Answer:
8.0356 * 10^-5 moles of NaHCO3
Explanation:
Sulphuric acid = H2SO4
Sodium bicarbonate = NaHCO3
The reaction between both compounds is given by;
2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
In the reactin above;
2 mol of NaHCO3 neutralizes 1 mol of H2SO4
At stp, 1 mol occupies 22.4 L;
1 mol = 22.4 L = 22400 mL
x mol = 0.9 mL
x = 0.9 / 22400 = 4.0178 * 10^-5 moles of H2SO4
Since 2 mol = 1 mol from the equation;
x mol = 4.0178 * 10^-5
x mol = 2 * 4.0178 * 10^-5
x = 8.0356 * 10^-5 moles of NaHCO3
The following reaction is second order in [A] and the rate constant is 0.025 M-1s-1: The concentration of A was 0.65 M at 33 s. The initial concentration of A was ________ M.
Select one:
a. 0.24
b. 1.2 � 10-2
c. 0.27
d. 2.4
e. 1.4
Answer:
e. 1.4.
Explanation:
Hello,
In this case, for a second-order reaction, the integrated rate law is:
[tex]\frac{1}{[A]} =kt+\frac{1}{[A]_0}[/tex]
In such a way, for the given data, we compute the initial concentration as shown below in molar units (M):
[tex]\frac{1}{[A]_0} =\frac{1}{[A]}-kt=\frac{1}{0.65M} -\frac{0.025}{M*s}*33s \\\\\frac{1}{[A]_0} =\frac{0.713}{M}[/tex]
[tex][A]_0=\frac{1M}{0.713} =1.4M[/tex]
Therefore, answer is e. 1.4 M.
Regards.
The initial concentration of the reaction is 1.4 M.
Using the formula;
1/[A] = kt + 1/[A]o
Where;
[A] = concentration at time t = 0.65 M
t = time = 33 s
k = rate constant = 0.025 M-1s-1
[A]o = initial concentration = 0.025 M-1s-1
Hence;
1/ 0.65 = (0.025 × 33) + 1/[A]o
1/[A]o = (0.65)^-1 - (0.025 × 33)
1/[A]o = 1.54 - 0.83
[A]o = 1.4 M
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1. Which of the following is a physical property of hydrochloric acid?
Fizzes when it is dropped on limestone
b. Turns potassium yellow and chromate orange
is a good conductor of electricity
d. Causes corrosion of magnesium metal
Answer:
c: is a good conductor of electricity
Explanation:
One of the physical properties of hydrochloric acid is that it is a good conductor of electricity in an aqueous state.
The dissolution of hydrochloric acid in water causes a complete dissociation into [tex]H^+[/tex] and [tex]Cl^-[/tex]. The free-floating ions that result from the dissolution enable hydrochloric acid to be able to conduct electricity.
In aqueous solution, the attraction of the positive dipole of the hydrochloric acid to the negative dipole of water and that of the positive dipole of water to the negative dipole of the hydrochloric acid creates a dipole-dipole interaction that allows complete dissociation of the hydrochloric acid. Thus, electrolyte results.
Correct option: c
According to naming rules, the types of compound that use prefixes in their names are ________.
Answer:
Covalent compounds.
Explanation:
Hello,
In this case, when forming chemical bonds in order to form compounds, we say that if electrons are shared, covalent compounds are to be formed and they usually have subscripts that need prefixes to be named, for instance phosphorous pentachloride (PCl5), dichlorine heptoxide (Cl2O7), carbon tetrachloride (CCl4) and many others.
Regards.
The combustion of 0.295 kg of propane produces 712 g of carbon dioxide. What is the percent yield of carbon dioxide? ( Make sure to balance equation) C3H8 (g)+ 029) à co2 g H200 0
a. 124%
b. 41 .4%
c. 80.5%
d. 0.805 %
Answer:
Option C. 80.5%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
C3H8 + 5O2 —> 3CO2 + 4H2O
Next, we shall determine the mass of C3H8 that reacted and the mass of CO2 produced from the balanced equation.
This is illustrated below:
Molar mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44 g/mol
Mass of C3H8 from the balanced equation = 1 x 44 = 44 g
Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol
Mass of CO2 from the balanced equation = 3 x 44 = 132 g
From the balanced equation above,
44 g of C3H8 reacted to produce 132 g of CO2.
Next, we shall determine the theoretical yield of CO2.
This can be obtained as shown below:
From the balanced equation above,
44 g of C3H8 reacted to produce 132 g of CO2.
Therefore, 0.295 kg (i.e 295 g) will react to produce = (295 x 132)/44 = 885 g of CO2.
Therefore, the theoretical yield of CO2 is 885 g.
Finally, we shall determine the percentage yield of CO2 as follow:
Actual yield of CO2 = 712 g
Theoretical yield of CO2 = 885 g
Percentage yield of CO2 =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield of CO2 = 712/885 x 100
Percentage yield = 80.5%
Therefore, the percentage yield of CO2 is 80.5%.
A 200.0 gram piece of silver (Ag) is initially at 25°C. It is then heated by addition of 3.6 kilojoules of energy. What is the final temperature of the silver? The specific heat of silver is 0.240 J/(g·°C).
Answer:
Final temperature = 100°C
Explanation:
The general equation of a coffee cup calorimeter is:
Q = m×C×ΔT
Where Q is the heat added to the system (3.6kJ = 3600J), m is the mass of the substance (200.0g), C is specific heat (0.240J /g°C) and ΔT is change in temperature).
Computing the values:
3600J = 200.0g×0.240J/(g°C)×ΔT
75°C = ΔT
ΔT is the difference in temperature between final and initial temperature. As intital temperature was 25°C:
75°C = Final T - 25°C
Final temperature = 100°CThe following reaction proceeds at a rate such that 3 mole of A is consumed per minute. Given this, how many moles of C are produced per minute? 2A+2B→4C
Answer:
6 mol/min
Explanation:
2A+2B→4C
The relationship between the reactants and products of this equation is given by;
1/2 -d[A]/dt = 1/2 -d[B]/dt = 1/4 d[C]/dt
Our focus is on A and C
From the question;
d[A]/dt = 3mol/min
We have;
1/2 (3) = 1/4 d[C]/dt
d[C]/dt = 4/2 * 3 = 6 mol/min
Four research teams measured the length of a rare whale, and what each team wrote in its team notebook is shown in the table below.
Suppose a later and more reliable measurement gives 45.0 m for the length of the same whale. Decide which of the earlier measurements was the most
accurate, and which was the most precise.
team
what was written
in the notebook
most accurate
measurement
most precise
measurement
?
A
"55.m"
B
*50.0m +0.5m
с
*44.0m + 6.0%
D
o
"between 30 m and 40 m
Answer:
Explanation:
Precision relates to instruments of measurement or scale . A scale measuring upto cm is less precise than an instrument measuring upto mm .
Here the first measurement appears to have been taken with an instrument measuring up to metre and the second measurement appears to have been taken with an instrument measuring up to decimeter . So second measurement is most precise .
Third measurement is closest to the actual measurement that is 45.0 m , so it is most accurate measurement .
Separate this redox reaction into its balanced component half‑reactions. Use the symbol e− for an electron. Cl2+2Li⟶2LiCl
Answer:
Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation
2Li + 2e⁻ ⟶ 2Li⁺ Reduction
Explanation:
The question requests to split the equation below into half equations;
Cl2+2Li⟶2LiCl
In redox chemistry, splitting into half equations simply means highlighting the reduction and oxidation reactions of the reaction.
Before proceeding, we hav to split the ionic compound; LiCl into it's component ions. So we have;
Cl₂ + 2Li ⟶ 2Li⁺Cl⁻
This leaves us with;
Cl₂ ⟶ 2Cl⁻ ............................... i
2Li ⟶ 2Li⁺ .............................. ii
Oxidation reactions can be identified by the increase in oxidation number and decrease in the case of reduction.
in reaction i, there is a decrease in oxidation number from 0 to -1. This is the reduction half equation,
in reaction ii, there is an increase in oxidation number from 0 to +1. This is the oxidation half equation
In terms of electrons, we have to even the charge;
Oxidation = Loss of electrons
Reduction = Gain of electrons
The half equations are given as;
Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation
2Li + 2e⁻ ⟶ 2Li⁺ Reduction
Separation of the redox reaction into its balanced component half-reactions is as follows;
Oxidation half-reaction;
2Li ⟶ Li²+ + 2e-Reduction half-reaction;
Cl2 + 2e- ⟶ 2Cl-By definition;
Oxidation is simply characterized by an increase in oxidation number of a reacting entity.
Reduction is simply characterized by a decrease in oxidation number of a reacting entity.
A reaction in which oxidation and reduction occur simultaneously is termed a Redox reaction as in the given reaction.
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In this simulation what do you think the bread and cheese separately represent, the atom or the molecule?
Answer:
the atom
Explanation:
The bread and cheese separately represent the atom because both bread and cheese are different from one another and we cannot assume it as a molecule because molecules formed when the group of atoms combine together by making bonds with each other and we know that bread and cheese did not make bonds with each other, they only attached, so we called them atoms not molecule.
choose the best answer. The 6.0 M HCl solution must be added to the crucible in the fume hood because
a) the wet hydrogen gas released can cause explosion.
b) the HCl used is very concentrated.
c) it is aafer to do the experiment in the fume hood rather on the bench.
d) the HCl can react quickly with the zinc solid in the fume hood.
Answer:
The answer is A
Explanation:
When nH2/nO2 =3/4 => it can cause explosion
7. A pair of shared valence electrons is referred to as a(n): hydrophobic effect. van der Waals interaction. ionic bond. hydrogen bond. covalent bond.
Answer:
I believe the answer is a covalent bond
when bisecting an angle, one of the steps is to draw an arc centered on a point on a ray of the angle. in particular this angle is to be drawn on the interior of the angle, why must The Arc be drawn in the interior of the angle?
Answer:
the answer is c
Explanation:
Tells how many of each atom there are in a formula
Answer:
[tex]\large \boxed{\mathrm{subscripts}}[/tex]
Explanation:
Subscripts are numbers written after elements, subscripts indicate how many of each atom there are in a compound.
The atoms of each element in an compund is written in the subscript. The number of atoms present in any compound whether made up of only same element or different element is termed as Atomicity.
☃️ For Example:Given compound = [tex]\boxed{\sf{H_2O}}[/tex]
Here, there are two atoms of hydrogen and one atom of oxygen, So ratio of the atoms of elements in Water molecule = 2 : 1
━━━━━━━━━━━━━━━━━━━━
For each of the reactions, calculate the mass (in grams) of the product formed when 15.93 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.
1) 2K(s)+Cl2(g)/15.93G→2KCl(s)
2) 2K(s)+Br2(l)/15.93→2KBr(s)
3) 4Cr(s)+3O2(g)/15.93→2Cr2O3(s)
4) 2Sr(s)/15.93+O2(g)→2SrO(s)
Answer:
1. 33.43 g of KCl
2. 23.70 g of KBr
3. 50.45 g of Cr₂O₃
4. 18.82 g of SrO
Explanation:
Molar mass of the elements and compounds in each of the reactions:
K = 39.0 g, Cl = 35.5 g, KCl = 74.5 g, Br = 80.0 g, KBr = 119.0 g, Cr = 52.0 g, O = 16.0 g, Cr₂O₃ = 152.0 g, Sr = 88.0 g, SrO = 104.0 g
1) 2K(s)+Cl2(g)/15.93G→2KCl(s)
From the mole ratio of the reaction above, 2 moles of K reacts with 1 mole of Cl₂ to give 2 moles of KCl
78.0 g (2 * 39.0 g) of K reacts with 71.0 g (2*35.5) of Cl₂ to produce 149.0 g(2*74.5) of KCl, therefore, Cl₂ is the limiting reactant.
15.93 g of Cl₂ will react to produce (149/71) * 15.93 of KCl = 33.43 g of KCl
2) 2K(s)+Br2(l)/15.93→2KBr(s)
From the mole ratio of the reaction, 2 moles of K reacts with 1 mole of Br₂ to give 2 moles of KBr
78.0 g (2 * 39.0 g) of K reacts with 160.0 g (2*80) of Br₂ to produce 238.0 g(2*119.0) of KBr, therefore, K is the limiting reactant which though is in excess.
15.93 g of Br₂ will react to produce (238/160) * 15.93 of KBr = 23.70 g of KBr
3) 4Cr(s)+3O2(g)/15.93→2Cr2O3(s)
From the mole ratio of the reaction, 4 moles of Cr reacts with 3 moles of O₂ to give 2 moles of Cr₂O₃
208.0 g (4 * 52.0 g) of Cr reacts with 96.0 g (3*2*16) of O₂ to produce 304.0 g (2*152.0) of Cr₂O₃, therefore, O₂ is the limiting reactant.
15.93 g of O₂ will react to produce (304/96) * 15.93 of Cr₂O₃ = 50.45 g of Cr₂O₃
4) 2Sr(s)/15.93+O2(g)→2SrO(s)
From the mole ratio of the reaction, 2 moles of Sr reacts with 1 mole of O₂ to give 2 moles of SrO
176.0 g (2* 88.0 g) of Sr reacts with 32.0 g (2*16) of O₂ to produce 208.0 g (2*104.0) of SrO, therefore, O₂ is the limiting reactant which though is in excess.
15.93 g of Sr will react to produce (208/176) * 15.93 of SrO = 18.82 g of SrO
If the components of a solution are in different states, which one is a solvent?
Answer:
The solvent is the substance which typically determines the physical state of the solution (solid, liquid or gas). The solute is the substance which is dissolved by the solvent.
How many moles of urea (60. g/mol) must be dissolved in 66.6g of water to give a 2.4 m solution?
a. 1.4 × 102 mol
b. 2.4 mol
c. 0.0024 mol
d. 0.15 mol
e. 9.0 × 102 mol
Answer:
d. 0.15 mol
Explanation:
Step 1: Given data
Mass of water (solvent): 66.6 gMolality (m): 2.4 mMoles of urea (solute): ?Step 2: Convert the mass of the solvent to kilograms
We will use the relationship 1 kg = 1,000 g.
66.6 g × (1 kg/1,000 g) = 0.0666 kg
Step 3: Calculate the moles of solute
The molality is equal to the moles of solute divided by the kilograms of solvent.
m = moles of solute / kilograms of solvent
moles of solute = m × kilograms of solvent
moles of solute = 2.4 mol/kg × 0.0666 kg
moles of solute = 0.15 mol
Calculate the mass of MgCO3 (84.31 g/mol) precipitated by mixing 10.0 mL of a 0.300 M Na2CO3 solution with 6.00 mL of 0.0400 M Mg(NO3)2 solution.
Answer:
[tex]m_{MgCO_3}=0.0202molMgCO_3[/tex]
Explanation:
Hello,
In this case, for this purpose we first have to write the undergoing chemical reaction:
[tex]Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3[/tex]
Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:
[tex]n_{Na_2CO_3}=0.010L*0.300\frac{molNa_2CO_3}{1L}=0.003molNa_2CO_3 \\\\n_{Na_2CO_3}^{consumed}=0.006L*0.0400\frac{molMg(NO_3)_2}{1L}*\frac{1molNa_2CO_3}{1molMg(NO_3)_2} =0.00024molNa_2CO_3[/tex]
In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:
[tex]m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3} \\\\m_{MgCO_3}=0.0202molMgCO_3[/tex]
Regards.
A neutralization of a strong acid by a strong base give out 4590 J of heat energy. Determine the enthalpy change (in kJ) for the reaction. Give your answer to three significant figures.
Answer:
A neutralization of a strong acid by a strong base gave out 5590 J of heat energy.
Determine the enthalpy change (in kJ) for the reaction.
Explanation:
A strong acid being neutralized by a strong base releases 5590 J of heat energy.
What is neutralization?A chemical reaction in which an acid and a base interact quantitatively is known as neutralization or neutralisation. By neutralizing a reaction in water, surplus hydrogen or hydroxide ions are removed from the solution.
Neutralization is a sort of chemical reaction in which an acid and a base combine to produce salt and water. For instance: Reaction of sodium hydroxide and hydrochloric acid (HCl) Sodium hydroxide reacts with hydrochloric acid to form their salt sodium chloride (NaCl) and water.
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What would be the major product obtained from hydroboration–oxidation of the following alkenes?
a. 2-methyl-2-butene
b. 1-methylcyclohexene
Answer:
a. 3-methylbutan-2-ol
b. 2-methylcyclohexan-1-ol
Explanation:
For this reaction, we must remember that the hydroboration is an "anti-Markovnikov" reaction. This means that the "OH" will be added at the least substituted carbon of the double bond.
In the case of 2-methyl-2-butene, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore the "OH" will be added to carbon three producing 3-methylbutan-2-ol.
For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore the "OH" will be added to carbon 2 producing 2-methylcyclohexan-1-ol.
See figure 1
I hope it helps!
What is the molarity of a solution in
which 6.9mol of potassium chloride
(KCI) is dissolved in water to a final
volume of 1.8L?
Answer:
3.83 mol/ L
Explanation:
Molarity is the amount of moles of a substance in a liter of solution.
1.8L ----- 6.9 mol of KCl
1L ----- 6.9 ÷1.8= 3.83 moles of KCl (3 s.f.)
Thus the molarity is 3.83 mol/L.
Answer: 3.8
Explanation:
Original answer is 3.83 but it needs significant figures so you need to round it to 2 significant figures
If 20.6 grams of ice at zero degrees Celsius completely change into liquid water at zero degrees Celsius, the enthalpy of phase change will be positive. TRUE FALSE
Answer:
True
Explanation:
The radioactive atom R 88 210 a is an alpha emitter. What nucleus does it produce?
Answer:
X 86 206
Explanation:
Radioactive atoms are nuclei that can under go disintegration to emit either an alpha particle, beta particle or gamma radiation. The process could be spontaneous or stimulated.
When a radioactive atom R 88 210 emits alpha particle, it would produce an element with atomic number 86 and mass number 206 i.e X 86 206. An alpha particle is usually a helium nucleus.
[tex]R^{210}[/tex] ⇒ [tex]x^{206}[/tex] + [tex]He^{4}[/tex] + energy
Please use your periodic table of elements to assist you. How many protons does Flourine have? Question 3 options: 8 18 9 10
Answer:
9 protons
Explanation:
An element's atomic number is equal to the amount of protons in the nucleus - otherwise, the element's chemical identity would be different and it would be a completely different element.
Referencing any periodic table, we notice that fluorine has the atomic number of 9.
Because the atomic number is the same as the number of protons in the nucleus, fluorine has 9 protons.
We study science to be able to answer questions about the world around us. Write down at least three questions
that you think might be answered by studying Earth and space science.
Answer:
1) If life exists in other planets
2) What type of living organisms exists on other planets
3) Predominant atoms/mixture/compound that exists on other planets
Explanation:
Earth and space science is the study of the earth, the space and other planets within the universe.
The following questions can be answered by studying earth and space science
1) If life exists in other planets
Over the years, there has been several reports suggesting this but a deep inquiry into this with scientific data can assist in putting the matter to rest.
2) What type of living organisms exists on other planets
If there is/are lives on other planets, what type of life/living organisms exist there. Are they plants? Are they animals? Or do we have a new classification for them?
3) Predominant atom/mixture/compound that exists on other planets
What is the most popular substance in each of these planets? The role of this substance in the survival of the planet or it's inhabitants also needs to be studied
what cells form enamel?
Answer:enamel is formed by differentiated dental epithelial cells known as ameloblasts.
Explanation:
422000 scientific notation form:
With all the complexity in the world or the universe , there are ____ . Kind of like leggos
3. Calculate the amount of C2H5OH in moles present in 5.0 g of ethanol. Calculate the number of C atoms present in the same amount of glucose. g
Answer:
- [tex]n_{etOH}=0.11mol\ et OH[/tex]
- [tex]atoms_C=1.0x10^{23} atoms\ C[/tex]
Explanation:
Hello,
In this case, since ethanol has a molar mass of 46 g/mol, the moles in 5.0 g are:
[tex]n_{etOH}=5.0g\ etOH*\frac{1mol\ etOH}{46 g\ etOH} =0.11mol\ et OH[/tex]
Moreover, since the formula of glucose is C₆H₁₂O₆, its molar mass is 180 g/mol and six moles of carbon are in one mole of glucose (based on carbon's subscript), the atoms are computed by using the 6:1 mole ratio and the Avogadro's number as shown below:
[tex]atoms_C=5.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molC}{1molC_6H_{12}O_6} *\frac{6.022x10^{23}atoms\ C}{1molC} \\\\atoms_C=1.0x10^{23} atoms\ C[/tex]
Regards.
What nuclide, when bombarded by an α particle, will generate carbon-12 and a neutron?
Answer:
Be
Explanation:
[tex]^94Be + ^42He \rightarrow ^{12}6C + ^10x[/tex]
Therefore, the element symbols is Be, which when bombarded by an α particle, will generate carbon-12 and a neutron.
Use the Debye-Huckel Equation to calculate the activity coefficient of Ce4+ at μ = 0.070 M.
english please so i can help