Answer:
time taken for one-half of the BrO⁻ ion to react is t= 27.45 secs
Explanation:
equation of reaction
3BrO⁻(aq) → BrO₃⁻(aq) + 2Br⁻(aq) (second order reaction)
given
the slope of the graph is 0.056M⁻¹s⁻¹ = k(constant)
initial concentration [A]₀ = 0.65M
for second order reaction,we can calculate the time taken for one-half of the BrO- ion to react using:
[tex]\frac{1}{[A]}[/tex] =[tex]\frac{1}{[A]}[/tex]₀ ⁺ k × t
where initial concentration [A]₀ = 0.65M
[A] = [A]₀÷2 = 0.325M
[tex]\frac{1}{0.325M}[/tex] = [tex]\frac{1}{0.65M}[/tex] + 0.056M⁻¹s⁻¹ × t
3.077= 1.54 + 0.056t
3.077-1.54=0.056t
1.537=0.056t
t= 27.45 secs
A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.0990 M NaOH solution. If 20.52 mL mL of the NaOH solution is required to neutralize completely 11.13 mL of the malonic acid solution, what is the molarity of the malonic acid solution
Answer:
0.0913 M
Explanation:
We'll begin by writing the balanced equation for the reaction.
This is given below:
H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O
From the balanced equation above, we obtained the following:
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 2
Data obtained from the question include:
Molarity of base, NaOH (Mb) = 0.0990 M
Volume of base, NaOH (Vb) = 20.52 mL
Volume of acid, H2C3H2O4 (Va) = 11.13 mL
Molarity of acid, H2C3H2O4 (Ma) =..?
The molarity of the acid, H2C3H2O4 can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 11.13 / 0.0990 x 20.52 = 1/2
Cross multiply
Ma x 11.13 x 2 = 0.0990 x 20.52 x 1
Divide both side by 11.13 x 2
Ma = (0.0990 x 20.52)/ (11.13 x 2)
Ma = 0.0913 M
Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M
Solid iron(II) oxide reacts with oxygen gas to produce solid iron(III) oxide. Balance the equation for this reaction (in lowest multiple integers). Write the unbalanced equation for this reaction.
Answer
Hello
I think the reaction is like this FeO+OFe²O³
And the balance reaction is 2Fe+OFe²O³
Explanation:
At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.
Finally our reaction balanced.
Good luck
Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K I O 3 with a solution of N a 2 S 2 O 3 of unknown concentration. The endpoint was observed to occur at 15.51 mL . How many moles of K I O 3 were titrated
Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
The heat of vaporization delta Hv of dichloromethane (Ch2CL2) is 28.0 kJ/mol . Calculate the change in entropy delta S when 473 g of dichloromethane boils at 39.8 degree.
Answer:
16 J/K.mol
Explanation:
From the question,
ΔS = ΔH/T............... Equation 1
Where ΔH = Heat change, T = Temperature
But,
ΔH = n(Hv).................. Equation 2
Where n = number of mole, Hv = heat of vaporization.
Given: Hv = 28.0 kJ/mol, n = 473/85 = 5.59 mole.
Substitute these values into equation 2
ΔH = 28/5.59
ΔH = 5.01 kJ.
Also: T = 273+39.8 = 312.8 J
Substitute into equation 1
ΔS = 5.01/312.8
ΔS = 0.016 kJ/K
ΔS = 16 J/K.mol
some students believe that teachers are full of hot air. If I inhale 3.5 liters of gas at a temperature of 19 degrees Celsius and it heats to a temperature of 58 degrees celsius in my lungs. what is the new volume of the gas?
Answer:
3.97 L
Explanation:
Data obtained from the question include the following:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 19 °C
Final temperature (T2) = 58 °C
Final volume (V2) =..?
Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:
Temperature (K) = temperature (°C) + 273
T (K) = T (°C) + 273
Initial temperature (T1) = 19 °C
Initial temperature (T1) = 19 °C + 273 = 292 K
Final temperature (T2) = 58 °C
Final temperature (T2) = 58 °C + 273 = 331 K
Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:
Initial volume (V1) = 3.5 L
Initial temperature (T1) = 292 K
Final temperature (T2) = 331 K
Final volume (V2) =..?
V1 /T1 = V2 /T2
3.5 /292 = V2 /331
Cross multiply
292 x V2 = 3.5 x 331
Divide both side by 292
V2 = (3.5 x 331) / 292
V2 = 3.97 L
Therefore, the new volume of the gas is 3.97 L.
Find the percentage composition of each element in the compound having 9.8 grams of nitrogen,0.7 grams of hydrogen and 33.6 grams of oxygen
Answer: The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.
Explanation:
Percentage composition is defined as the ratio of mass of substance to the total mass in terms of percentage.
Percentage composition=[tex]\frac{\text {mass of the element}}{\text {Total mass of the substance}}\times 100\%[/tex]
a) [tex]{\text {percentage composition of nitrogen}}=\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of nitrogen}}=\frac{9.8g}{9.8+0.7+33.6}\times 100\%=22.2\%[/tex]
b) [tex]{\text {percentage composition of hydrogen}}=\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of hydrogen}}=\frac{0.7}{9.8+0.7+33.6}\times 100\%=1.59\%[/tex]
c) [tex]{\text {percentage composition of oxygen}}=\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%[/tex]
[tex]{\text {percentage composition of oxygen}}=\frac{33.6}{9.8+0.7+33.6}\times 100\%=76.2\%[/tex]
The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm
Answer:
3.00 L
Explanation:
PV = nRT
(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K
V = 0.00300 m³
V = 3.00 L
Considering that catalysts are not consumed in a reaction, how do you think increasing the amount of catalyst would affect the reaction rate for the decomposition of hydrogen peroxide?
a. increase
b. decrease
c. no effect
Answer:
a. increase
Explanation:
Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.
By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.
What may be expected when K < 1.0? Choose the THREE correct statements. The concentration of one or more of the reactants is small. The concentration of one or more of the products is small. The reaction will not proceed very far to the right. The reaction will generally form more reactants than products.
Answer:
The concentration of one or more of the products is small.
The reaction will not proceed very far to the right.
The reaction will generally form more reactants than products
Explanation:
We often write
K =[Products]/[Reactants]
Thus, if K is small
We have fewer products than reactants We have more reactants than products The position of equilibrium lies to the leftA. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.
What's the difference between velocity time graph and distance time graph
Explanation:
Hi there!
I attached a photo of a unit summary that states the difference between s-t and v-t graph.
Hope this helps ;) ❤❤❤
Draw the major product(s) obtained when the following compounds are treated with bromine in the presence of iron tribromide.
a. Bromobenzene
b. ortho-Xylene
c. Benzene sulfonic acid
d. Benzaldehyde
e. meta-Nitrotoluene
f, para-Dibromobenzene
g. Nitrobenzene tert-Butylbenzene
h. Benzoic acid
i. Dibromobenzene
Answer:
The halogens are the ortho and para directing groups. Whenever they react with other benzene compounds they will attach to the ortho or para positions of the benzene ring.
Major products which are obtained by reacting these given compounds are given in attached pictures with complete reactions.
HBr will always be the side product of the bromine reactions along with the major compound.
Explanation:
Which of the following reagents should be used to convert to Question 26 options: A) H2, Lindlar's catalyst B) Na, liquid NH3 C) H2 / Nickel D) H2SO4, H2O
Answer: The question is is not complete...here is the complete question.
Which of the following reagents should be used to convert 3-Hexyne to E-3-hexene
Option B.
Na, liquid NH3.
Explanation:
3-Hexyne to E-3-hexene can be converted with by using the reagent of Na, NH3 (birch reduction) and this can be done by hdrogenation of H2.
The reagents NaNH3 convert 3-Hexyne to E-3-hexene because it is a reducing agents that convert or has the ability to reduce alkynes to trans alkenes.
3 Hexyne is an alkynes and it is converted to E- 3 hexene Na and NH3.
Phosphorus pentafluoride, PF5, acts as a __________ during the formation of the anion PF−6. Select the correct answer below: A. Lewis acid B. Lewis base C. catalyst D. drying agent
Answer:
Lewis acid
Explanation:
In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.
If we look at the formation of PF6^-, the process is as follows;
PF5 + F^- -----> PF6^-
We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.
Hence in the formation of PF6^-, PF5 acts a Lewis acid.
The ionization constant of lactic acid ch3ch(oh) co2h am acid found in the blood after strenuous exercise is 1.36×10^-4 If 20.0g of latic acid is used to make a solution with a volume of 1.00l what is the concentration of hydronium ion in the solution
Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
The concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter .
Ionization of lactic acid can be represented as:
CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻ + H⁺
Given:
ionization constant = 1.36 x 10⁻⁴
mass= 20.0 g
Now, Molecular weight of lactic acid = 90 g
[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]
It is dissolved in 1.00L so molar concentration of lactic acid formed will be
C = 0.22M
Consider "n" to be the fraction of moles ionized
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionization constant Ka
[tex]K_a =\frac{nC*nC}{C-nC}[/tex]
[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )
On substituting the values we will get:
[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]
To find the concentration of hydronium ion in the solution,
[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]
So, concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per liter.
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What is it’s molecular formula for C5H4 if it’s molar mass is 128.17g/mol
✔ C5H4 has a molecular molar mass of :
M(C5H4) = 5 x M(C) + 4 x M(H)
M(C5H4) = 5 x 12 + 4 x 1 M(C5H4) = 60 + 4 M(C5H4) = 64 g/mol✔ The molecular mass of C5H4 is therefore 64 g/mol.
But, 128/64 = 2 This is double the molar mass of C5H4, this molecule has the formula 2C5H4.Answer:
C10H8
Explanation:
I clicked on that answer, and it is correct.
Which of the following is true regarding a voltaic (or galvanic) cell? a) It can only be used with hydrogen. b) It produces electrical current spontaneously. c) It consumes electrical current to drive a nonspontaneous chemical reaction.
Answer:
b) It produces electrical current spontaneously.
Explanation:
Cells capable of converting chemical energy to electrical energy and vice versa are termed Electrochemical cells. There are two types of electrochemical cells viz: Galvanic or Voltaic cells and Electrolytic cells. Voltaic cell is an elctrochemical cell capable of generating electrical energy from the chemical reaction occuring in it.
The voltaic cell uses spontaneous reduction-oxidation (redox) reactions to generate ions in a half cell that causes electric currents to flow. An half cell is a part of the galvanic cell where either oxidation or reduction reaction is taking place. Hence, the spontaneous production of electric currents is true about Voltaic/Galvanic cells.
if your acetic acid buret was still wet inside with deionized water when you filled it with acetic acid?
Answer:
The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.
Which Group is in the second column of the periodic table?
A. Noble gases
B. Halogens
C. Alkali metals
D. Alkaline earth metals
Answer:
Hey there!
That would be the alkaline earth metals.
Hope this helps :)
Answer: alkaline earth metals
Explanation:
If a 275 mL gas container had pressure of 732.6 mm Hg at -28°C and the gas was condensed into a liquid with a mass of 1.95 g, what is the molar mass of the gas?
Answer:
THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE
Explanation:
Using PV = nRT
n = Mass / molar mass
P = 732.6 mmHg = 1 atm = 760 mmHg
So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm
P = 0.964 atm
V = 275 mL = 275 *10 ^-3 L
R = 0.082 Latm/ mol K
T = -28 C = 273 - 28 K = 245 K
mass = 1.95 g
molar mass = unknown
Having known the other variables in the formula, the molar mass of the gas can be obtained.
PV = m R T/ molar mass
Molar mass = m RT / PV
Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3
Molar mass = 39.1755 / 265.1 *10^-3
Molar mass = 39.1755 / 0.2651
Molar mass = 147.78 g/mol
The molar mass of the gas is 147.78 g/mol
Daniel has a sample of pure copper.its mass is 89.6 grams (g),and its volume is 10 cubic centimeters (cm3). whats the destiny of the sample.
Answer:
8.96g \ cm3
Explanation:
(89.6\ 10) (g\ cm3) = 8.96g\cm3
Describe the process of scientific inquiry ?
Answer:
It usually consists of six steps: question, observation or investigation, hypothesis, experiment, analysis of data (reviewing what happened during the experiment), and conclusion. Scientific inquiry, on the other hand, is non-linear, which means it does not follow a consistent step-by-step process.
Explanation:
Hope it helps
A 40.80 gram sample of copper is heated in the presence of excess sulfur. A metal sulfide is formed with a mass of 51.09 g. Determine the empirical formula of the metal sulfide.
Answer:
Cu₂S
Explanation:
From the question,
Cu S
Mass: 40.80 g 51.09-40.80 = 10.29 g
Mole ratio: 40.80/63.5 10.29/32.1
0.64 : 0.32
Divide by the smallest,
0.64/0.32 : 0.32/0.32
2 : 1
Therefore,
Empirical formula = Cu₂S.
Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. 204 hours 568 hour 590 hours 291 hours 281 hours
Answer:
Use 62% - the equation is for the amount present at a given time. 0.62 = (1) e-kt -> ln(0.62)=-kt -> k = -ln(0.62)/t. I get k = .00117 hr-1 t(half) = 0.693/k = 590 hr.
HOPE THIS HELPS AND PLSSS MARK AS BRAINLIEST AND THNXX :)
The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hr. Thus, option C is correct.
What is half-life?Half-life is the time required by a substance to get reduced to half of its initial concentration. The half-life of the substance can be determined by the rate constant.
Given,
The initial quantity of substance (A₀) = 100
Remaining quantity (At) = 10 - 38 = 62
Time elapse (t) = 407 hours
The rate constant (k) is calculated as:
ln (At ÷ A₀) = - kt
ln (62 ÷ 100) ÷ 407 hour = - k
-0.47803580094 ÷ 407 = - k
k = 0.00117453513
Now, half-life from rate constant (k) is calculated as:
[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ k
[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ 0.00117453513
[tex]\rm t ^{\frac{1}{2}}[/tex] = 590 hours
Therefore, option C. 590 hours is the half-life of the substance.
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A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass of the gas is 0.0961 g/mol
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of [tex]H_{2}[/tex] effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]
where
[tex]R_{h}[/tex] = rate of effusion of hydrogen gas
[tex]R_{g}[/tex] = rate of effusion of unknown gas
[tex]M_{h}[/tex] = molar mass of H2 gas
[tex]M_{g}[/tex] = molar mass of unknown gas
substituting values, we have
[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587
[tex]\sqrt{M_{g} }[/tex] = 0.31
[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol
The molar mass of the unknown gas will be "0.0961 g/mol".
Given:
Effusion rate of unknown gas,
[tex]R_g = 11.1 \ min[/tex]Effusion rate of [tex]H_2[/tex],
[tex]R_h = 2.42 \ min[/tex]Molar mass of hydrogen,
[tex]M_h = 1\times 2[/tex][tex]= 2 \ g/m[/tex]
According to the Graham's law, we get
→ [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]
By substituting the values, we get
→ [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]
→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]
→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]
[tex]\sqrt{M_g} = 0.31[/tex]
[tex]M_g = 0.0961 \ g/mol[/tex]
Thus the above solution is right.
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There are parts of a standing wave that do not move at all. These parts are called ___________.
Answer:
They are called nodes.
Explanation:
Answer:
i guess this is the ans nodes
hope this helps
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If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?
The Average atomic mass of phosphorus is 29.9.
What is Average atomic mass ?The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.
Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.
An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.
The versions of an element with different neutrons have different masses and are called isotopes.
The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,
Average atomic mass of P = ∑(Isotope mass) (its abundance)
∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)
Abundance of isotope = % of the isotope / 100.
∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.
Hence , The Average atomic mass of phosphorus is 29.9.
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Name the following alkanes, please need answer for f,g,h?!
Answer:
f is =2,2-dimethyl butane
g is = 2,2-dimethyl propane
h is = 3,3-diethyl pentane
Explanation:
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Which of the following metals has a low melting point?
2 A. Rubidium
B. Potassium
C. Calcium
D. Sodium
Answer:
Rubidium
Explanation:
1L of bleach has a mass of 1,100 grams, 7.25% of the mass of bleach is NaClO, 1 mol of NaClO has a mass of 74.44 grams. What is the molarity (mol/L) of NaClO in the bleach? A.0.097 B.0.93 C. 1.07 D.79.75
Answer:
C. 1.07 M.
Explanation:
Hello,
In this case, we can define the molarity of the bleach as shown below:
[tex]M=\frac{moles_{NaClO}}{V_{solution}}[/tex]
In such a way, given the mass of bleach in a 1-L solution, we can compute the density:
[tex]\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL[/tex]
In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:
[tex]V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L[/tex]
Afterwards, using the by-mass percent of bleach we compute the mass:
[tex]m_{NaClO}=100g*0.0725=7.25g[/tex]
And the moles:
[tex]n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol[/tex]
Therefore, the molarity turns out:
[tex]M=\frac{0.097mol}{0.091L}\\ \\M=1.07M[/tex]
Thus, answer is C. 1.07 M.
Regards.