Solution :
We know that :
[tex]$\Delta T_f = k_f.m$[/tex] and [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]
Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex] ..................(1)
Where,
[tex]w_1[/tex] = amount of solvent (in kg)
[tex]w_2[/tex] = amount of solute (in kg)
[tex]m_2[/tex] = molar mass of solute (g/mole)
[tex]m[/tex] = molality of solution (mole/kg)
Given :
[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex], [tex]k_f= 5.12\ ^\circ C/m[/tex]
[tex]=5.12 \ ^\circ C/mole/kg[/tex]
[tex]=5.12 \ ^\circ C \ kg/mole[/tex]
[tex]w_1[/tex] = 0.250 kg, [tex]w_2[/tex] = 24.3 g
Then putting this values in the equation is (1),
[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]
[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]
[tex]m_2= 158.49[/tex] g/mole
So, the molar mass of the unknown compound is 158.49 g/mole.
Avogradro's number is the number of particles in one gram of carbon- 12 atom true or false?
Answer:
True
Explanation:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
What is represented by a straight line on a graph?
o the sum of the independent and dependent variables
O only the independent variable
O only the dependent variable
o the relationship between independent and dependent variable
1 2
3
4
5
Answer:
the relationship between independent and dependent variable
Explanation:
A straight line or linear graph is one of the ways to represent a given data. It shows the relationship between two given set of data; one called the independent variable is plotted on the x-axis (horizontal) while the other called the dependent variable is plotted on the y-axis (vertical).
The straighter the line is, the stronger the relationship between the two variables and vice versa. Hence, the straight line in the graph represents the relationship between independent and dependent variable.
Methane (CH4) is the major component of natural gas. 40.0 grams of methane were placed in a commercial calorimeter and subjected to a combustion reaction. The reaction released 2800 kJ of energy.
1. Compare this energy value to the energy values of paraffin and isopropanol. Is methane a good choice as a fuel?
Based on comparison of energy produced per kilogram, a given mass of methane produces more energy than similar masses of either paraffin or isopropanol, therefore;
Methane is a good choice as a fuel
The reason for the above comparison conclusion is as follows:
The given information:
The details of the combustion of the methane gas, CH₄, are as follows;
The mass of the methane gas placed in the calorimeter, m = 40.0 g
The amount of heat released from the combustion of the 40.0 grams of methane = 2,800 kJ
The data from online resources of paraffin and isopropanol includes
1. The energy value of paraffin = 46 MJ/kg
The energy value of isopropanol = 33.6 MJ/kg
The energy produced from 1 kilogram of methane gas is given as follows;
40.0 g of methane gas produces 2,800 kJ of energy, therefore;
1 kg = 1,000 g of methane gas will produce, 2,800kJ/(40.0 g) × 1,000 g = 70,000,000 J
Therefore;
1 kg of methane produces 70,000,000 J = 70 MJ of energy
Therefore, energy produced from methane = 70 MJ/kg
Given that methane produces more than twice the amount of energy that
is produced from similar mass of isopropanol and more than one and half
times the amount of energy that is produced from the same mass of
paraffin, methane is a good choice as a fuel for energy
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What is a reaction rate?
Answer:
A reaction is the time that is required for a chemical reaction to go essential to completion
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
A 1 liter solution contains 0.370 M hypochlorous acid and 0.493 M sodium hypochlorite. Addition of 0.092 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.
The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:
The addition of barium hydroxide will raise the pH slightly because the buffer still working.
The initial moles of those species are:
Hypochlorous acid:
[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]
Sodium hypochlorite:
[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]
Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:
Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O
For a complete reaction of 0.092 moles of barium hydroxide are required:
[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]
As there are 0.370 moles, the moles of HClO after the reaction are:
0.370 moles - 0.184 moles = 0.186 moles of HClO will remain
As you still have hypochlorite and hypochlorous acid you still have a buffer.
Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.
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Why do we need Chemistry in Nursing?
Answer:
We need chemistry in nursing because it deals with various kinds of drugs and the reactions of these drugs on the human body as well as with each other.
Which of the following is a reduction half-reaction?
Solution : An oxidation reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between tow species an oxidaion reductin reaction is any chemical reaction in which the oxidation number of a molecule atom or ion changes by gaining or losing an electron
A compound with a molecular weight of about 64.47 g/mol was found to be 18.63 % of C, 1.56 % of H, 24.82 % of O, and 54.99 % of Cl by mass. Determine the molecular formula and draw the Lewis structure showing an accurate 3-D perspective. *Show your calculations
Answer:
See detailed explanation.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly calculating the moles of each element, assuming those percentages are masses, so that we divide by their molar masses:
[tex]C=\frac{18.63}{12.01}=1.55\\\\H=\frac{1.56}{1.01} =1.55\\\\O=\frac{24.82}{16.00}=1.55\\\\Cl=\frac{54.99}{35.45}=1.55[/tex]
Then, we divide all of them by 1.55 to realize the empirical formula is:
[tex]CHOCl[/tex]
Whose molar mass is 64.47 g/mol, and therefore, since the molar mass of these two is the same, we infer the molecular formula is also CHOCl.
The Lewis structure is shown on the attached document, whereas, the central atom is C and it does complete its octet as well as both O and Cl.
Regards!
why does D2O
have a higher boiling point than H2O
Answer:
deuterium is heavier isotope of hydrogen need
high weight of atom will lead to higher boiling point
How is the compound NH3 classified?
A. As a salt
B. As a base
C. As an acid
D. As ionic
Answer:
B
Explanation:
Ammonia is considered a base as it's pH is 11
Answer from Gauthmath
The compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.
What is Ammonia (NH3)?Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.
Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.
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Balance the redox reaction Al(s) + MnO4^- (aq) --> MnO2 (s) + Al(OH)4^- (aq) in aqueous basic solution
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
1.Q= {n: 7 <n<31}, list the members of the set Q
Q={x:x[tex]\epsilon[/tex]n,7<n<31}
[tex]\\ \sf\longmapsto Q=\left\{8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30\right\}[/tex]
You can write it like this too
[tex]\\ \sf\longmapsto Q=\left\{8,9......30,31\right\}[/tex]
The elements present in a group of periodic table have
Similar chemical properties) Give reason and a
suitable element?
Answer:
group 1 elements(hydrogen,sodium,etc)
Explanation:
bexause if noticed all the element in the same group have the same eletron in thr outer most shell for example the group 1 elements are said to have 1 outermost elect ron which make them react so the same
When 1 mole of CO(g) reacts with H2O(l) to form CO2(g) and H2(g) according to the following equation, 2.80 kJ of energy are absorbed. CO(g) + H2O(l)CO2(g) + H2(g) Is this reaction endothermic or exothermic? _________ What is the value of q? kJ
State what would be observed when the following pairs of reagents are mixed in a test tube.
C6H2COOH and Na2CO3(aq)
(ii) CH3CH2CH2OH and KMnO4 /H
(iii) CH3CH2OH and CH3COOH + conc. H2SO4 (iv) CH3CH = CHCH3 and Br2 /H2O
Answer:
(i). C6H2COOH and Na2CO3(aq)
observation: Bubbles of a colourless gas (carbon dioxide gas)
(ii) CH3CH2CH2OH and KMnO4 /H
observation: The orange solution turns green.
[This is because oxidation of propanol to propanoic acid occurs]
(iii) CH3CH2OH and CH3COOH + conc. H2SO4
observation: A sweet fruity smell is formed.
[This is because an ester, diethylether is formed]
(iv) CH3CH = CHCH3 and Br2 /H2O
observation: a brown solution is formed.
A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to the gas by heat. kJ (b) Calculate the increase in its internal energy. kJ (c) Calculate the work done on the gas. kJ
Answer:
a) The energy transferred is 6.91 kJ
b) The internal energy is 4.90 kJ
c) The work done on the gas is - 2.01 kJ
Explanation:
Step 1: Data given
Number of moles of hydrogen gas = 2.00 moles
Pressure = constant
Temperature is heated from 294 K to 414 K
Molar heat capacity of hydrogen gas = 28.8 J/mol*K
Step 2: Calculate the energy transferred to the gas by heat.
Q = n* Cp * ΔT
⇒with Q =the energy transferred
⇒with n = the number of moles = 2.00 moles
⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K
⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
Q = 2.00 * 28.8 * 120
Q = 6912 J = 6.91 kJ
Step 3: Calculate the increase in its internal energy.
ΔEint = n*Cv*ΔT
⇒with ΔEint = the increase in its internal energy.
⇒with n = the number of moles = 2.00 moles
⇒with Cv = The constant volume = 20.4 J/mol*K
⇒with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K
ΔEint = 2.00 * 20.4 * 120
ΔEint =4896 J = 4.90 kJ
Step 4: Calculate the work done on the gas.
Work done on the gas = -Q + ΔEint
W = -6.91 kJ + 4.90 kJ
W = -2.01 kJ
Để xác định hàm lượng Cu trong hợp kim Cu-Zn người ta làm như sau: Hòa
tan hoàn toàn 2,068g mẫu hợp kim Cu-Zn trong lượng dư axit HNO3, thu được dung
dịch X. Đun đuổi axit dư, điều chỉnh tới pH 3 thu được 100mL dung dịch Y. Lấy
10mL dung dịch Y, thêm KI dư, rồi chuẩn độ dung dịch tạo thành bằng dung dịch
Na2S2O3 0,1M thì thấy hết 15,0 mL. Viết các phương trình phản ứng xảy ra. Tính
hàm lượng Cu trong mẫu hợp kim trên.
At elevated temperatures, hydrogen iodide may decompose to form hydrogen gas and iodine gas, as follows:
2HI(g) ⇌ H2 (g) + I2 (g)
In a particular experiment, the concentrations at equilibrium were measured to be [HI] = 0.85 mol/L, [I2] = 0.60 mol/L, and [H2] = 0.27 mol/L. What is Kc for the above reaction?
Explanation:
Since Kc is
[tex]k = \frac{(products)}{(reactants)} [/tex]
You can insert the Hydrogen and Iodine gas on top, and Hydrogen Iodide in the denominator.
Note: you can only include gases and aqueous species in an equilibrium expression, and all the species in this reaction are gaseous so you're good.
Inserting their molarity at equilibrium into their places, and you can solve. Don't forget to make the coefficient of HI turn into a power.
81.5 g of metal was heated from 11 degrees Celsius to 69 degrees Celsius. If 6739 joules of heat energy were used, what is the specific heat capacity of the metal?
Answer:
the metal become red hot
All of the following are characteristics of metals except: Group of answer choices good conductors of heat malleable ductile often lustrous tend to gain electrons in chemical reactions
Answer:
Hence the correct option is the last option that is tends to gain electrons in chemical reactions to become anions.
Explanation:
Metals tend to donate electrons in chemical reactions to become cations.
An ion of a single pure element always has an oxidation number of ________.
A. None of these
B. magnitude equal to its atomic number
C. 1
D. 0
Answer:
0
Explanation:
pure elements will always possess an oxidation number of 0, regardless of their charge.
Answer:
D.) 0
Explanation:
I got it correct on founders edtell
Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol
isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol
Answer:
[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:
[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]
Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.
Regards!
Given the following reaction:
CO (g) + 2 H2(g) <==> CH3OH (g)
In an experiment, 0.42 mol of CO and 0.42 mol of H2 were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.29 mol of CO remaining. Keq at the temperature of the experiment is ________.
A) 2.80
B) 0.357
C) 14.5
D) 17.5
E) none of the above
Answer:
Option D. 17.5
Explanation:
Equiibrium is: CO + 2H₂ ⇄ CH₃OH
1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.
Initially we have 0.42 moles of CO and 0.42 moles of H₂
If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.
So in the equilibrium we may have:
0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂
Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen
Finally 0.13 moles of methanol, are found after the equilibrium reach the end.
Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²
0.13 / (0.29 . 0.16²)
Kc = 17.5
Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid H3PO4 with ammonia NH3. What mass of ammonium phosphate is produced by the reaction of 5.5g of phosphoric acid
Answer:
8.3 g
Explanation:
Step 1: Write the balanced equation
H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄
Step 2: Calculate the moles corresponding to 5.5 g of H₃PO₄
The molar mass of H₃PO₄ is 97.99 g/mol.
5.5 g × 1 mol/97.99 g = 0.056 mol
Step 3: Calculate the moles of (NH₄)₃PO₄ produced
The molar ratio of H₃PO₄ to (NH₄)₃PO₄is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.056 mol = 0.056 mol.
Step 4: Calculate the mass corresponding to 0.056 moles of (NH₄)₃PO₄
The molar mass of (NH₄)₃PO₄ is 149.09 g/mol.
0.056 mol × 149.09 g/mol = 8.3 g
Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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e. Which of the following is a mixture? i. Water ii. Hydrogen iii. Air iv. Iron
water is known as the mixture
Answer:
iv. Iron
water is not a mixture
hydrogen is the simplest element
air is pure
What makes it possible
for a vascular plant to
be a long distance from
a water source?
A. long leaves
B. flowers
C. long roots
D. long stems
Answer:
I think long roots
Explanation:
Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.
Explanation:
here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula
^^^^Changes in state of matter are ALWAYS changes.
Answer:
physical
Explanation:
The change in the state of matter is always physical change, because it can be done with physical processes.
