Answer:
Cyanobacteria, Azotobactor and Azospirillum are the microbes which is required for nitrogen cycle.
Explanation:
Nitrogen Cycle is a type of cycle in which nitrogen moves from atmosphere to the earth surface and again return back to the atmosphere. Cyanobacteria, Azotobactor and Azospirillum are the microbes which is required for nitrogen cycle. If these microbes is vanished from the environment, the nitrogen cannot be converted into other forms. Carbon is present in the atmosphere in carbondioxide form. This CO2 is used by the plants from atmosphere and make food from it. When these plants are eaten by animals, this CO2 is again release in the atmosphere. Due to global warming, the microbes present in the soil die due to increase in temperature of the soil. Levels of CO2 and CH4 increases if the soil warms up. With the increase in temperature, soil respiration increases which leads to more emission of CO2 and as a result more global warming occurs.
The partial pressure of CO2 gas above the liquid in a carbonated drink is 0.45 atm. Assuming that the Henry's law constant for CO2 in the drink is that same as that in water, 3.7 x 10-2 mol/L atm, calculate the solubility of carbon dioxide in this drink.
Answer:
[tex]M_{CO_2}=0.01665M[/tex]
Explanation:
Hello,
In this case, the Henry's law allows us to relate the molar concentration and partial pressure of a solute (carbon dioxide) in a solution (solvent is water) by:
[tex]M_{CO_2}=H_{CO_2}p_{CO_2}[/tex]
Whereas we introduce the Henry constant, therefore, we can easily compute the molar solubility by:
[tex]M_{CO_2}=p_{CO_2}*H_{CO_2} =0.45atm*3.7x10^{-2}\frac{M}{atm}\\\\M_{CO_2}=0.01665M[/tex]
Regards.
Why can't we destroy bothersome pollutants by just dissolving in ocean
Answer:
Bothersome pollutants are pollutants that persist in water.
Explanation:
Bothersome pollutants are those pollutants that have deleterious effects on human health which are consistently lodged into water bodies. Bothersome pollutants include petroleum products, heavy metals, and toxic chemicals etc.
Many of these bothersome pollutants are non biodegradable. They accumulate at the bottom of the ocean thereby causing problems for aquatic organisms and disrupting the aquatic ecosystems in the ocean.
These bothersome pollutants are not soluble in water hence we can not just destroy them by dissolving them in water. They persist in the aquatic ecosystem and can even migrate from place to place thereby spreading pollution.
What are three things that
living things need to live?
Answer:
Water, air, and food.
Explanation:
All living things need air water and food to live.
Answer:
the first the Question is what do you need to live then that your answer
food,water,air ;]
Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)→Cu3(PO4)2(s)+6NaCl(aq)2Na3PO4(aq)+3CuCl2(aq)→Cu3(PO4)2(s)+6NaCl(aq) Part A What volume of 0.174 MNa3PO4MNa3PO4 solution is necessary to completely react with 90.8 mLmL of 0.102 MCuCl2MCuCl2? Express your answer in milliliters.
Which of the following characteristics is the kinetic molecular theory based on? Matter does not have motion. All matter is in continual random motion. All matter has mass. All matter is made up of atoms and molecules that have mass.
Answer:
All matter is in continual random motion.
Explanation:
Because Kinetic Molecular theory is based on a principle idea that tiny particles such as gas particles are always in motion.
Answer:
All matter is in continual random motion.
Explanation:
Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds
The given question is incomplete, the complete question is:
Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?
Answer:
The lowest whole-number mass ratio in the two compounds is 1:2.
Explanation:
There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.
The no. of moles can be determined by using the formula,
moles = mass/molecular mass
moles = 2.98 g/207.2 g/mol
= 0.0144 moles
The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.
moles = 0.461 g /16 g/mol
= 0.0288 moles
The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2
On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,
moles = 9.89 g / 207.2 g/mol
= 0.0477 moles
The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,
moles = 0.763 g / 16 g
= 0.0477 moles
The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1
Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.
In all measurements, you are looking for values that are accurate. Pretend that a pebble with a known mass of 0.567 g was massed four times where the following masses were obtained: 0.256 g, 0.723 g, 0.554 g and 0.354 g. Those masses are an example of:__________.
a) being precise but not accurate.
b) being both precise and accurate.
c) being neither precise nor accurate.
d) being accurate but not precise.
Answer:
The correct answer is option c, that is, being neither precise nor accurate.
Explanation:
Obtaining the experimental values that come almost close to the true value is termed as accuracy. On the other hand, precision is obtaining experimental values continuously, which may come either far away or near to the true value.
Based on the given information, it is clear that 0.554 grams, that is, the third value obtained from the experiment is showing some closeness to the true value, which is 0.567 grams. While, neither any of the other values are close or is coming near to the true value. This shows that the measurements are an illustration of being neither precise nor accurate.
When the equation
Ca3(PO4)2+HNO3→Ca(NO3)2+H3PO4
is balanced, the smallest whole number coefficient for HNO3 is
Answer:
The least whole number coefficient for HNO₃ is 6
Explanation:
The chemical equation above is the reaction between calcium orthophosphate and nitric acid.
To balance a chemical equation, we have to consider law of conservation of matter which states that matter can neither be created nor destroyed.
What this law implies is that, whatever we have at the reactant side must be equal to whatever is obtainable at the product side.
The above equation is
Ca₃(PO₄)₂ + HNO₃ → Ca(NO₃)₂ + H₃PO₄
To balance the equation, we'll have to check the number of atoms at each side and possibly balance the equation with the number of moles.
The balanced equation is
Ca₃(PO₄)₂ + 6HNO₃ → 3Ca(NO₃)₂ + 2H₃PO₄
From the balanced equation above, we can see that the number of calcium (Ca), Phosphorus (P), Oxygen(O), Nitrogen(N) and hydrogen (H) are balanced at both sides of the equation.
The least number coefficient for HNO₃ is 6
For the reaction A+B↽−−⇀C+D, assume that the standard change in free energy has a positive value. Changing the conditions of the reaction can alter the value of the change in free energy (ΔG). Classify the conditions as to whether each would decrease the value of ΔG, increase the value of ΔG, or not change the value of ΔG for the reaction. For each change, assume that the other variables are kept constant.
A. Adding a catalystb.
B. Increasing [C] and [D]
C. Coupling with ATP hydrolysis
D. Increasing [A] and [B]
Explanation:
a. Adding a catalyst
no effect .( Catalyst can only change the activation energy but not the free energy).
b. increasing [C] and [D]
Increase the free energy .
c. Coupling with ATP hydrolysis
decrease the free energy value .
d.Increasing [A] and [B]
decrease the free energy.
What are the oxidation numbers of the atoms in this reaction?
Check all that apply.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
-4 for C in CH4
+4 for C in CO2
-2 for O in all substances
+1 for H in both CH4 and H2O
+4 for O in H2O
Answer:
A) -4 for C in CH4
B) +4 for C in CO2
D) +1 for H in both CH4 and H2O
Hope this helped! :D
A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C) Express your answer to three significant figures and include the appropriate units.
Answer:
25.0 grams is the mass of the steel bar.Explanation:
Heat gained by steel bar will be equal to heat lost by the water
[tex]Q_1=-Q_2[/tex]
Mass of steel= [tex]m_1[/tex]
Specific heat capacity of steel = [tex]c_1=0.452 J/g^oC[/tex]
Initial temperature of the steel = [tex]T_1=2.00^oC[/tex]
Final temperature of the steel = [tex]T_2=T=21.50^oC[/tex]
[tex]Q_1=m_1c_1\times (T-T_1)[/tex]
Mass of water= [tex]m_2= 105 g[/tex]
Specific heat capacity of water=[tex]c_2=4.18 J/g^oC[/tex]
Initial temperature of the water = [tex]T_3=22.00^oC[/tex]
Final temperature of water = [tex]T_2=T=21.50^oC[/tex]
[tex]Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))[/tex]
On substituting all values:
[tex](m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}[/tex]
25.0 grams is the mass of the steel bar.Answer:
[tex]m_{steel}=24.9g[/tex]
Explanation:
Hello,
In this case, since the water is initially hot, the released heat by it is gained by the steel rod since it is initially cold which in energetic terms is illustrated by:
[tex]\Delta H_{water}=-\Delta H_{steel}[/tex]
That in terms of mass, specific heat and temperature change is:
[tex]m_{water}Cp_{water}(T_f-T_{water})=-m_{steel}Cp_{steel}(T_f-T_{steel})[/tex]
Thus, we simply solve for the mass of the steel rod:
[tex]m_{steel}=\frac{m_{water}Cp_{water}(T_f-T_{water})}{-Cp_{steel}(T_f-T_{steel})} \\\\m_{steel}=\frac{105mL*\frac{1g}{1mL}*4.18\frac{J}{g\°C}*(21.50-22.00)\°C}{-0.452\frac{J}{g\°C}*(21.50-2.00)\°C} \\\\m_{steel}=24.9g[/tex]
Best regards.
Which of the following statements regarding hydrogen bonding in secondary structures is true? Group of answer choices Both α-helices and β-sheets only use intrachain hydrogen bonds. Both α-helices and β-sheets only use interchain hydrogen bonds. α-helices only use intrachain hydrogen bonds and β-sheets can use either intrachain or interchain hydrogen bonds. α-helices can use either intrachain or interchain hydrogen bonds and β-sheets only use interchain hydrogen bonds.
Answer:
The correct answer is "α-helices only use intrachain hydrogen bonds and β-sheets can use either intrachain or interchain hydrogen bonds".
Explanation:
Hydrogen bonding are essential forces for the formation of the secondary structures of proteins. The alpha helix is a coiled secondary structure that is stabilized only by intrachain hydrogen bonds, which are formed between the NH and CO groups of the main chain. On the other hand, β-sheets can use either intrachain or interchain hydrogen bonds for its stabilization. Adjacent chains in β-sheets can run either in the same or opposite directions, which are stabilized by interchain hydrogen bonds.
The statement regarding hydrogen bonding in secondary structures which is true is: C. α-helices only use intrachain hydrogen bonds and β-sheets can use either intrachain or interchain hydrogen bonds.
Hydrogen bond can be defined as a weak chemical bond existing between a partially positively charged hydrogen atom and an electronegative chemical atom.
Protein is a macromolecule which is formed as a result of the union of many amino acids.
A hydrogen bond is an essential intermolecular force of attraction that is required for the formation of the secondary structures of proteins.
Generally, there are two (2) main types of secondary structure in proteins and these include:
Alpha (α) helix: it's a right-handed coiled strand. It is only stabilized or maintained by intrachain hydrogen bonds, which are typically formed between amide hydrogens (NH) and carbonyl oxygens (CO) of the peptide backbone.Beta (β) sheets: it can be stabilized or maintained by either intrachain or interchain hydrogen bonds.In conclusion, α-helices can only use intrachain hydrogen bonds while β-sheets can use either intrachain or interchain hydrogen bonds.
Read more: https://brainly.com/question/14681125
A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.582 g CO2 and 3.922 g H2O. What percent by mass of oxygen is contained in the original sample?
Answer:
[tex]\% O=27.6\%[/tex]
Explanation:
Hello,
In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:
- Moles of carbon are contained in the 9.582 grams of carbon dioxide:
[tex]n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2} =0.218molC[/tex]
- Moles of hydrogen are contained in the 3.922 grams of water:
[tex]n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH[/tex]
- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:
[tex]m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO[/tex]
Finally, we compute the percent by mass of oxygen:
[tex]\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%[/tex]
Regards.
Express your answer to three significant figures.
This balanced equation shows the reaction of sodium hydroxide and sulfuric acid:
2NaOH + H2SO4 - Na2SO4 + 2H20.
In a laboratory experiment, a student mixes 355 grams of sulfuric acid with an excess of sodium hydroxide. What is the theoretical mass of
sodium sulfate produced? Refer to the periodic table and the polyatomic ion resource.
The theoretical mass of sodium sulfate is
grams.
Answer: The theoretical mass of [tex]Na_2SO_4[/tex] is, 514 grams.
Explanation : Given,
Mass of [tex]H_2SO_4[/tex] = 355 g
Molar mass of [tex]H_2SO_4[/tex] = 98 g/mol
First we have to calculate the moles of [tex]H_2SO_4[/tex].
[tex]\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}[/tex]
[tex]\text{Moles of }H_2SO_4=\frac{355g}{98g/mol}=3.62mol[/tex]
Now we have to calculate the moles of [tex]Na_2SO_4[/tex]
The balanced chemical equation is:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]H_2SO_4[/tex] react to give 1 mole of [tex]Na_2SO_4[/tex]
So, 3.62 mole of [tex]H_2SO_4[/tex] react to give 3.62 mole of [tex]Na_2SO_4[/tex]
Now we have to calculate the mass of [tex]Na_2SO_4[/tex]
[tex]\text{ Mass of }Na_2SO_4=\text{ Moles of }Na_2SO_4\times \text{ Molar mass of }Na_2SO_4[/tex]
Molar mass of [tex]Na_2SO_4[/tex] = 142 g/mole
[tex]\text{ Mass of }Na_2SO_4=(3.62moles)\times (142g/mole)=514g[/tex]
Therefore, the theoretical mass of [tex]Na_2SO_4[/tex] is, 514 grams.
Consider the following reaction where Kc = 0.159 at 723 K. N2(g) + 3H2(g) 2NH3(g) A reaction mixture was found to contain 1.97×10-2 moles of N2(g), 3.82×10-2 moles of H2(g) and 5.27×10-4 moles of NH3(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc, equals 9.63x10^4 . The reaction b A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.
Answer:
The reaction moves to the left. To the reactants side
Explanation:
The reaction given by the problem is:
[tex]N_2_(_g_)~+~3H_2_(_g_)~<=>~2NH_3_(_g_)[/tex]
Therefore the equilibrium constant expression would be:
[tex]K_e_q=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
With the values equilibrium concentration values we can calculate the quotient "Qc", so:
[tex]Q_c=\frac{[5.27X10^-^4]^2}{[1.97X10^-^2][3.82X10^-^2]^3}=0.251[/tex]
In this case Qc>Kc we will have more amount of product. Therefore, the reaction will go to the left to reach the equilibrium.
I hope it helps!
What is the temperature, in degrees Celsius, of a substance with a temperature of 49K? –322°C –224°C 224°C 322°C
Answer:
-224ºC
Explanation:
Answer:
-224c
Explanation:
The salt used to malt snow on a walkway outside a house is calcium chloride and is sold in 20kg bags. How many molecules of calcium chloride are in the bag?
Acid-Base Reactions Custom Learning Goal: To learn to classify acids and bases and to predict the products of neutralization reactions. Acids are substances that ionize to form free H ions in solution whereas bases are substances that combine with H ions. Strong acids and strong bases completely ionize but weak acid and weak bases only partially ionize. A salt is a term for an ionic compound such as NaCl or MgBrz When an acid and a base are mixed together, a neutralization reaction occurs. The products of an acid-base reaction do not have the chemical characteristics of either the acid or the base that originally reacted. Part A Classify each substance as a strong acid, strong base, weak acid, or weak base.
Answer:
HCL, H2SO4 and HNO3 are strong acids.
NaOH, KOH and LiOH are strong bases.
H3PO4, CH3COOH and HF are weak acids.
NH4OH, N2H4 and Zn(OH)2 are weak bases.
Explanation:
HCL, H2SO4 and HNO3 are strong acids because they completely converted into ions and gives H ion in the solution while H3PO4, CH3COOH and HF are weak acids due to their half ionization in the solution. NaOH, KOH and LiOH are strong bases completely dissociate in the solution and give OH ion while NH4OH, N2H4 and Zn(OH)2 are weak bases which cannot dissociate completely and yield less amount of OH ions. when acid and base combine together, they exchange their partners produces salt and water.
How does the A Hreaction relate to the A He of molecules involved in a reaction?
A. A Hreaction = A Hf, reactants
B. AHreaction = A Hf, products - AHf, reactants
C. A Hreaction = AHf, products
D. A Hreaction = A Ht, products + AHf, reactants
Answer:
B. ΔHreaction = ΔH°f reactants- ΔH°f products
Explanation:
Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction.
Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).
For example, in ΔH°f of H₂O, the equation is:
H₂(g) + 1/2O₂(g) → H₂O(g)
The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).
Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:
NaOH + HCl → H₂O + NaCl. ΔHr
The ΔH°f for each substance in the reaction is:
NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH (1)
HCl: 1/2H₂ + 1/2Cl₂ → HCl (2)
H₂O: H₂ + 1/2O₂ → H₂O (3)
NaCl: Na + 1/2Cl₂ → NaCl (4)
The algebraic sum of (3) + (4) is -(ΔH°f reactants):
H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants
This reaction - {(1)+(2)} ΔH°f products
NaOH + HCl → H₂O + NaCl.
ΔHr = ΔH°f reactants- ΔH°f products
In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:
B. ΔHreaction = ΔH°f reactants- ΔH°f productsA volume of 38.7 mL of H2O is initially at 28.0 oC. A chilled glass marble weighing 4.00 g with a heat capacity of 3.52 J/oC is placed in the water. If the final temperature of the system is 26.1 oC , what was the initial temperature of the marble? Water has a density of 1.00 g/mL and a specific heat of 4.18 J/goC. Enter your answer numerically, to three significant figures and in terms of oC.
Answer:
- 61.2°C = Initial T°
Explanation:
This is a calorimetry problem, where the heat from the water was gained by the marble.
Q = m . C . ΔT
where ΔT is final T° - initial T°, C is the specific heat and m, mass.
By the volume of water, we realize the mass (we apply density):
1 g/mL = mass / 38.7 mL
38.7 g = mass of water
Now, we need to find out the specific heat for the marble and we have Heat Capacity data
Heat Capacity = C . m
3.52 J/°C / 4 g = C → specific heat → 0.88 J/g °C
We make the equations for both heats:
m . C . ΔT from water = m . C . ΔT from the marble
38.7 g . 4.18 J/g°C ( 26.1°C - 28°C) = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)
307.35 J = 4 g . 0.88 J/g °C . (26.1°C - Initial T°)
- 307.35 is a negative value, because the was has decreased the temperature, is a loss of heat, but we have to work with the positive number.
307.35 J / (4 . 0.88 °C/J) = 26.1°C - Initial T°
87.32°C = 26.1°C - Initial T°
- 61.2°C = Initial T°
To identify a liquid substance, a student determined its density. Using a graduated cylinder, she measured out a 45-mL sample of the substance. She then measured the mass of the sample, find ing that it weighed 38.5 g. She knew that the substance had to be either isopropyl alcohol (density 0. 785 g/ mL) or toluene ( density 0.866g/ mL). What are the calculated density and the probable identity of the substance?
Answer:
Calculated density = 0.8555g/ mL
Probable identity = Toulene
Explanation:
First off, the substance choices are;
isopropyl alcohol (density 0. 785 g/ mL)
toluene ( density 0.866g/ mL)
So upon calculation, if the density of the liquid substance is equal to either of the substances in the choice, it is that substance.
The relationship between density, mass and volume is given as;
Density = Mass / Volume
Mass = 38.5g
Volume = 45ml
Density = 38.5 / 45 = 0.8555g/ mL
Calculated density = 0.8555g/ mL
Among both substances, it is closest to toluene ( density 0.866g/ mL).
Probable identity = Toulene
What is the wavelength of radiation that has a frequency of 6.912 × 10^15 s–1?
Answer:
Wavelength = 4.34x10⁻⁸m
Explanation:
In a wave, frequency is inversely proportional to wavelength where the constant is the speed of light. The equation is:
Wavelength = Speed of light / Frequency
The speed of light is 3.0x10⁸m/s
Replacing:
Wavelength = 3.0x10⁸m/s / 6.912x10¹⁵s⁻¹
Wavelength = 4.34x10⁻⁸mWhich food has the least amount of monounsaturated fat?
Answer:
Butter
Explanation:
Assume you have a table like the one below.
It tells you the mass of each type of fat in a 14 g sample of food.
Start at the top of the Monounsaturated Fat column.
Go straight down until you reach the smallest number in the column (3.7).
Then, move horizontally left until you reach the Food column.
The food with the least amount of monounsaturated fat is butter.
what is the maximum number of electrons in p&q shell
Answer:
6 electrons
Explanation:
:)
hope this helps :))))))))
Answer:
each p shell can hold max. 6 electorns
i dont know what a q shell is
A buffered solution has a pH of 7.5. What would happen to the pH if a small
amount of acid were added?
Answer:
Dear user,
Answer to your query is provided below
When small amount of acid was added to buffered solution, pH will change very less.
Explanation:
Buffer solution resists change in ph on adding small amount of acid or base but when we calculate the value of buffer capacity we take the change in ph when we add acid or base to 1 lit solution of buffer.This contradicts the definition of buffer solution.
A sample of carbon dioxide gas (CO2) contains 6 x 1022 molecules. How many moles of carbon dioxide does this represent?
Answer:
Explanation:
1 ST METHOD
GIVEN DATA:
number of molecules=6.022×10²²molecules
Avogodro's number Na=6.022×10²²
TO FIND:
number of moles
SOLUTION:
As we knom that number of moles=number of molecules/Na
number of moles=6.022×10²²/6.022×10²²
number of moles=1 moles
2ND METHOD:
AS WE KNOW THAT A MOLE OF A SUBSTANCE CONTAINS 6.022×10²² PARTICLES OF THAT SUBATANCE SO
1MOLE OF CO2=6.022×10²²MOLECULES OF CO2
Amino acids are the building blocks of proteins. The simplest amino acid is glycine . Draw a Lewis structure for glycine. (Hint: The central atoms in the skeletal structure are nitrogen bonded to carbon which is bonded to another carbon. The two oxygen atoms are bonded directly to the right-most carbon atom.
Answer:
Sorry for the lack of precision, if you have any questions you can consult me again.
Explanation:
Glycine is an amino acid, forms proteins and is also called in its molecular chemical formula as C2H5NO2
The pressure of sulfur dioxide (SO2) is 2.27 x 104 Pa. There are 418 moles of this gas in a volume of 57.9 m3. Find the translational rms speed of the sulfur dioxide molecules.
Answer:
The translational rms speed of the sulfur dioxide molecules. vrms = 52.8 m/s
Explanation:
The root-mean-square speed is the measure of the speed of particles in a gas. It is given by the formula below;
vrms=√3RT/M
where vrms is the root-mean-square of the velocity, M is the molar mass of the gas in kilograms per mole, R is the molar gas constant, and T is the temperature in Kelvin.
Molar mass of SO₂ in Kg/ mole = (64/1000) Kg/mol = 0.064 kg/mol
R = 8.314 m³.Pa/K*mol
T = ?
Using PV = nRT to find T
T = PV/nR
Substituting for T in the rms formula
vrms = √(3R*PV/nR)/M
vrms = √3PV/nM
vrms = √3 * 2.27 * 10⁴ * 57.9)/418 * 0.064
vrms = 52.8 m/s
Why are covalent substances gases and liquid rather than solids?
Covalent compounds are held together with an intra molecular attraction which is weaker than metallic bond
hence covalent compounds exist as liquids, gases and soft solids
Write the chemical reaction for hydrogen thiocyanate in water, whose equilibrium constant is Ka. Include the physical states for each species
Write the chemical reaction for thiocyanate ion in water, whose equilibrium constant is Kb. Include the physical states for each species
Answer:
HSCN (aq) + H₂O(l) ⇄ SCN⁻(aq) + H₃O⁺(aq) Ka
SCN⁻ (aq) + H₂O(l) ⇄ HSCN (aq) + OH⁻(aq) Kb
Explanation:
We identify the formula:
HSCN → hydrogen thiocyanate which is also known as Thiocyanic acid
HSCN (aq) + H₂O(l) ⇄ SCN⁻(aq) + H₃O⁺(aq) Ka
As an acid, it gives proton to the solution. It is a weak acid, because the Ka
Ka = [SCN⁻] . [H₃O⁺] / [HSCN]
As a weak acid, the thiocyanate ion, will be the conjugate strong base. In water It can make hydrolisis:
SCN⁻ (aq) + H₂O(l) ⇄ HSCN (aq) + OH⁻(aq) Kb
As a base, it takes a proton from water.
Kb = [HSCN] . [OH⁻] / [SCN⁻]
H-S-C-N is the structure of hydro-thi-ocyanate.
H-S-C-N structure:1. combines with water, it produces the ion thiocyanate and the ion hydronium.
2. When thiocyanate combines with water, it produces the ion hydrogen thiocyanate and the ion hydroxy.
Find out more information about 'Thiocyanate'.
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