Answer:
The correct answer is 0.06 M
Explanation:
The concentration of the dilute solution (Cd= dilute concentration) can be calculated from the concentrated solution (Cc) and the volumes (Vi and Vf) as follows:
Cc x Vc = Cd x Vd
Cd = (Cc x Vc)/Vd
We have:
Cc= 6 M
Vc = 10 ml (volume of concentrated solution)
Vd= 1 L= 1000 ml (volume of dilute solution)
⇒Cd= (6 M x 10 ml)/1000 ml
Cd = 0.06 M
Which statement is true about the total mass of the reactants during a chemical change?
Answer:
Equal to the total mass
Explanation:
just took the test
Relative and average atomic mass both describe properties of an element related to its different isotopes. Out of these two Relative atomic mas is more accurate. Therefore, the correct option is option C.
What is mass?Mass defines the quantity of a substance. It is measured in gram or kilogram. Average mass is the mass of atoms of an element that are isotopes. It can be calculated by multiplying mass of a isotope to natural abundance of that isotope. In chemistry relative mass is equal to the mass of one-twelfth the mass of C-12 isotope .
Average atomic mass = (mass of first isotope× percent abundance of first isotope)+(mass of second isotope× percent abundance of second isotope)
The total mass of the reactants during a chemical change is equal to the total mass of the product.
Therefore, the correct option is option C.
To learn more about mass, here:
https://brainly.com/question/28704035
#SPJ5
Complete and balance this neutralization reaction: HNO3 + Al(OH)3
Answer:
Al(OH)3 + HNO3 = Al(NO3)3 + H2O - Chemical Equation Balancer.
Explanation:
If you made 6 moles of NO2 how many grams of N2 did you use?
Answer:
3
Explanation
N is 14, so multiply it by 6 (because 6 moles). Since it's N2, divide it by 2, then divide it by 14 (N in grams).
Will a reaction occur in KCl +Br2= KBr + Cl2
You are given the reaction Cu + HNO3 Cu(NO3)2 + NO + H2O complete the final balanced equation based on half-reactions
The balanced equation based on half-reactions is 2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O.
How to determine the balanced equation based on half-reactionsTo complete the balanced equation for the given reaction Cu + HNO3 -> Cu(NO3)2 + NO + H2O using half-reactions, we need to break down the overall reaction into separate oxidation and reduction half-reactions.
1. Oxidation Half-Reaction:
Cu -> Cu2+ + 2e-
In this step, copper (Cu) is oxidized, losing two electrons to form copper(II) ions (Cu2+).
2. Reduction Half-Reaction:
HNO3 + 3H+ + 2e- -> NO + 2H2O
In this step, nitric acid (HNO3) is reduced, gaining two electrons to form nitric oxide (NO) and water (H2O).
Now, to balance the half-reactions, we need to make sure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. In this case, we can achieve this by multiplying the oxidation half-reaction by two.
Balanced Half-Reactions:
Oxidation: 2Cu -> 2Cu2+ + 4e-
Reduction: HNO3 + 3H+ + 2e- -> NO + 2H2O
Finally, we can combine the balanced half-reactions to obtain the balanced equation for the overall reaction:
2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O
Therefore, the balanced equation based on half-reactions is 2Cu + 8HNO3 + 6H+ -> 2Cu(NO3)2 + 2NO + 4H2O.
Learn more about oxidation at https://brainly.com/question/25551544
#SPJ6
Which of the following is the correct definition of conduction?
A.
the transmission of heat across empty space
B.
the electromagnetic radiation from the surface of an object which is due to the object's temperature
C.
the transfer of heat by currents within a liquid or gas
D.
the transmission of heat across matter
Answer:
the transmission of heat across matter
Explanation:
Conduction is the transmission of heat across matter. Even though heat flows across matter during conduction, the matter itself does not flow.
Calculate the molarity of a nitric acid solution if 38 ml of the solution is neutralized by 16 ml of 0.25 M barium hydroxide solution. The balanced equation for the reaction is
Answer:
l
Explanation:
mention two instruments that could be used to transfer sodium hydroxide solution into conical flask
Answer:
Pipette and Burettes
Explanation:
The two instruments that can be used for
a) Pipette – It is used to dispense measured volume of any liquid into any other equipment. Pipette works on the concept of creating a partial vacuum to take in the liquid and then release the volume to dispense the liquid.
b) Burettes – It has a stopcock that regulate the release of liquid. It is larger than the pipette and works as a graduated cylinder.
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions take place and before equilibrium is established), was 5.56. On the following graph, plot the points representing the initial concentrations of all three gases. Label each point with the formula of the gas.
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
[tex]Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}[/tex]
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.
The initial concentration of the HI at the time 0 sec, has been 0.05 mol/L.
The balanced chemical equation for the reaction has been:
[tex]\rm H_2\;+\;I_2\;\rightarrow\;2\;HI[/tex]
The given initial concentration of hydrogen has been 0.030M, and the given concentration of iodine has been 0.015 M.
The reaction quotient for the following reaction has been:
Reaction quotient = [tex]\rm \dfrac{[HI]^2}{[H_2]\;[I_2]}[/tex]
The concentration of HI for the reaction quotient 5.56 has been:
5.56 = [tex]\rm \dfrac{[HI]^2}{[0.030]\;[0.015]}[/tex]
[tex]\rm [HI]^2[/tex] = 0.0025 M
HI = 0.05 mol/L or 0.05 M.
The concentration of the HI at the initial concentration has been 0.05 mol/L.
The graph attached has been plotted as the concentration as the function of time.
The initial concentration of hydrogen, iodine, and hydrogen iodide have been plotted as the function of time.
For more information about the reaction quotient, refer to the link:
https://brainly.com/question/8205004
Given the equation representing a solution equilibrium: BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq)
What occurs when Na2SO4(s) is added to this system, increasing the concentration of SO4 2- (aq)?
1) The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases
2)The equilibrium shifts to the left, and the concentration of Ba2+(aq) increases
3)The equilibrium shifts to the right, and the concentration of Ba2+(aq) decreases
4)The equilibrium shifts to the right, and the concentration of Ba2+(aq) increases
Answer:
The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases
Explanation:
Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in solution decreases significantly.
In this case, both BaSO4 and Na2SO4 both possess the SO4^2- anion. Hence SO4^2- anion is the common ion. Given the equilibrium;
BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq), addition of Na2SO4 will decrease the solubility of BaSO4 due to the presence of a common SO4^2- anion compared to pure water.
This implies that the equilibrium will shift to the left, (more undissoctiated BaSO4) hence decreasing the Ba^2+(aq) concentration.
1) The equilibrium shifts to the left, and the concentration of [tex]Ba^{2+}[/tex](aq) decreases
Conditions for equilibrium:Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in the solution decreases significantly.
In this case, both BaSO₄ and Na₂SO₄ both possess the [tex]SO_4^{2-}[/tex] anion. Hence [tex]SO_4^{2-}[/tex] anion is the common ion.
Given the equilibrium;
[tex]BaSO_4(s) < = > Ba^{2+} (aq) + SO_4^{2-} (aq)[/tex],
The addition of Na₂SO₄ will decrease the solubility of BaSO₄ due to the presence of a common [tex]SO_4^{2-}[/tex] anion compared to pure water.
This implies that the equilibrium will shift to the left, (more undissociated BaSO₄ ) hence decreasing the [tex]Ba^{2+}[/tex]aq) concentration.
Find more information about Equilibrium here:
brainly.com/question/19340344
What is the molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl?
Answer:
The molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl, is 3.42 [tex]\frac{moles}{L}[/tex]
Explanation:
Molarity is a unit of concentration based on the volume of a solution and is defined as the number of moles of solute per liter of solution. Then, the molarity of a solution is calculated by dividing the moles of the solute by the liters of the solution.
[tex]Molarity (M)=\frac{number of moles of the solute}{volume of the solution}[/tex]
Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex]).
Then you must know the amount of moles of the NaCl solute. For that it is necessary to know the molar mass. Being:
Na: 23 g/moleCl: 35.45 g/molethe molar mass of NaCl is: 23 g/mole + 35.45 g/mole= 58.45 g/mole
Then a rule of three applies as follows: if 58.45 grams are present in 1 mole of NaCl, 10 grams in how many moles will they be?
[tex]moles=\frac{10 grams*1 mole}{58.45 grams}[/tex]
moles= 0.171
So you know:
number of moles of solute= 0.171 molesvolume= 50 mL= 0.05 LReplacing in the definition of molarity:
[tex]Molarity (M)=\frac{0.171 moles}{0.05 L}[/tex]
Solving:
Molarity= 3.42 [tex]\frac{moles}{L}[/tex]
The molarity of a 50.0 ml aqueous solution containing 10.0 grams of table sal, Nacl, is 3.42 [tex]\frac{moles}{L}[/tex]
Could u please balance this equation
Explanation:
I don't know how to explain
How many moles of O2 are needed to react completely with 35.0 mol of 10 points
FeCl3? *
4FeCl3 + 302 — 2Fe2O3 + 3Cl2
A) 26.3 mol
B) 46.7 mol
C) 23.3 mol
D) 10.0 mol
if the [H+] = 0.01 M, what is the pH of the solution, and is the solution a strong acid, weak acid, strong base, or weak base?
12, strong base
2, weak acid
12, weak base
2, strong acid
Answer:
2, strong acid
Explanation:
Data obtained from the question. This includes:
[H+] = 0.01 M
pH =?
pH of a solution can be obtained by using the following formula:
pH = –Log [H+]
pH = –Log 0.01
pH = 2
The pH of a solution ranging between 0 and 6 is declared to be an acid solution. The smaller the pH value, the stronger the acid.
Since the pH of the above solution is 2, it means the solution is a strong acid.
does temperature and pressure affect the solubilty of any solute
Answer:
Yes very much so, especially gases
Explanation:
Solubility of gases decrease with increasing pressure. Be familiar with Henry's Law.
- Hope that helps! Please let me know if you need further explanation.
Answer:
Yes, it does.
What is an oxidizing agent ?
Answer:
A substance that tends to bring about oxidation by being reduced and gaining electrons
Explanation:
Answer:
An oxidizing agent is a substance that takes negatively charged electrons "away" from a substance or object; usually taking these electrons for itself it can also transport the electrons from an object and "give" them to another substance or object. The object that has been oxidized will then have a positive charge after losing the negative electrons and the substance receiving the electrons will become negative.
Explanation:
Examples include halogens, potassium nitrate, and nitric acid.
The teacher made a 0.5M solution. How is this number read/said?
Answer:
It reads as follows: 0.5 moles of solute per liter of solution.
Explanation:
Molarity is the most frequent way of expressing the concentration of solutions in chemistry, and it indicates the number of moles of solute dissolved per liter of solution; is represented by the letter M.
A gas sample is stored in a 2 liter container under 3 atmospheres of pressure at 28 C. If the gas is placed into a 5 liter container at 39 C, what will the pressure be?
Answer:
28c
Explanation:
formula of silver nitrite
Answer:
AgNO₃
Explanation:
Ag = Silver
NO₃ = Nitrate
In ΔDEF, the measure of ∠F=90°, FE = 16, DF = 63, and ED = 65. What ratio represents the tangent of ∠E?
Answer:
63/16
Explanation:
tanE=
adjacent
opposite
=
16
63
Answer:
63/16
Explanation:
this for delta math?
The enthalpies of formation of the compounds in the combustion of methane, , are CH4 (g): Hf = –74.6 kJ/mol; CO2 (g): Hf = –393.5 kJ/mol; and H2 O(g): Hf = –241.82 kJ/mol. How much heat is released by the combustion of 2 mol of methane? Use .
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Answer: answer is C
Explanation:
Just took the test
Which is not a reason that water is used to store spent fuel rods from nuclear power plants?
Water increases the speed of the chain reaction in the fuel rods.
Water protects nuclear power plant workers from the high temperature and radiation of the fuel rods.
Water acts as a radiation shield to reduce the radiation levels.
Water cools the spent rods.
Answer:
I believe the answer would be A
5. How is a cell wall different from a cell membrane?
Consider the equation for acetic acid plus water HC2H2O2 + H2O --> C2HcO2- + H3O+ a)compare the strengths of the two acids in the equation. Do the same for the two bases. b) Determine which direction -- forward or reverse -- is favored in the reaction.
Considering the two acids and bases involved, the reverse reaction is favored.
Reversible reactionThe term reversible reaction refers to a reaction that can go either forward or backwards depending on the conditions of the reaction.
We must recall that acetic acid is a weak acid hence the equilibrium position would always lie towards the left hand side of the equation. Hence considering the two acids and bases involved, the reverse reaction is favored.
Learn more about reversible reaction:https://brainly.com/question/8592296
When will the simping end
Answer:
Nvr XD
Explanation:
Answer:
the world may never know
Explanation:
Can someone help me with this?
Answer:
D) condensing C) boiling
In Humans, the liquid waste
filtered by the kidneys is stored
in the ......?
A) Stomach
B) Bladder
C) Lungs
D) I'm not sure. I need
more help with this
topic.
A 5.72-L football is filled with air at 1.85 atm. At the same temperature, the volume of the football is reduced to 2.13 L. What is the pressure of air in the ball?
Answer:
4.97 atm
Explanation:
P2=V1P1/V2
I know this is correct because P and V at constant T have an inverse relationship.
- Hope that helps! Please let me know if you need further explanation.
Sulfuric acid (H2SO4) has a molar mass of 98.1 g/mol. How many oxygen atoms are found in 75.0 g of H2SO4?
A. 1.15 × 10^23
B. 1.84 × 10^24
C. 4.61 × 10^23
D. 7.87 × 10^23
Answer:
B. 1.84 × 10^24
Explanation:
[tex]75.0 g H2SO4(\frac{1 mol H2SO4}{98.1 g})(\frac{4 mol O}{1 mol H2SO4})(\frac{6.023 x 10^23 atoms}{1 mol O} )[/tex] = 1.84 x 10^24 atoms
ipt Which of the following is true about semipermeable membranes?
A. A semipermeable membrane allows passage of solute particles but not solvent particles.
B. A semipermeable membrane allows passage of solvent particles but not solute particles.
C. A semipermeable membrane allows passage of both solvent and solute particles.
D. A semipermeable membrane does not allow passage of either solvent or solute particles.