Answer:
These batteries are constructed of several single cells connected in series each cell produces approximately 2.1 volts. ... A battery cell consists of two lead plates a positive plate covered with a paste of lead dioxide and a negative made of sponge lead, with an insulating material (separator) in between.
Consider a galvanic (voltaic) cell that has the generic metals X and Y as electrodes. If X is more reactive than Y (that is, X more readily reacts to form a cation than Y does), classify the following descriptions by whether they apply to the X or Y electrode.
i. anode
ii. cathode
iii. electrons in the wire flow toward
iv. electrons in the wire flow away
v. cations from salt bridge flow toward
vi. anions from salt bridge flow toward
vii. gains mass
viii. loses mass
Answer:
X
anode
electrons in the wire flow away
anions from salt bridge flow toward
loses mass
Y
cathode
electrons in the wire flow toward
cations from salt bridge flow toward
gains mass
Explanation:
In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.
Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.
At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.
Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.
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:
A saturated solution of potassium iodide contains, in each 100 mL, 100 g of potassium iodide. The solubility of potassium iodide is 1 g in 0.7 mL of water. Calculate the specific gravity of the saturated solution
Answer:
Specific gravity of the saturated solution is 2
Explanation:
The specific gravity is defined as the ratio between density of a solution (In this case, saturated solution of potassium iodide, KI) and the density of water. Assuming density of water is 1:
Specific gravity = Density
The density is the ratio between the mass of the solution and its volume.
In 100mL of water, the mass of KI that can be dissolved is:
100mL * (1g KI / 0.7mL) = 143g of KI
That means all the 100g of KI are dissolved (Mass solute)
As the volume of water is 100mL, the mass is 100g (Mass solvent)
The mass of the solution is 100g + 100g = 200g
In a volume of 100mL, the density of the solution is:
200g / 100mL = 2g/mL.
The specific gravity has no units, that means specific gravity of the saturated solution is 2
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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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.
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
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
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
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
Why must beta particles be used to detect leaks in a pipe?
Why must beta particles be used to detect leaks in a pipe?
A. Beta particles will not be absorbed by the soil like gamma radiation, but won't pass through the pipe before reaching the leak like alpha radiation would.
B. Beta particles will not cause an electrical discharge like alpha particles when they interact with the metal pipe, or contaminate the water like gamma radiation.
C. Beta particles are not used, only alpha particles are used because they are not harmful to humans.
D. Beta particles will not be absorbed by the soil like alpha particles, but won't pass through the pipe before reaching the leak like gamma radiation would.
Should be D, because alpha particles are absorbed by soil and gamma isn't
Beta particles will not be absorbed by the soil like alpha particles, but won't pass through the pipe before reaching the leak.
What is Beta particle?This type of particle is a high-speed electron and is derived from the process of beta decay.
It is used to detect leaks in pipe because it will not be absorbed by the soil like alpha particles, but won't pass through the pipe before reaching the leak like gamma radiation would.
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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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A certain polytomic ion contains 49 protons and 50 electrons. What's the net charge of this ion?
Answer:
the charge is -1
Explanation:
because the charge of proton is +and electron -
charge = +49 +(-50)
= -1
Answer:
Net charge is -1
Explanation:
[tex]{ \sf{net \: charge = p {}^{ + } + {e}^{ - } }} \\ = { \sf{49 + ( - 50)}} \\ = - 1[/tex]
A sample of nitrogen gas occupies 117 mL at 100°C. At what
temperature would it occupy 234 mL if the pressure does not
change? (express answer in K and °C)
47
Page
8 I 8
- Q +
Answer:
The new temperature of the gas is 746 K.
Explanation:
Given that,
The volume of the gas, V₁ = 117 mL
Temperature, T₁ = 100°C = 373
Final volume of the gas, V₂ = 234 mL
We need to find the final temperature. The relation between temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{373\times 234}{117}\\\\T_2=746\ K[/tex]
So, the new temperature of the gas is 746 K.
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
PLEASE HELP!!
this is on USAtestprep
a)
b)
c)
d)
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.
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
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.
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.
Given the reactants of the chemical reaction that will take place in Part D (construction of a lead concentration cell) prior to the assembly of the cell, determine the type of chemical reaction it is. Hint: Determine the products of the reaction.
Answer:
hi
Explanation:
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]
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!
Low-density polyethylene is formed because _______ polymerization is very unpredictable and difficult to control.
dehydration-condensation
anionic-initiated
radical-initiated
esterification
Answer:
radical-initiated
Explanation:
Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.
What is a reaction rate?
Answer:
A reaction is the time that is required for a chemical reaction to go essential to completion
^^^^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.
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
How many grams of CO2 are formed if 44.7 g C5H12 is mixed with 108 g O2?
Explanation:
here's the answer to your question
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!
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
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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