Answer:
entropy increases when a glass of water is set in the sun until the water evaporates.
Hope it will help you :)
Explanation:
In the water cycle, evaporation occurs when sunlight warms the surface of the water. The heat from the sun makes the water molecules move faster and faster, until they move so fast they escape as a gas. ... When it is cool enough, the water vapor condenses and returns to liquid water.
In photosynthesis, carbon dioxide and water are converted into glucose and oxygen in the presence of chlorophyll and light energy. If 1.0 mole of water is available with an excess of carbon dioxide, how many moles of glucose (C6H12O6) can be produced
Answer:
0.17 mole of glucose is formed.
Explanation:
Step 1:
The equation for the reaction. This is given below:
CO2 + H2O —> C6H12O6 + O2
Step 2:
Balancing the equation.
The equation can be balanced as follow:
CO2 + H2O —> C6H12O6 + O2
There are 6 atoms of C on the right side and 1 atom on the left side. It can be balance by putting 6 in front of CO2 as shown below:
6CO2 + H2O —> C6H12O6 + O2
Therefore are 12 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 6 in front of H2O as shown below:
6CO2 + 6H2O —> C6H12O6 + O2
There are a total of 8 atoms of O the right side and a total of 18 atoms on the left. It can be balance by putting 6 in front of O2 as shown below:
6CO2 + 6H2O —> C6H12O6 + 6O2
Now the equation is balanced.
Step 3:
Determination of the number of mole of glucose (C6H12O6) produced by 1 mole of water.
This is illustrated below:
6CO2 + 6H2O —> C6H12O6 + 6O2
From the balanced equation above,
6 moles of H2O produced 1 mole of C6H12O6.
Therefore, 1 mole of H2O will produce = 1/6 = 0.17 mole of C6H12O6.
Answer:
IF 1.0 mol of water is available, 0.167 moles of glucose (C6H12O6) will be produced
Explanation:
Step 1: Data given
Number of moles of water = 1.0 moles
Carbon dioxide is in excess
Step 2: The balanced equation
6CO2 + 6H2O → C6H12O6 + 6O2
Step 3: Calculate moles of glucose
For 6 moles CO2 we need 6 moles H2O to produce 1 mol glucose and 6 moles O2
For 1.0 moles of water we'll have 1.00 / 6 = 0.167 moles of glucose
IF 1.0 mol of water is available, 0.167 moles of glucose (C6H12O6) will be produced
__________ reactions are used to detect antibodies for relatively large pathogens
Answer:
The question is incomplete, no worries I got you.
Here is the complete question;
__________ reactions are used to detect antibodies for relatively large pathogens, such as bacteria. For these tests, the antigen is mixed with the test sample at various dilutions. Reaction mixes are then monitored for the formation of visible aggregates.
Explanation:
AGGLUTINATION is the reaction used.
Agglutination is the reaction in which there is the clumping of particles. A agglutination reaction is the visible clumping of the bacterial cells as an antigen reacts with its corresponding antibody. This type is often used as an initial confirmation of specific pathogens. Agglutination tests is is used to detect antibody or antigen and it involves the agglutination of bacteria, red cells, or antigen- or antibody-coated latex particles. It is therefor used for large pathogens like bacteria. In this reaction, antigens are introduced into various dilutions of antibodies in test tubes or surfaces of glass slides, visible clumping is observed which depends on the size of the antibodies, amount and acidity of the antibody molecule, time of incubation and as well as the environment of the reaction which includes optimum pH, protein concentration among others.
If I have 5.6 liters of gas in a piston at a pressure of 150 kPa and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? round to 2 decimal places
Answer:
P1V1 = P2V2
(1.5 atm)(5.6 L) = (x)(4.8 L) x = 1.8 atm
If we have 5.6 liters of gas in a piston at a pressure of 150 kPa and compress the gas until its volume is 4.8 L, the new pressure inside the piston will be 175 kPa.
What is combined gas law?The combined gas law is the law of of gaseous state which is made by combination of Boyle's law, Charle's law, Avogadro's law and Gay Lussac's law.
It is a mathematical expression that relates Pressure, Volume and Temperature.
(P1 × V1)÷T1 = (P2 × V2)÷T2
Here temperature doesn't change, hence-
(P1 × V1) = (P2 × V2)
P1 = 150 kPa
V1 = 5.6 litres
P2 = ?
V2 = 4.8 litres
P2 = 175 kPa
Therefore, If we have 5.6 liters of gas in a piston at a pressure of 150 kPa and compress the gas until its volume is 4.8 L, the new pressure inside the piston will be 175 kPa.
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In the diagram, which letter represents the activation energy?
A
B
C
D
The letter that represents the activation energy from the diagram is letter A.
A chemical reaction occurs only when there is collision between the particles of reactants.
These colliding particles become activated with increased kinetic energy.
Activation energy is the energy barrier that must be overcome before a reaction takes place.
From the diagram,
C is the reactants
D is the products
B is the heat of reaction
While A is the activation energy.
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Name this compound please
Answer:
The answer to your question is butanal
Explanation:
To name this compound we must consider:
1.- Identify the functional group. The functional group of this molecule is the first carbon to the right and its oxygen.
When carbon is attached to oxygen is a border, this functional group is called Aldehyde.
2.- Count the total number of carbons starting from the right. This molecule has 4 carbons.
3.- Name the compound
An organic molecule with 4 carbons is called butane but change the ending for al, then the name will be butanal
What is a base used in toothpaste?
sorbitol
O carrageenan
O magnesium hydroxide
sodium lauryl sulfate
Answer: Option:C
Explanation:
Examine the reaction C 3 H 8 +5O2—>3CO 2 +4H 2 O. If the reaction starts with 2.5g C 3 H 8 what is the theoretical yield of water ? If the reactant yields 75 % of the theoretical yield of water, how many grams of water were produced?
Answer:
Theoretical yield = 4.09 g water.
75% yield = 3.07g water.
Explanation:
C3H8 + 5O2 ---> 3CO2 + 4H2O
Using atomic masses:
(12 *3 + 8 * 1.008) g C3H8 yield (theoretically) (4* (2*1.008 + 16) g water
44.06g C3H8 yields 72.06g water
So 2.5 g yields (72.06/44.06) * 2.5
= 4.09 g water.
If the yield is 75% then it is 4.09 * 0.75
= 3.07 g water.
If I correct, answer is C, 75%.
Of the reactions below, which one is not a combination reaction?Immersive Reader (2 Points) 2N2 + 3H2 → 2NH3 2Mg + O2 → 2MgO 2CH4 + 4O2 → 2CO2 + 4H2O C + O2 → CO2 CaO + H2O → Ca(OH)2
Answer: [tex]2CH_4+4H_2O\rightarrow 2CO_2+4H_2O[/tex] is not a combination reaction.
Explanation:
Synthesis reaction or combination reaction is defined as the reaction where two or more substances combine to form a single product.
[tex]2N_2+3H_2\rightarrow 2NH_3[/tex]
[tex]2Mg+O_2\rightarrow 2MgO[/tex]
[tex]CaO+H_2O\rightarrow Ca(OH)_2[/tex]
Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.
[tex]2CH_4+4H_2O\rightarrow 2CO_2+4H_2O[/tex]
Among the provided reactions, the reaction 2CH4 + 4O2 → 2CO2 + 4H2O is not a combination reaction, but rather a combustion reaction.
Explanation:In the given reactions, the reaction 2CH4 + 4O2 → 2CO2 + 4H2O is not a combination reaction. In a combination reaction, two or more substances (either elements or compounds) combine to form a single compound. However, the mentioned reaction is actually an example of a combustion reaction where methane (CH4) undergoes combustion in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O).
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How many dozens of doughnuts are in 48 doughnuts
Final answer:
There are 4 dozens in 48 doughnuts, as a dozen is a group of 12 objects. This quantity would require four boxes with a volume of 324 cubic inches each to hold all the doughnuts.
Explanation:
To find out how many dozens of doughnuts are in 48 doughnuts, we can simply divide the total number of doughnuts by the number in a dozen. Since a dozen refers to 12 objects, we divide 48 by 12.
48 ÷ 12 = 4
Therefore, there are 4 dozens in 48 doughnuts. When buying items like doughnuts, it is common to acquire them in groups, like a dozen, because it is more convenient and efficient.
Let's consider the dimensional aspect for a moment. If you have a 9x9x4 inch box for a dozen doughnuts, this box would have a volume of 324 cubic inches. For 48 doughnuts, which we have established are 4 dozens, we would need four of these boxes, equating to 4 × 324 in3 or 1296 in3 in total to hold all of them.
A solution containing 1.0 M NiCl2 and 1.0 M SnBr2 undergoes electrolysis by passing a current between two platinum electrodes. What are the most likely first products to be formed? Cl2 (aq) + 2e- → 2Cl- (aq) ℰ°= 1.36 V Br2 (aq) + 2e- → 2Br- (aq) ℰ° =1.08 V Sn2+ (aq) + 2e- → Sn(s) ℰ° = -0.14 V Ni2+ (aq) + 2e- → Ni(s) ℰ°= -0.24 V
Answer:
Sn2+ (aq) + 2e- → Sn(s) ℰ° = -0.14 V
Explanation:
A close look at all the options shows that the most feasible first reaction is the reduction of tin II ion to ordinary metallic tin.
Given the two half cells, nickel is oxidized in one half cell to Ni II while in the second half cell, tin II ion is reduced to metallic tin. The platinum electrodes simply act as electron conduits in the cell.
How many moles are contained in 2.0 L of N2 at standard temperature and pressure.
0.091 moles are contained in 2.0 L of N2 at standard temperature and pressure.
Explanation:
Data given:
volume of the nitrogen gas = 2 litres
Standard temperature = 273 K
Standard pressure = 1 atm
number of moles =?
R (gas constant) = 0.08201 L atm/mole K
Assuming nitrogen to be an ideal gas at STP, we will use Ideal Gas law
PV = nRT
rearranging the equation to calculate number of moles:
PV = nRT
n = [tex]\frac{PV}{RT}[/tex]
putting the values in the equation:
n = [tex]\frac{1X2}{0.08201 X 273}[/tex]
n = 0.091 moles
0.091 moles of nitrogen gas is contained in a container at STP.
Final answer:
To find how many moles are in 2.0 L of N2 at standard temperature and pressure, we use the fact that 1 mole of any gas occupies 22.4 L at STP. By dividing 2.0 L by 22.4 L, we find that there are approximately 0.089 moles of N2.
Explanation:
The question asks, How many moles are contained in 2.0 L of N2 at standard temperature and pressure? To answer this, it's essential to know that at standard temperature and pressure (STP), which is defined as 0°C (273K) and 1 atmosphere (atm) of pressure, 1 mole of any ideal gas occupies 22.4 liters of volume. This is a key concept in understanding gas laws and mol calculations in chemistry.
Given that we have 2.0 liters of N2 gas at STP, we can use the proportionality given by the molar volume at STP to find the number of moles. Since 22.4 liters is equivalent to 1 mole of a gas, we can set up a simple calculation to find the moles of N2 in 2.0 liters:
Moles of N2 = Volume of N2 (L) / Volume of 1 mole of gas at STP (L)
= 2.0 L / 22.4 L
= 0.089 moles of N2
This calculation demonstrates the direct application of molar volume at STP to determine the number of moles in a given volume of gas.
An aqueous feed solution of 1000 kg/h containing 23.5 wt % acetone and 76.5 wt % water is being extracted in a countercurrent multistage extraction system using pure methylisobutyl ketone solvent at 298–299 k. the outlet water raffinate will contain 2.5 wt % acetone. use equilibrium data from
Question:
The question is incomplete. What is required to calculate was not added.The equilibrium data was not also added. Below is the additional questions and the answers.
1. Calculate the minimum solvent that can be used.
2.Using a solvent rate of 1.5 times the minimum, calculate the number of
theoretical stages.
Answer:
1. Minimum solvent = 411.047
2. N = 5
Explanation:
See the attached files for explanations.
An aqueous feed solution.
The aqua solution is a solvent on water that has a chemical equation of aq. The example includes the solution of the salt table and NaCl. Having positive and negative ions. The solution feed id of 1000 KG/hand consists of 23.5 wt% acetone and has a 76.5 wt% of water by volume which is being used.
Thus the answer is minimum solvent = 411.047 and the N equals 5.
The 76.5wt of water is extracted in a concurrent multistage of the extraction system and is used with the methyl isobutyl. With the ketone solvent at the temperature of 298-299K. The outer or output of 2.5 wt 5 acetone which is a colorless liquid gets dissolved.Hence the answer is a mini solvent of 411.0 and value of N as 5.Learn more about the aqueous feed solution.
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An equivalent is:___________
a) an equivalent is the amount of ion that has a 1+ charge.
b) 1 mole of an ionic compound.
c) 1 mole of any ion. the amount of ion that carries 1 mole of electrical charge.
d) the amount of ion that has a 1- charge.
Answer:
The amount of ion that carries 1 mole of electrical charge
Explanation:
One equivalent: In reaction stoichiometry, the amount of one substance that reacts with one mole of another substance. This will often (but not always) be a 1:1 mole ratio.
An equivalent is the amount of ion that carries 1 mole of electrical charge.
An equivalent is the amount of a substance that reacts with one mole of another substance in a given chemical reaction.
It is a unit of measurement in Stoichiometry that helps to describe the amount of ion that carries 1 mole of electrical charge.
Examples include:
1 mole of Na⁺ = 1 Eq1 mole of Cl⁻ = 1 Eq1 mole of Ca²⁺ = 2 EqThus, we can conclude that an equivalent is the amount of ion that carries 1 mole of electrical charge.
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Based on the enthalpy changes given for the reactions, which reactions are endothermic?
2H2O → 2H2 + O2, ΔH = 484 kJ
C + O2 → CO2, ΔH = -394 kJ
H2 + Br2 → 2HBr, ΔH = -73 kJ
2NH3 → N2 + 3H2, ΔH = 92 kJ
Answer:
the first and last
Explanation:
the enthalpy Change is positive.
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Answer:
One (2H2O → 2H2 + O2, ΔH = 484 kJ )
Four (2NH3 → N2 + 3H2, ΔH = 92 kJ)
Explanation:
An endothermic process is any process with an increase in the enthalpy H of the system. So if it has a positive result then the heat is going up which means it is endothermic .
What is the only force that acts between particles of helium gas?
Answer: The only force that acts between particles of helium gas is Vanderwaal forces.
Explanation:
When molecules are uncharged in nature but still combined together due to the dipole moment of elements then this force holding the molecules together is known as Vanderwaal forces.
This force is weak in nature.
For example, particles of helium gas has Vanderwaal force between its atoms.
Thus, we can conclude that the only force that acts between particles of helium gas is Vanderwaal forces.
A student determines that according to the reaction N2(g) + 3H2(g) --> 2NH3(g), if 34g of nitrogen gas is reacted with excess hydrogen, 41g of ammonia can be produced. When the actual reaction was complete, only 38g of ammonia formed. Determine the percent yield for the reaction.
Answer:
The percent yield of this reaction is 92.7 %
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 34.0 grams
Mass of ammonia (NH3 produced = 41.0 grams
Molar mass of N2 = 28.0 g/mol
Molar mass of NH3 = 17.02 g/mol
Actual yield of ammonia = 38 grams
Step 2: The balanced equation
N2(g) + 3H2(g) → 2NH3(g)
Step 3: Calculate moles
Moles = mass / molar mass
Moles N2 = 34.0 grams / 28.0 g/mol
Moles N2 = 1.214 moles
Step 4: Calculate moles NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 1.214 moles N2 we'll have 2* 1.214 = 2.428 moles NH3
Step 5: Calculate mass NH3
Mass NH3 = moles * molar mass
Mass NH3 = 2.428 moles * 17.02 g/mol
Mass NH3 = 41 grams
Step 6: Calculate percent yield for the reaction
Percent yield = (actuald yield / theoretical yield) * 100 %
Percent yield = (38 grams / 41 grams ) * 100 %
Percent yield = 92.7 %
The percent yield of this reaction is 92.7 %
Answer:
[tex]Y=92\%[/tex]
Explanation:
Hello,
In this case, by considering the given chemical reaction, with given mass of nitrogen, one could compute the theoretical yield of ammonia as shown below and considering their 1 to 2 molar relationship in the chemical reaction:
[tex]m_{NH_3}^{theoretical}=34gN_2*\frac{1molN_2}{28gN_2}*\frac{2molNH_3}{1molN_2}*\frac{17gNH_3}{1molNH_3} \\m_{NH_3}^{theoretical}=41.3gNH_3[/tex]
In such a way, the percent yield is obtained as shown below:
[tex]Y=\frac{m_{NH_3}^{actual}}{m_{NH_3}^{theoretical}} *100\%=\frac{38g}{41.3g} *100\%\\\\Y=92.0\%[/tex]
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A student obtains a 10.0g sample of a white powder labeled as BaCl2. After completely dissolving the powder in 50.0mL of distilled water, the student adds excess Na2SO4(s), which causes a precipitate of BaSO4(s) to form, as represented by the equation above. The student filters the BaSO4(s), rinses it, and dries it until its mass is constant. Which of the following scientific questions could best be answered based on the results of the experiment? Is the Na2SO4(s) used in the experiment pure? A Is the BaCl2(s) used in the experiment pure? B What is the molar solubility of BaCl2 in water? C What is the molar solubility of BaSO4 in water?
Answer:
(B.) What is the molar solubility of barium chloride, BaCl2 in water?
Explanation:
Molar solubility is the number of moles of a substance that can dissolve in a liter of solution to the point of the solution's saturation. It can be calculated stoichiometrically from a substance's solubility product constant in mol/L.
Since all the [tex]BaCl_{2}[/tex] reacted all the [tex]Na_{2} SO_{4}[/tex] from the information, we can easily assume all the substances were consumed in the reaction, and hence account for their purity. Furthermore, [tex]BaSO_{4}[/tex] is insoluble in water, the most probable scientific query would be the molar solubility of the [tex]BaCl_{2}[/tex] used in the experiment.
Answer:
The best question is A Is the BaCl2(s) used in the experiment pure?Explanation:
Step 1: Data given
Mass of the BaCl2 sample = 10.0 grams
Volume of water = 50.0 mL
We add excess Na2SO4
A precipitate BaSO4 will be formed
Step 2: The balanced equation
BaCl2(aq) → Ba^2+(aq) + 2Cl-(aq)
Ba^2+(aq) + SO4^2-(aq) →BaSO4(s)
Step 3: Calculate moles BaCl2
Moles BaCl2 = 10.0 grams / 208.23 g/mol
Moles BaCl2 = 0.048 moles
Step 4: Calculate moles Ba^2+
For 1 mol BaCl2 we have 1 mol Ba^2+
For 0.048 moles BaCl2 we have 0.048 moles Ba^2+
Step 5: Calculate mass Ba^2+
Mass Ba^2+ = moles Ba^2+ * molar mass Ba^2+
Mass Ba^2+ = 0.048 moles * 137.33 g/mol
Mass Ba^2+ = 6.59 grams
After measuring the mass of barium in BaSO4 we can determine if the BaCl2 was pure or not.
If the mass = 6.59 grams the BaCl2 was pure
If the mass <6.59 grams the BaCl2 wasn't pure
The best question is A Is the BaCl2(s) used in the experiment pure?
A 5.45 L sample at 10.0°C and a pressure of 64.7 kPa is allowed to expand to a volume of 18.50L. The final pressure of the gas is 58.0 kPa. What is the final temperature of the gas in degrees Celsius?
Answer: [tex]481^0C[/tex]
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 64.7 kPa
[tex]P_2[/tex] = final pressure of gas = 50.8 kPa
[tex]V_1[/tex] = initial volume of gas = 5.45 L
[tex]V_2[/tex] = final volume of gas = 18.50 L
[tex]T_1[/tex] = initial temperature of gas = [tex]10^oC=273+10=283K[/tex]
[tex]T_2[/tex] = final temperature of gas = ?
Now put all the given values in the above equation, we get:
[tex]\frac{64.7kPa\times 5.45}{283K}=\frac{50.8kPa\times 18.50}{T_2}[/tex]
[tex]T_2=754K=(754-273)^0C=481^0C[/tex]
Thus final temperature of the gas in degrees Celsius is 481
"11. Barium nitrate reacts with aqueous sodium sulfate to produce solid barium sulfate and aqueous sodium nitrate. Abigail places 20.00 mL of 0.500 M barium nitrate in a flask. She has a 0.225M sodium sulfate solution available. What volume of this solution must she add to her flask of barium nitrate so she has no excess reactant left over?"
Answer:
44 mL of Na2SO4
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)
Step 2:
Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:
Molarity of Ba(NO3)2 = 0.5 M
Volume of solution = 20 mL = 20/1000 = 0.02 L
Mole of solute (Ba(NO3)2) =?
Molarity = mole /Volume
0.5 = Mole of Ba(NO3)2 / 0.02
Cross multiply to express in linear form
Mole of Ba(NO3)2 = 0.5 x 0.02
Mole of Ba(NO3)2 = 0.01 mole
Step 3:
Determination of the number of mole of Na2SO4 that reacted.
Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)
From the balanced equation above,
1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.
Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.
Step 4:
Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:
Mole of Na2SO4 = 0.01 mole
Molarity of Na2SO4 = 0.225M
Volume =?
Molarity = mole /Volume
0.225 = 0.01 / volume
Cross multiply to express in linear form
0.225 x Volume = 0.01
Divide both side by 0.225
Volume = 0.01/0.225
Volume of Na2SO4 = 0.044 L
Converting 0.044 L to mL, we have
Volume of Na2SO4 = 0.044 x 1000
Volume of Na2SO4 = 44 mL
Therefore, 44 mL of Na2SO4 is needed for the reaction
Answer:
She has to add 44.44 mL of Na2SO4
Explanation:
Step 1: Data given
Volume of barium nitrate = 20.00 mL = 0.020 L
Molarity of barium nitrate = 0.500 M
Molarity of sodium sulfate = 0.225 M
Step 2: The balanced equation
Ba(NO3)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2 NaNO3 (aq)
Step 3: Calculate volume of the sodium sulfate solution needed
For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol basO4 and 2 moles NaNO3
C1*V1 = C2*V2
⇒with C1 = the molarity of Ba(NO3)2 = 0.500 M
⇒with V1 = the volume of Ba(NO3)2 = 0.020 L
⇒with C2 = the molarity of Na2SO4 =0.225 M
⇒with V2 = the volume of Na2SO4 = TO BE DETERMINED
0.500 M * 0.020 L = 0.225 M * V2
V2 = (0.500 M * 0.020 L) / 0.225 M
V2 = 0.04444 L = 44.44 mL
She has to add 44.44 mL of Na2SO4
1. What is the kinetic energy of 44 kg cheetah running at 31m/s speed?
Explanation:
Given
mass (m) = 44 kg
velocity (v) = 31 m/s
Now
Kinetic energy
= 1/2 mv²
= 1/2 * 44 * 31²
= 0.5 * 44 * 961
= 21142 joule
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Creatine phosphate
a. is produced by the process of anaerobic respiration.
b. can replace ATP in binding to myosin molecules during contraction.
c. acts as an energy reserve in muscle tissue.
d. is only formed during strenuous exercise.
e. cannot transfer its phosphate group to ADP.
Answer: C. Acts as energy reserve in muscle tissue.
Explanation:
Creatine Phosphate is a high energy compound that can be found in muscle tissue.
It serves as an extra energy storage, since the normal metabolism cannot supply the high energy requirement of a muscle at work.
Creatine Phosphate is quickly transfered to ADP, for the production of ATP needed for muscle contraction.
What is the main purpose of using NaOH in the saponification reaction . NaOH works as the catalyst b. NaOH solution is the solvent for the reaction c. NaOH acts as a reactant in the reaction d. NaOH facilitates the soap to salt out from the solution
Answer:
C. NaOH acts as a reactant in the reaction
Explanation:
Because during the saponification process, Na+ replaces the H+ in the fatty acid been used for the saponification process
Specific heat is used to explain why different substances
a) sink or float
b) change temp. at different rates
c) vaporize or condense at different temp.’s
d) melt and freeze at the same temp.
Answer:
change temperature at different rates
The specific heat is used to explain why different substances change the temperature at different rates.
Specific Heat:
It is the amount of heat that is required to increase the temperature of 1 kg of a substance by 1^oc. It is measured in the [tex]\bold {J/g/^oC }[/tex].
The specific heat capacity of water is [tex]\bold { 4.184 J/g/^oC.}[/tex]If the heat capacity is it takes more time to heat or cool it down. If specific heat capacity of a substance is low, then it takes less time to heat or cool it down.Therefore, the specific heat is used to explain why different substances change the temperature at different rates.
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Which of the following groups contain elements that are gaseous at room temperature?
a) alkali metals and alkaline earth metals
b) alkali metals and transition metals
c) noble gases and transition metals
d) noble gases and halogens
Final answer:
The group that contains elements that are gaseous at room temperature are noble gases and halogens. Noble gases include helium, neon, argon, krypton, xenon, and radon, and halogens like fluorine and chlorine are gases under standard conditions. So the correct option is D.
Explanation:
The question at hand involves identifying which groups of elements are gaseous at room temperature. Of the options provided, d) noble gases and halogens contain elements that are gaseous at room temperature. Noble gases, which include elements like helium, neon, argon, krypton, xenon, and radon, are all gases under standard conditions. Halogens such as fluorine and chlorine are also gaseous at room temperature. Alkali metals and alkaline earth metals are solid at room temperature. Furthermore, most transition metals are also solid at room temperature, with mercury being an exception as it is liquid.
To define the subject for the questions based on the examples given, we can use the keywords: alkali metal, halogen, noble gas, and alkaline earth metal as they relate to the classifications on the periodic table. The given examples classify elements like lithium as an alkali metal (Group 1), argon as a noble gas (Group 18), and chlorine as a halogen (Group 17), which are key aspects of elementary Chemistry covered in high school.
Test your knowledge of water's ability to dissolve
substances by matching the following statements
with the correct answers.
when water breaks apart an ionic compound into
cations and anions.
when water surrounds substances that have been
"broken apart":
when water breaks down substances and
surrounds individual molecules or particles:
Answer: 1 dissociation. 2. Hydration. 3. Dissolving.
Explanation:
Predict the effect of adding a non competitive inhibitor to the reaction mixture on the rate of reaction at a high substrate concentration. Support your prediction by describing how a non competitive inhibitor affects the structure and function of an enzyme.
Answer:
A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time
Explanation:
this is because It changes the conformation of an enzyme as well as its active site, which makes the substrate unable to bind to the enzyme effectively so that the efficiency of the enzyme decreases. A noncompetitive inhibitor binds to the enzyme away from the active site, altering/distorting the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively and most of the time also the inhibitor is reversible
A student dissolved 3.00 g of Co(NO 3) 2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the solution then diluted it with water to give 275 mL of a final solution. How many grams of NO 3 - ion are there in the final solution?
Answer:
0.0812 grams of nitrate ions are there in the final solution.
Explanation:
Mass of cobalt (II) nitrate = 3.00 g
Moles of cobalt(II) nitrate = [tex]\frac{3.00 g}{183 g/mol}=0.0164 mol[/tex]
Volume of the solution = 100 mL = 0.100 L
1 mL = 0.001 L
[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]
Molarity of the solution = [tex]\frac{0.0164 mol}{0.100 L}=0.164 M[/tex]
Cobalt (II) nitrate in its aqueous solution gives 1 mole of cobalt(II) ion and 2 moles of nitrate ions.
[tex][NO_3^{-}]=2\times [Co(NO_3)_2]=2\times 0.164 M=0.328 M[/tex]
Molarity of the nitrate ion before solution = [tex]M_1=0.328 M[/tex]
Volume of the nitrate ion before solution = [tex]V_1=4.00 mL[/tex]
Molarity of the nitrate ion after solution = [tex]M_2=?[/tex]
Volume of the nitrate ion after solution = [tex]V_2=275 mL[/tex]
[tex]M_1V_1=M_2V_2[/tex] ( Dilution)
[tex]M_2=\frac{0.328 M\times 4.00 mL}{275 mL}=0.00477 M[/tex]
Moles of nitrate ions in 275 ml = n
Molarity of the nitrate ion after solution =0.00477 M
volume of the final solution = 275 mL = 0.275 L
[tex]n=0.00477 M\times 0.275 L=0.00131 mol[/tex]
Mass of 0.00131 moles of nitrate ions:
0.00131 mol × 62 g/mol = 0.0812 g
0.0812 grams of nitrate ions are there in the final solution.
Why is the use of pesticides on soils and
crops harmful for humans?
A. Pesticides are carried across the surface
in runoff which can contaminate aquifers.
B. Pesticides kill organisms that can harm
plants.
C. Pesticides pollute wells that people drink
from.
D. Pesticides are evaporated with water and
contaminate the atmosphere.
Answer:
A.
Once it rains the pesticide goes down the runoff and pollute the water
Acetic acid has a Ka of 1.8 * 10-5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: acetic acid ten times greater than acetate, acetate ten times greater than acetic acid, and acetate=acetic acid.
Match each buffer to the expected pH. pH = 3.74 ; pH = 4.74 ; pH = 5.74
Part B: How many grams of dry NH4Cl need to be added to 2.30 L of a 0.600 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.71? Kb for ammonia is 1.8 * 10-5.
Express your answer with the appropriate units.
Answer:
A) acetic acid ten times greater than acetate pH = 3.74
B) acetate ten times greater than acetic acid pH = 5.74
C) For solution 3: acetate=acetic acid pH = 4.74
Mass NH4Cl = 40.30 grams
Explanation:
Step 1: Data given
Ka is: 1.8 * 10^-5
Solution 1: acetic acid ten times greater than acetate
Solution 2: acetate ten times greater than acetic acid
Solution 3: acetate=acetic acid
Step 2: The pH formula
pH = pKa + log[CH3COO-]/[CH3COOH]
For solution 1: acetic acid ten times greater than acetate this means ) [[CH3COO-]/[CH3COOH]) has a value of 1/10
pH = pKa + log[CH3COO-]/[CH3COOH]
pH = -log(1.8*10^-5) + log(1/10)
pH = 4.74 -1
pH = 3.74
For solution 2: acetate ten times greater than acetic acid
pH = pKa + log[CH3COO-]/[CH3COOH]
pH = -log(1.8*10^-5) + log(10)
pH = 4.74 + 1
pH = 5.74
For solution 3: acetate=acetic acid
pH = pKa + log[CH3COO-]/[CH3COOH]
pH = -log(1.8*10^-5) + log(1)
pH = 4.74 + 0
pH = 4.74
Part B: Calculate molarity
pOH = pKb + log [H
5.29= -log (1.8*10^-5) + log [BH+]/[0.600 M]
5.29 = 4.744 + log [BH+]/[0.600 M]
0.546 = [BH+]/0.600
[BH+] = 0.3276 M
Moles NH4+ = 0.3276 M * 2.30 L
Moles NH4+ = 0.75348 moles
Moles NH4Cl = 0.75348 moles
Mass NH4Cl = 0.75348 moles * 53.49 g/mol
Mass NH4Cl = 40.30 grams
A patient excretes a large volume of very dilute urine on a continuing basis. This is may be due to
Answer:
arterial hypertension
Explanation:
When increasing the arterial pressure, in the renal glomerulus the filtration of liquids increases since the liquids diffuse from zones of greater pressure to less pressure.
In this way, the liquid content of the urine increases and it is more diuluid because this does not apply to proteins or solutes, since these are affected by discrepancies in concentrations and not in pressures.
Excessive excretion of dilute urine, or polyuria, can be caused by conditions such as diabetes insipidus or diabetes mellitus, as well as the excessive use of diuretics, kidney disease, and excessive water intake. Diabetes insipidus reduces the number of water channels leading to the loss of water, while diabetes mellitus results in glucose in the urine which also causes the loss of water.
Explanation:If a patient is excreting a large volume of very dilute urine on a continuing basis, this can be due to a condition known as polyuria. Polyuria is characterized by urine production in excess of 2.5 L per day, and it can be caused by a number of potential factors.
One such factor could be diabetes insipidus, a condition caused by an insufficient release of the pituitary hormone known as antidiuretic hormone (ADH), or insufficient numbers of ADH receptors in the kidneys. This results in an insufficient number of water channels, known as aquaporins, which reduces water absorption and leads to high volumes of very dilute urine.
Another potential cause is diabetes mellitus, a condition where blood glucose levels exceed the number of available sodium-glucose transporters in the kidneys. This causes glucose to appear in the urine, and its osmotic properties attract water, leading to its loss in urine. Other potential causes of polyuria can also include the excessive use of diuretics, kidney disease, and excessive water intake.
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