Answer:
False
Explanation:
An object with a density less than the density of water will float.
And an object that is denser than water (more than 1.00g/mL) will sink when placed in water.
Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
When determining the amount of oxidant present by titration, you can use iodine/starch as an indicator. First, the oxidant, like hypochlorite, oxidizes Choose... When starch and iodine are both present, the solution is Choose... During the titration, a titrant like thiosulfate reduces the
The question is incomplete, the complete question is;
When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.
First, the oxidant, like hypochlorite, oxidizes
Choose...
neutral iodine into iodide ion
iodide ion into neutral iodine
iodate polyatomic ion into iodide ion
When starch and iodine are both present, the solution is
Choose...
blue-black
brownish yellow
clear
During the titration, the titrant, like thiosulfate, reduces the
Choose...
iodide ion into iodate polyatomic ion
neutral iodine into iodide ion
iodide ion into neutral iodine
When the iodine has completely reacted at the endpoint of the titration, the solution should become
Choose...
clear
blue-black
brownish yellow
Answer:
1. iodide ion into neutral iodine
2. blue-black
3. neutral iodine into iodide ion
4. clear
Explanation:
Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;
ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)
When iodine is added, the colour of the starch solution immediately changes to blue-black.
A reduction reaction occurs when the titrant, thiosulfate is added as follows;
I2 + 2S2O32- → 2I- + S4O62-
The solution at end point is found to become clear again.
How long do spent fuel rods remain dangerously radioactive?
Answers
A.
The rods are no longer radioactive because the radioisotopes are used up.
B.
Spent fuel rods remain radioactive for several years after the fuel is exhausted.
C.
It takes tens of thousands of years for the radioisotopes in the rods to decay to safe levels.
D.
It is impossible to determine how long it will take for the radioisotopes to decay because they last too long.
Answer:
c
Explanation:
it takes 10,000 years to just reduce down the decay
Which redox reaction would most likely occur if zinc and copper metal were
added to a solution that contained zinc and copper ions?
Click for a reduction potential chart
A. Cu + Zn → Cu2+ + Zn2+
B. Cu + Zn2+
Cu2+ + Zn
C. Cu2+ + Zn → Cu + Zn2+
D. Cu2+ + Zn2+ → Cu + Zn
Answer:
C
Explanation:
b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex] is the redox reaction. Hence, option C is correct.
What is Redox Reaction?A chemical reaction taking place between an oxidizing substance and a reducing substance.
The oxidizing substance is used to lose electrons in the reaction, and the reducing substance is used to gain electrons.
On the reduction potential chart, zinc is a stronger oxidizing agent than, Copper (Cu), which is a reducing agent as compared to silver
The redox reaction most likely occurs if silver and copper metal were added to a solution that contained silver and copper ions is ;
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex]
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The structure of the compound CuI is best described as a cubic closest packed array of iodide ions with the copper ions in tetrahedral holes. What percent of the tetrahedral holes are occupied in this solid
Answer:
12.5 %
Explanation:
In CCP, the effective number of anion is 4
That is there are 4 I- present in 1 unit cell
Number of tetrahedral void = 2*effective number of anion
= 2*4
= 8
In ZrI4, for every 4 anion, there are only 1 Zr atom.
So, one tetrahedral void is occupied per unit cell out of 8
% tetrahedral void occupied = 1*100/8
= 12.5 %
Answer: 12.5 %
According to Arrhenius, NH4+ an acid or a base? Write an equation to support
According to Arrhenius definition of acids, [tex]NH4^+[/tex] is an acid.
According to Arrhenius definition of acids and bases, acid is any substance that produces hydrogen ion in solution as its only positive ion.
Following this definition, let us now consider what happens when [tex]NH4^+[/tex] is introduced into a water;
[tex]NH4^+[/tex](aq)-------> NH3(aq) + [tex]H^+[/tex](aq)
Hence, according to Arrhenius definition of acids, [tex]NH4^+[/tex] is an acid.
For another example of Arrhenius acids, see
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A chemical equation is shown below.
FeC13 + NaOH -> NaCl + Fe(OH)3
What are the coefficients that should be added to balance this question?
Explain how this chemical reaction demonstrates the conservation of mass.
Answer:
Explanation:
Balance the OHs first. There's 3 on the right.
FeC13 + 3NaOH -> NaCl + Fe(OH)3
Now balance the Nas. There are 3 on the left.
FeC13 + 3NaOH -> 3NaCl + Fe(OH)3
The equation is now balanced
Chemical Left Right
Fe 1 1
Na 3 3
Cl 3 3
OH 3 3
The standard entropy change of a reaction has a positive value. This reaction results in: Select the correct answer below: a decrease in entropy. an increase in entropy. no entropy change. neither an entropy increase nor decrease.
Explanation:
The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.
Positive entropy means the system has increased its degree of disorderness.
A student was given a solid containing a mixture of nitrate salts. The sample completely dissolved in water, and upon addition of dilute HCl , no precipitate formed. The pH was lowered to about 1 and H2S was bubbled through the solution. No precipitate formed. The pH was adjusted to 8 and H2S was again bubbled in. This time, a precipitate formed. Which compounds might have been present in the unknown?
a. Ca(NO3)2
b. AgNO3
c. Fe(NO3)3
d. Cr(NO3)3
e. Cu(NO3)2
f. KNO3
g. Bi(NO3)2
Answer:
Fe(NO3)3, Cr(NO3)3, Co(NO3)3
Explanation:
According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.
Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.
The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.
The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.
Please comment the chart!!!!!
thanks!!
Answer:
Exothermihic chart
Explanation:
A tree is an example
of a vascular plant that
is
because it
has deep roots.
A. tall
B. tiny
C. small
Dyshort
There are _______ alkanes with molecular formula C10H22
a. 74
b. 75
c. 76
d. 77
Which species is the conjugate base of H2SO3
Explanation:
As you know, the conjugate base of an acid is determined by looking at the compound that's left behind after the acid donates one of its acidic hydrogen atoms.
The compound to which the acid donates a proton acts as a base. The conjugate base of the acid will be the compound that reforms the acid by accepting a proton.
In this case, sulfurous acid has two protons to donate. However, the conjugate base of sulfurous acid will be the compound left behind after the first hydrogen ion is donated.
Titanium is a metal often used as an alloying agent to provide materials that are strong, lightweight, and temperature-resistant Which of the following represents the correct ground-state configuration for a neutral atom of titanium?
A) 1s 2s 2p 3s 3p 48°30°
B) 1s 2s 2p 3s 3p 4s3d
C) 15*2s2p 3s 3p 4s
D) 15°2s 2p 3s 3p 3d
Answer:B) 1s 2s 2p 3s 3p 4s 3d
Explanation:
The ground state electron configuration shows how the electrons in the atomic orbitals of an atom are in their lowest , most stable energy arrangements and since Electrons must be filled following the Aufbau's principle(electrons fill lowest energy shells first)
Now, Titanium lies in period IV and group 4 of the periodic table with 22 as its atomic number
Thus, the ground-state electron configuration of a neutral atom of titanium is 1s²2s²2p⁶3s²3p⁶4s²3d².
Why does the temperature stop rising while ice melts into water?
A. The temperature does not stop rising.
B. The electrons are increasing in energy levels.
C. Because no more heat is being added to the system.
D. The energy is being absorbed to separate the particles.
Answer:
When you heat ice, its temperature rises, but as soon as the ice starts to melt, the temperature stays constant until all the ice has melted. This happens because all the heat energy goes into breaking the bonds of the ice's crystal lattice structure.
Explanation:
The temperature stop rising while ice melts into water because, the energy is being absorbed to separate the particles. This is because of latent heat of fusion.
What is latent heat of fusion?The amount of energy needed to convert the solid substance into a liquid substance by modifying its physical effects. It exists also named enthalpy of fusion. When heat exists supplied to ice, it begins melting and heat is used to increase temperature initially. But after the occasional temperature of ice does not vary and the extra heat exists utilized to melt the ice by cracking bonds between crystal lattice of ice.
The temperature stops increasing while the ice melts into the water because the energy exists being absorbed to divide the particles. This exists because of the latent heat of fusion.
Therefore, (D) option is the correct answer.
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write the chemistry of Epsom salt
Use the reaction: 2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2HNO3(aq) What volume (mL) of 0.568 M AgNO3(aq) is needed to form 0.21 g of Ag2SO4(s)
Answer:
The mole ratio of AgNO3 to Ag2SO4 IS 2:1 .0.657 g Ag2SO4 x 1 mol / 312 g = 0.00211 mol Ag2SO4.
0.00211 mol Ag2SO4 x 2 mol AgNO3 / 1 mol Ag2SO4 = 0.00421 mol AgNO3
0.00421 mol AgNO3 x 1 L / 0.123 mol AgNO3 = 0.0342 L = 34.2 mL of AgNO3 solution.Therefore,34.2ml of 0.123M AgNO3 will be required.
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
how many moles of KF are present in 46.5 grams of KF
Explanation:
here's the answer to your question
Answer:
0.8017
Explanation:
Find the molar Mass of KF
K = 39
F = 19
Total = 58
Note: these numbers are approximate. Use your periodic table to get the exact numbers.
mols = given mass / molar mass
given mass = 46.5
molar mass = 58
mols = 46.5 / 58
mols = 0.8017
What is the main reason for using a data table to collect data?
A. To interpret the possible meaning of the data
B. To find the possible errors that were made in recording the data
C. To organize the information so that it is easier to understand
O
D. To make an experimental journal more attractive
Answer:
c
Explanation:
table of data help us to understand and present our work better
I did it and got it right, it's c
What is the percentage by mass of carbon in CH3(CH2)5COOH?
A. 48.6%
B. 9.2%
C. 55.4%
D. 64.6%
Answer:
F 64.6 percent of carbon may be
What is the molecule shown below?
A. Pentane
B. Trimethylethane
C. 2,2-dimethylpropane
D. 3-dipropane
Q2
Answer:
C
Explanation:
if we were to followw the IUPAC
LION
If 3.0L of helium at 20°C is allowed to expand to 4.4L, with pressure remain the same
Answer:
This question is asking to find the new temperature
The answer for the final temperature is 429.73K
Explanation:
Using Charles law equation as follows:
V1/T1 = V2/T2
Where;
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question;
V1 = 3.0L
V2 = 4.4L
T1 = 20°C = 20 +273 = 293K
T2 = ?
Using V1/T1 = V2/T2
3/293 = 4.4/T2
Cross multiply
293 × 4.4 = 3 × T2
1289.2 = 3T2
T2 = 1289.2 ÷ 3
T2 = 429.73K
Calculating the expected pH of the buffer solution: Given that the pKa for Acetic Acid is 4.77, calculate the expected pH of the buffer solutions using the Henderson-Hasselbalch equation and the concentrations of Acetic Acid and Acetate added to the 250 ml Erlenmeyer flask: pH
Answer:
[tex]pH=4.77[/tex]
Explanation:
From the question we are told that:
pKa for Acetic Acid [tex]pK_a= 4.77[/tex]
Therefore
For Equal Concentration of acetic acid and acetatic ion
[tex]CH_3COOH=CH_3COO^-[/tex]
Generally the Henderson's equation for pH value is mathematically given by
[tex]pH=pK_a+log\frac{base}{acid}[/tex]
[tex]pH=4.77+log\frac{CH_3COO^-}{CH_3COOH}[/tex]
[tex]pH=4.77+log1[/tex]
[tex]pH=4.77[/tex]
Chlorine radicals perform the first propagation step:
a. in comparison to bromine radicals.
b. radicals form easily in the presence of chlorine radicals.
c. Subsequently, the resulting radicals can react with bromine in a second propagation step to yield monobrominated products.
Answer:
b. radicals form easily in the presence of chlorine radicals.
Explanation:
Chlorine radicals perform the first propagation step: because "radicals form easily in the presence of chlorine radicals."
This is because the first propagation step consumes a CHLORINE RADICAL while the second propagation step regenerates a CHLORINE RADICAL. In this way, a chain reaction occurs, whereby one CHLORINE RADICAL can ultimately cause thousands of molecules of methane to be converted into chloromethane with C12 present.
Hence, in this case, the correct answer is that "radicals form easily in the presence of chlorine radicals."
What separates the inner planets from the outer planets?
a. Main asteroid belt
b. Main comet belt
c. Kuiper belt
d. Outer planet belt
please help this is for SCIENCE test I need help
Answer:
main asteroid belt separates the inner planets from the outer planets
Which of the following metals will liberate hydrogen from dilute HCL? A. Ag B.Au C.Hg D.Sn
Answer:
ag and au are sure not to react. but hg and sn might or might not
The metal sample suspected of being aluminum is warmed and then submerged into water, which is near room temperature. The final temperature of the water and the metal is given below. The specific heat capacity of water is 4.18 J/g.oC. Calculate the specific heat capacity of the metal based on the data below. Remember heat lost = heat gained.
Type of metal used:
Trial 1 Trial 2 Trial 3
Mass of metal, g 2.746 g 2.750 g 2.900 g
Mass of water, g 15.200 g 15.206 g 15.201 g
Initial Temp. of Water, oC 24.7 oC 24.6 oC 24.5 oC
Initial Temp. of Metal, oC 72.1 oC 72.2 oC 71.9 oC
Final Temp of Water & Metal,oC 26.3 oC 26.2 oC 24.7 oC
ΔT for water, oC ______ ______ ______
ΔT for metal, oC ______ ______ ______
Specific heat capacity of metal, J/g.oC ______ ______ ______
Average specific heat capacity, J/g .oC ______ (use two significant figures due to ΔT of water)
Answer:
Average specific heat capacity of metal = 0.57 J/g°C
Explanation:
Heat lost = Heat gained
Heat energy gained or lost, H = mcΔT
where m = mass of substance, c = specific heat capacity, ΔT = temperature change
Trial 1:
Heat lost by metal = -[2.746 g × c × ΔT]
ΔT = (26.3 - 72.1) °C = -45.8 °C
Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C
Heat gained by water = 15.200 × 4.18 × ΔT
ΔT = (26.3 - 24.7) = 1.6 °C
Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J
From Heat lost = Heat gained
c × (125.7688)g°C = 101.6576 J
c = 101.6576 J / 125.7688 g°C
c = 0.8083 J/g°C
Trial 2:
Heat lost by metal = -[2.750 g × c × ΔT]
ΔT = (26.2 - 72.2)°C] = - 46 °C
Heat lost by metal = -[2.750 g × c × (-46 °C)
Heat lost by metal = c × (126.5) g°C
Heat gained by water = 15.206 × 4.18 × ΔT
ΔT = (26.2 - 24.6) = 1.6 °C
Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J
From Heat lost = Heat gained
c × (126.5)g°C = 101.6977 J
c = 101.697728 J / 126.5 g°C
c = 0.8039 J/g°C
Trial 3:
Heat lost by metal = -[2.900 g × c × ΔT]
ΔT = (24.7 - 71.9)°C] = - 47.2 °C
Heat lost by metal = -[2.900 g × c × (- 47.2 °C)
Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C
Heat gained by water = 15.201 × 4.18 × ΔT
ΔT = (24.7 - 24.5) = 0.2 °C
Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J
From Heat lost = Heat gained
c × (136.88)g°C = 12.708036 J
c = 12.708036 J / 136.88 g°C
c = 0.0928 J/g°C
Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3
Average specific heat capacity of metal = 0.57 J/g°C
A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at .Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to significant digits.
Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:
[tex]M=58g[/tex]
Explanation:
From the question we are told that:
Heat Capacity [tex]H=0.897[/tex]
Mass of water [tex]M=200g[/tex]
Initial Temperature of Aluminium [tex]T_a=85.6[/tex]
Initial Temperature of Water [tex]T_{w1}=16.0[/tex]
Final Temperature of Water [tex]T_{w2}=16.0[/tex]
Generally
Heat loss=Heat Gain
Therefore
[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]
[tex]M=58g[/tex]
