases?An acid can be defined as any substance that is the ability to donate a proton in an aqueous solution. A base is a molecule or ion that is capable to donate a hydroxide ion in an aqueous solution.
Acidic substances are generally characterized by sour taste and can donate an H⁺ ion. Bases are characterized by a bitter taste and when bases react with acids, they give salts and water as products. Acids turn blue litmus into red while bases turn red litmus into blue.
When the potassium hydroxide is base when dissolved in water it gives potassium cation (K⁺) and hydroxide ion (OH⁻). When the barium hydroxide (Ba(OH)₂) is dissolved in water it gives the barium (Ba²⁺) and two ions of hydroxide.
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The molar volume of a gas at STP, in liters, is ___ .
You can use the molar volume to convert 2 mol of any gas to ___ L.
You can also use the molar volume to convert 11.2 L of any gas to ____ mol. Avogadro’s law tells you that 1.2 L of O2(g) and 1.2 L of NO2(g) are ______-numbers of moles of gas.
Answers
Let us consider the following points
a) The volume of one mole of an ideal gas at STP (273.15 K and 1 atm) is 22.4 L.
b) the number of moles of in any given volume of gas can be calculated as:
[tex]Moles=\frac{22.4L}{Volumegiven}[/tex]
Solutions:
1) The molar volume of a gas at STP, in liters, is 22.4 L.
2) You can use the molar volume to convert 2 mol of any gas to 44.8 L
Volume = moles X 22.4L
Volume = 2 X 22.4 = 44.8L
3) You can also use the molar volume to convert 11.2 L of any gas to 0.5 mol.
As mentioned above
[tex]mole=\frac{volume}{22.4}=\frac{11.2}{22.4}=0.5mol[/tex]
4) Avogadro’s law tells you that 1.2 L of O2(g) and 1.2 L of NO2(g) are ______-numbers of moles of gas.
The moles of oxygen will be:
[tex]mole=\frac{volume}{22.4}=\frac{1.2}{22.4}=0.0536mol[/tex]
The moles of nitrogen dioxide will be:
[tex]mole=\frac{volume}{22.4}=\frac{1.2}{22.4}=0.0536mol[/tex]
Total moles = 0.0536+0.0536 = 0.1072 moles
The molar volume of a gas at STP is 22.4 L. Therefore, 2 moles of any gas equals 44.8 L and 11.2 L of any gas equals 0.5 moles. Avogadro's law states that 1.2 L of O2(g) and 1.2 L of NO2(g) have the same number of moles.
Explanation:The molar volume of a gas at Standard Temperature and Pressure (STP) is approximately 22.4 liters. Therefore, if you have 2 moles of any gas at STP, this would occupy a volume of 2 x 22.4 = 44.8 liters. Conversely, if you have 11.2 liters of any gas at STP, this equates to 11.2 / 22.4 = 0.5 moles of gas. According to Avogadro's law, equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. So, 1.2 liters of O2(g) and 1.2 liters of NO2(g) contain the same number of moles of gas.
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Caustic soda is 19.1 M NaOH and is diluted for household use. What is the household concentration if 10 mL of the concentrated solution is diluted to 400 mL?
Answers
To calculate the household concentration of NaOH we need to use the dilution formula, clearing for M2, as you can see in the equation below:
[tex]M1*V1=M2*V2 \\ \\ V2= \frac{M1*V1}{V2} [/tex]
Now, we input the values from the data we have onto this equation. M1=19,1 M; V1=10 mL; V2=400 mL, and solve the equation to get the result:
[tex]V2= \frac{M1*V1}{V2}= \frac{19,1M*10mL}{400 mL}=0,48M [/tex]
So, the household concentration of NaOH will be 0,48 M
Have a nice day!
To calculate the household concentration of NaOH after dilution, use the formula M1V1 = M2V2, which results in a final molarity of 0.4775 M when diluting 10 mL of 19.1 M NaOH to a total volume of 400 mL.
Explanation:Calculating the Household Concentration of Diluted NaOH
When diluting a concentrated solution of NaOH for household use, the concentration of the solution changes. To find the new concentration after dilution, we apply the principle of conservation of moles, which states that the moles of solute before dilution are equal to the moles of solute after dilution. The formula for this is M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
In this case, we have:
Initial molarity (M1) = 19.1 M (concentrated NaOH)
Initial volume (V1) = 10 mL (the amount of concentrated NaOH being diluted)
Final volume (V2) = 400 mL (the total volume after dilution)
To find the final molarity (M2), simply rearrange the equation to solve for M2:
M2 = (M1V1) / V2
Substitute the known values:
M2 = (19.1 M * 10 mL) / 400 mL = 0.4775 M
Thus, the household concentration of NaOH after dilution is 0.4775 M.
when butane reacts with oxygen, the temperature of the surrounding area
Answers
When butane reacts with oxygen, the temperature of the surrounding area increases.
The chemical equation for this reaction is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
This is a combustion reaction. Combustion reactions are highly exothermic, meaning that there is a high amount of energy that is released from the breaking of the chemical bonds and the oxidation of butane to carbon dioxide. This energy is released in the form of heat, increasing the temperature of the surroundings.
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Answer:
increases.
Explanation:
How do impulses travel from one neuron to another?
Answers
How many grams of sodium iodide Nai must be used to produce 51.9 G of iodine I2
Answers
61,30 grams of Sodium Iodide NaI must be used to produce 51,9 g of Iodine I₂
The reaction for the production of Iodine from NaI is the following:
2NaI(aq) + Cl₂ → I₂ + 2 NaCl
To calculate how many grams of NaI are needed to produce 51,9 g of Iodine, we use a conversion factor to go backward from grams of iodine to grams of sodium iodide. The coefficients of the reaction will help us, as well as the molar masses of each compound.
[tex]51,9gI_2* \frac{1 mol I_2}{253,809 g I_2}* \frac{2 mol NaI}{1 mol I_2} * \frac{149,89 g NaI}{1 mol NaI}=61,30 g NaI [/tex]
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Final answer:
To produce 51.9 grams of iodine (I2), approximately 61.2 grams of sodium iodide (NaI) must be used. This calculation is based on stoichiometry and the molar masses of NaI and I₂.
Explanation:
To calculate the amount of sodium iodide (NaI) needed to produce a given amount of iodine (I₂), we will use stoichiometry based on the balanced chemical equation of the reaction. The molecular weight of I₂ (iodine) is approximately 253.8 g/mol (126.9 g/mol for each iodine atom). Let's calculate how many moles of I₂ are in 51.9 grams.
Step 1. Convert from grams to moles of I₂ using the molar mass of I₂ in the unit conversion factor.
51.9 g I₂ × (1 mol I₂ / 253.8 g I₂) = 0.204 moles of I₂.The balanced chemical equation for the production of iodine from sodium iodide may look like this, depending on the reacting substances involved:
2 NaI + Cl₂ → 2 NaCl + I₂From the balanced equation, we see that the molar ratio between NaI and I₂ is 2:1. This means it takes 2 moles of NaI to produce 1 mole of I₂.
Step 2. Use the stoichiometry to find the moles of NaI needed.
0.204 moles I₂ × (2 moles NaI / 1 mole I₂) = 0.408 moles of NaI needed.Step 3. Convert moles of NaI to grams.
The molar mass of NaI (22.99 g/mol for Na + 126.9 g/mol for I) is approximately 149.89 g/mol.0.408 moles NaI × (149.89 g NaI / 1 mole NaI) = 61.2 grams of NaI.Therefore, to produce 51.9 grams of iodine (I₂), approximately 61.2 grams of sodium iodide (NaI) must be used.
Bromophenol blue is the indicator used in detecting the endpoint for the antacid analysis in this experiment. what is the expected color change at the endpoint?
Answers
For the antacid analysis, the chemical reactions that occur in the titration are the following ones:
First, the antacid (composed of weak bases and carbonates) is completely neutralized by the H⁺ ions in the HCl
2HCl + CaCO₃ → CO₂ + H₂O + 2CaCl₂
HCl + OH⁻ → H₂O + Cl⁻
The titration reaction consists in titrating the excess H⁺ ions that are left in the solution, by the following reaction:
H⁺ + NaOH → H₂O + Na⁺
So, when the equivalence point is reached, the solution will go from acid to basic. As bromophenol blue is yellow in acidic solution and blue in basic solution, you'll expect the indicator to change from yellow to blue.
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The expected color change at the endpoint using bromophenol blue as an indicator in an antacid analysis experiment is from yellow to blue-violet, indicating a pH rise above 4.
Explanation:In an antacid analysis experiment using bromophenol blue as the indicator, the expected color change at the endpoint is from yellow to blue-violet. Bromophenol blue turns yellow below a pH of about 3 and changes to blue-violet above a pH of about 4. Therefore, when the color change occurs, it is indicating that the pH has risen above 4, signifying the endpoint where neutralization has been achieved in the titration process.
This distinct color transition provides a visual cue for researchers to precisely determine the endpoint of the antacid titration, ensuring accurate measurement and analysis of the reaction's neutralization point.
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What is the relationship between environmental health and our own health
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The relationship between environmental health and our own health is intrinsic and multifaceted. Environmental health encompasses the aspects of human health that are determined by physical, chemical, biological, social, and psychosocial factors in the environment. It also refers to the theory and practice of assessing, correcting, controlling, and preventing those factors in the environment that can potentially affect adversely the health of present and future generations.
Here are several key points that illustrate the relationship between environmental health and personal health:
1. Exposure to Pollutants: The quality of air, water, and soil can directly impact health. For example, exposure to air pollutants like particulate matter, ozone, nitrogen dioxide, and sulfur dioxide can lead to respiratory and cardiovascular diseases. Similarly, contaminated water and soil can lead to a variety of health issues, including gastrointestinal illnesses and heavy metal poisoning.
2. Chemical Safety: The use of pesticides, herbicides, and industrial chemicals can have toxic effects on humans. Exposure to these chemicals, whether through direct contact or through the food chain, can increase the risk of cancer, reproductive issues, and developmental problems.
3. Climate Change: Changes in climate can affect health in numerous ways, including an increase in heat-related illnesses, the spread of vector-borne diseases (such as malaria and Lyme disease), and the impact of extreme weather events on mental health.
4. Natural Resources: Access to clean water, nutritious food, and natural spaces for recreation and relaxation contributes to good health. Conversely, a lack of these resources can lead to malnutrition, dehydration, and increased stress levels.
5. Built Environment: The design of our communities, including housing, transportation, and recreational facilities, influences physical activity levels, exposure to noise and light pollution, and opportunities for social interaction, all of which affect health outcomes.
6. Occupational Health: Workplace environments can expose individuals to hazardous conditions, dangerous materials, and stressful situations, which can lead to occupational diseases and injuries.
7. Social Determinants of Health: Environmental health also includes social and economic factors, such as access to education, healthcare, and safe neighborhoods. These factors can have a profound impact on health disparities and life expectancy.
8. Mental Health: The environment can influence mental health through factors such as noise pollution, overcrowding, and lack of green spaces, which can contribute to stress, anxiety, and depression.
In summary, environmental health is a critical determinant of the health of individuals and populations. By understanding and addressing the environmental factors that affect health, we can work towards creating healthier living conditions and reducing the burden of disease. Public health initiatives that focus on improving environmental quality, such as reducing pollution, promoting sustainable practices, and ensuring access to clean water and nutritious food, are essential for enhancing the health and well-being of current and future generations.
Write formulas for these hydrates sodium sulfate decahydrate
Answers
Sodium sulfate decahydrate - Na₂SO₄×10H₂O. Sodium has oxidation number +1 and sulfate -2.
Calcium chloride dihydrate - CaCl₂×2H₂O. Calcium has oxidation number +2 and chlorine -1.
Barium hydroxide octahydrate Ba(OH)₂×8H₂O. Barium has oxidation number +2 and hydroxide -1.
Magnesium sulfate heptahydrate MgSO₄×7H₂O. Magnesium has oxidation number +2.
Final answer:
The chemical formulas for the hydrates are Na₂SO₄⋅10H₂O (sodium sulfate decahydrate), CaCl₂⋅2H₂O (calcium chloride dihydrate), and Ba(OH)₂⋅8H₂O (barium hydroxide octahydrate).
Explanation:
To write the chemical formulas of hydrates, you start by writing the formula for the anhydrous compound (the compound without water) followed by a dot, then the number of water molecules, represented as H₂O. Each number of water molecules is prefaced by a prefix that indicates how many molecules of water are included.
Sodium sulfate decahydrate, which has 10 water molecules, is written as Na₂SO₄⋅10H₂O.Calcium chloride dihydrate, which has 2 water molecules, is written as CaCl₂⋅2H₂O.Barium hydroxide octahydrate, which has 8 water molecules, is written as Ba(OH)₂⋅8H₂O.These formulas indicate the precise number of water molecules associated with each ionic compound.
A 595 mL sample of chlorine gas at 24.7°C is held at constant pressure while it is heated and the volume of the gas expands to 876mL. What is the new temperature in Kelvin?
A) 83.7 K
B) 202 K
C) 365 K
D) 438 K
Answers
[tex] \frac{V1}{T1} = \frac{V2}{T2} [/tex]
Parameters for the first instance is given on the left side and parameters for the second instance are given on the right side of the equation.
temperature should be given in K.
temperature in Kelvin = temperature in celcius + 273
24.7 °C + 273 = 297.7 K
595 mL / 297.7 K = 876 mL / T
T = 438.2 K rounded off is 438 K
Answer is D
A sample of gas in which [h2s] = 5.76 m is heated to 1400 k in a sealed vessel. after chemical equilibrium has been achieved, what is the value of [h2s]? assume no h2 or s2 was present in the original sample.
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Final answer:
To calculate the equilibrium concentration of H₂S after heating to 1400 K, changes in the initial concentration are compared to that at equilibrium. If the change is less than 2%, it's negligible, following similar principles found in other chemical equilibria.
Explanation:
The question concerns the calculation of the equilibrium concentration of H₂S after heating a sample to 1400 K. In the original sample, there was only H₂S present with no H₂ or S₂. When the equilibrium is reached after heating, we compare the initial concentration with the concentration at equilibrium. A change of less than 2% is considered negligible. Therefore, the assumption that 2x is negligibly small compared to the initial concentration h₂s is confirmed. This principle follows the pattern found in other equilibrium scenarios such as the decomposition of PCl₅ into PCl₃ and Cl₂ where the remaining concentrations can compute the equilibrium constant.
How many orbitals are there in the third shell (n=3)? express your answer numerically as an integer?
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Principal quantum number (n) describes the size of the orbital. There are one s orbital (angular quantum number l=0), three p orbitals (angular quantum number l=1 and magnetic quantum number m = -1,0,+1) and five d orbitals (angular quantum number l=2 and magnetic quantum number m = -2,-1,0,+1,+2).
There are three subshells (s, p, and d) and a total of nine orbitals in the third shell (n=3) of an atom.
The question asks about the number of subshells and orbitals in the third principal shell (n=3) of an atom. In the third shell, there are three subshells, which are designated as 3s, 3p, and 3d. The subshell with l=0 is called s and has 2(0)+1 = 1 orbital, the subshell with l=1 is called p and has 2(1)+1 = 3 orbitals, and the subshell with l=2 is called d and has 2(2)+1 = 5 orbitals.
Total number of orbitals for n=3 = 1 + 3 + 5 = nine orbitals. These orbitals can accommodate a maximum number of electrons using the formula 2n², which gives us 2(3)² = 18 electrons for the third shell.
Hannah just finished building a house of cards that stands four stories high. She is worried that it will fall down. Which of the following statements must be true? A. A house of cards is too fragile and must fall down eventually due to gravity. B. If Hannah adds another card to the house of cards, then it will fall down. C. As long as nobody touches the house of cards, it will remain standing. D. If no unbalanced force acts upon the house of cards, then it will remain standing forever.
Answers
Comment the correct answer.
the answer is if no unbalanced force acts upon the house of cards then it will remain standing forever.
explanation:Newton's first law of motion states that an object at rest will remain at rest unless acted upon by an unbalanced force.
Even if nobody touches the house of cards, another type of unbalanced force (like wind) could knock the cards down, or if Hannah adds another card to the house of cards, it may or may not fall down.
However, if no unbalanced force acts upon the house of cards, then it will remain standing forever.
What is the oxidation state of each element in k2cr2o7?
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K is always 1+
And O is always 2-, except for in peroxide where it is 1-.
So now you must remember that a compound is always neutral. Your net charge has to be 0.
K2Cr2O7.
K2 (1+ x 2) = 2+
Cr (2x) = 2x
O7 (2- ×7) = 14-
+2 +2x + (-14) = 0
•Simple algebra.
2x -12 = 0
2x = 12
x = +6.
Each Cr atom has an oxidation number of +6.
The oxidation number is the charge when the bonds are ionic in the atom. The oxidation state of potassium is +1, oxygen is -2 and chromium is +6.
What is the oxidation state?The oxidation state or the number is the total of the electron gained or lost by the atom to form the chemical bond.
Potassium is always +1, and oxygen is -2 except in some cases.
The state can be shown as:
[tex]\rm K_{2} (+1 \times 2) = +2[/tex]
Cr (2x) = 2x
[tex]\rm O_{7} (-2 \times 7) = -14[/tex]
When the compound is neutral and the net charge is 0 then,
[tex]\begin{aligned} \rm +2 +2x + (-14) &= 0\\\\\rm 2x -12& = 0\\\\\rm x &= +6\end{aligned}[/tex]
Therefore, the oxidation number of chromium is +6.
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What is the overall nuclear fusion reaction in the sun?
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Which type of map projection uses a cone-shaped piece of paper to depict the Earth?
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The type of map projection that uses a cone-shaped piece of paper to depict the Earth is the Conical projection.
To understand Conical Projection, you'll need to imagine a cone placed over the traditional globe like a hat. Now you imagine that the cone is cut and is extended over a table to form a map. There are some maps that use this kind of projection, which includes Albers Equal Area Conic, Equidistant Conic and Lambert Conformal Conic projections.
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Describe a consequence of overpopulation of deer in the forest areas of the northeastern United States