Answer:
gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata
Explanation:
vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs
Name the following molecule
Answer:
It is a Biological Molecule
وزن الملي مكافئ لحامض الخليك
Answer:
hshssytdtctdyeheb
Explanation:
yye6d66d6d6dududyydydydyehwj2
Write the separation scheme for the isolation of triphenylmethanol from the reaction mixture once the reaction is complete. The separation begins after the addition of HCl and water to the reaction and includes the column chromatography procedure to further purify crude triphenylmethanol isolated in the day 1 procedure.
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The absorption spectrum of neon has a line at 633 nm. What is the energy of this line? (The speed of light in a vacuum is 3.00 x 108 m/s, and Planck's constant is 6.626 x 10-34 J·s.)
Answer:
B
Explanation:
E = hc/[tex]\lambda[/tex]. Remember, it is in meters not nanometers so you have to convert. You end up with B.
Gaseous ammonia chemically reacts with oxygen gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of nitrogen monoxide produced by the reaction of 1.5 moles of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Explanation:
here's the answer to your question
3. A mixture of gases- oxygen, carbon dioxide and nitrogen - exerts a total pressure of 107.6 kPa. If the oxygen
exerts 45.8 kPa and the carbon dioxide exerts 37.1 KPA, then what is the partial pressure of the nitrogen?
Answer: 23.7kPa
Explanation: Given:
P total= Sum of partial P of all gases in the mix P total = PO2 +PN2 +PCO2 107.6 = 45.8 + PN2 +37.1 PN2= 107.6- (45.8+ 37.1) = 107.6- 83.1 = 23.7kPA Therefore Partial pressure of Nitrogen is 23.7 kPAPredict the Normality of H2SO4 if 75 ml of 96.6 % pure H2SO4 added to 425 ml water. The density of H2SO4 is 1.83 g/cm3?
Explanation:
Normality is one of the concentration terms.
It is expressed as:
[tex]N=\frac{mass of the substance}{equivalent mass}* \frac{1}{volume of solution in L.}[/tex]
The volume of the solution is 425 mL.
Mass of sulfuric acid given is:
[tex]mass=volume * purity* density\\ = 75 mL * 0.966 * 1.83 g/mL\\\\=132.5 grams\\[/tex]
The equivalent mass of sulfuric acid is 49.0g/equivalents
Hence, the normality of the given solution is:
[tex]N=\frac{132.5g}{49.0g/equi.} *\frac{1000}{425mL} \\Normality=6.36N[/tex]
Answer is: 6.36N.
An intravenous solution was prepared by add-in 13.252 g of dextrose (C6H12O6) and 0.686 g of sodium chloride to a 250.0 mL volumetric flask and diluting to the calibration mark with water. What is the molarity of each component of the solution
Explanation:
Molarity(M) of a solution is defined as the number of moles of solute(n) present in one liter of solution(V).
[tex]M=\frac{n}{V}[/tex]
The number of moles(n) can be calculated as shown below:
[tex]n=\frac{mass of solute}{molar mass of solute}[/tex]
Molar mass of dextrose is 198.17 g/mol
Molar mass of NaCl is 58.5 g/mol.
Volume of the solution =250.0mL=0.250 L
The number of moels of dextrose([tex]n_{d}[/tex]) is:
[tex]n_{d} =\frac{13.252g}{198.17g/mol} \\=0.0669mol[/tex]
The number of moles of NaCl is:
[tex]n_{NaCl} =\frac{0.686 g}{58.5g/mol} \\\\=0.01177 mol[/tex]
Thus, the molarity of dextrose is:
[tex]M_d=\frac{n_d}{0.250 L} \\=0.0669mol/0.250L\\=0.268 M[/tex]
The molarity of NaCl is:
[tex]M_Na_Cl=\frac{n_d}{0.250 L} \\\\=0.0118mol/0.250L\\\\=0.0472 M[/tex]
Answer:
The molarity of dextrose is 0.268 M.
The molarity of NaCl is 0.0472 M.
Enzyme catalyzing breakdown of atp to adp
Answer:
ATP hydrolase
Explanation:
Enzymes are biological catalysts which perform diverse functions in the body. Enzymes are specific in their mode of action because an enzyme fits into its substrate as a key fits into a lock.
The particular enzyme that catalyzes the breakdown of ATP to ADP is ATP hydrolase. The phosphate released by the action of this enzyme is used in the phosphorylation of other compounds thereby making them more reactive.
You are given a metal sample that you are told is gold. Explain in a step-by-step procedure exactly how you could (a) determine if the metal is actually gold and (b) determine the purity of the gold if you know what other metals may be present. Write out your answer in a clear and well supported paragraph.
Answer:
The answer is provided below
Explanation:
To determine the metal is gold we will use the following steps
Calculate the density of the MetalTake the density of the pure goldCompare both densitiesTake a full water container
Place the metal in the container
Collect the water that spills out due to the placement of the metal
measure the mass of collected water.
Calculate the value in terms of the density of water, it will be the volume of metal.
Calculate the mass of the metal
Use the following formula to calculate the density of the metal
Density = Mass / Volume
Now compre the resulted density to the density of pure gold.
Why does increasing the temperature of two reactants in solution make a
reaction proceed more quickly?
Answer:
-The particles of the two reactants will gain kinetic energy and collide with one another more frequently and forcefully, which makes the reaction take place more quickly
How much BaSO4 can be formed from 196.0 g of H2SO4?
Answer:
a) You can form 466 g of BaSO₄.
Explanation:
a) Mass of BaSO4
196 g H₂SO4 × 1 mol H₂SO4
98.08 g H₂SO4
1 mol BaSO 1 mol H₂SO4 X X
466 g BaSO4
233.39 g BaSO4
1 mol BaSO4
How many molecules (or formula units) are in 138.56 g C4H10 Express your answer using four significant figures.
Answer:
dont buy cheap and off we went
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode where Co2 (aq) is reduced to Co (s) . Assume all aqueous solutions have a concentration of 1 mol/L.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
What is the oxidation state of nitrogen in N ?
Answer:
+5
Explanation:
The diagram above shows the hydrides of groups 14, 15, 16, and 17 elements. Why does H20, HF, and NH3 have much higher boiling points than the rest of the molecules in their groups?
Answer:
Hydrogen Bonding
Explanation:
Hydrogen Bonding occurs when a hydrogen atom is bonded to N, O, and F atoms.
The molecules H₂O, HF, and NH₃ all experience hydrogen bonding, which is a relatively strong IMF, causing the molecules to have stronger attraction to each other. Having a stronger attraction between molecules results in more energy required to separate them, thus these molecules will have a higher boiling point than the rest of the molecules in their group.
A researcher is attempting to produce ethanol using an enzyme catalyzed batch reactor. The ethanol is produced from corn starch by first-order kinetics with a rate constant of 0.05 hr-1. Assuming the concentration of ethanol initially is 1 mg/L, what will be the concentration of ethanol (in mg/L) after 24 hours
Answer:
The correct solution is "3.32 gm/L".
Explanation:
Given:
Rate constant,
[tex]K = 0.05 \ hr^{-1}[/tex]
Time,
[tex]t = 24 \ hours[/tex]
Concentration of ethanol,
[tex]C_o= 1 \ mg/L[/tex]
Now,
The concentration of ethanol after 24 hours will be:
⇒ [tex]C_o=C\times e^{-K\times t}[/tex]
By putting the values, we get
[tex]1=C\times e^{-0.05\times 24}[/tex]
[tex]1=C\times 0.30119[/tex]
[tex]C= 3.32 \ gm/L[/tex]
Suppose we have two rock samples, A and B. Rock A was subject to both physical and chemical weathering while rock B was subject to chemical weathering only. Which rock would experience more chemical weathering? Why? (2pts) (Hint: consider the effect of surface area on the rate of chemical weathering)
Answer:
Rock A will have far more chemical weathering than Rock B due to the rise in area effect
Explanation:
Rock A undergoes both Physical and Chemical weathering. So, thanks to physical weathering there'll appear cracks within the rock, which can, in turn, increase the area of rock on which weathering is occurring. So, Chemical weathering will happen much faster now as there's a rise in the area. within the case of Rock B, there's only chemical weathering therefore the increase in the area won't be that very much like compared to Rock A.
1. When 6.0 grams of zinc are dropped into excess hydrochloric acid, how many grams of zinc chloride will be produced?
2. When 45.0 grams of copper (II) carbonate are decomposed with heat, how many grams of carbon dioxide will be produced? (Teachers note: Other product is copper (II) oxide.)
Please explain as well if possible! Thanks.
Explanation:
here are the answers for your questions
I basically converted the given grams to moles, and then multiplied that by the product-to-reactant ratio in the equation, and then convert that to grams
Answer:
Explanation:
1.
The reaction can be represented by the equation: Zn + 2 HCl -> ZnCl2 + H2
From the equation, molar ratio of Zn and ZnCl2 is 1:1.
Molar mass of Zn = 65.38
Molar mass of ZnCl2 = 65.38 + 35.45*2 = 136.28
So 6.0 grams of Zn will produce 6 / 65.38 * 136.28 = 12.5 grams
2.
As the only other product is copper (II) oxide, the reaction can be represented by the equation: CuCO3 -> CuO + CO2
From the equation, molar ratio of CuCO3 and CuO is 1:1.
Molar mass of CuCO3 = 123.55
Molar mass of CuO = 79.55
So 45.0 grams of CuCO3 will produce 45 / 123.55 * 79.55 = 28.96 grams
calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation
Answer:
The pressure is 5.62 atm.
Explanation:
An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
P= ?V= 5.005 Ln= 1.255 molR= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 273.5 KReplacing:
P* 5.005 L= 1.255 mol* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] *273.5 K
Solving:
[tex]P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}[/tex]
P= 5.62 atm
The pressure is 5.62 atm.
A rigid, sealed container that can hold 26 L of gas is filled to a pressure of
5.97 atm at 374 °C. The pressure suddenly decreases to 3.64 atm. What is
the new temperature inside the container, in units of °C?
Answer:
121 °C
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 374 °C
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
NOTE: Volume = constant
Next, we shall convert 374 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 374 °C
Initial temperature (T₁) = 374 °C + 273
Initial temperature (T₁) = 647 K
Next, we shall determine the final temperature. This can be obtained as follow:
Initial pressure (P₁) = 5.97 atm
Initial temperature (T₁) = 647 K
Final pressure (P₂) = 3.64 atm
Final temperature (T₂) =?
P₁ / T₁ = P₂ / T₂
5.97 / 647 = 3.64 / T₂
Cross multiply
5.97 × T₂ = 647 × 3.64
5.97 × T₂ = 2355.08
Divide both side by 5.97
T₂ = 2355.08 / 5.97
T₂ = 394 K
Finally, we shall convert 394 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
Final temperature (T₂) = 394 K
Final temperature (T₂) = 394 – 273
Final temperature (T₂) = 121 °C
Thus, the new temperature is 121 °C
Why did Rutherford choose alpha particles in his experiment?
how does iron I differ from iron II
Answer:
Metals tend to form positive oxidation states. Here, Iron (I) has an oxidation state of +1 while Iron (II) has an oxidation state of +2. Similarly, Lead (I) has an oxidation state of +1 while Lead(II) has an oxidation state of +2. A change in oxidation state can rather cause significant changes in the compound.
how does lead resemble chromium?
A solution is made by dissolving 20 ml of acetic acid in 180ml of water. Calculate its volume concentration
Answer:
water is 9/10
chemical would be 1/10
Explanation:
180/200 would be water concentration in solution
and 20/200 would be chemical solution concentration in solution (if the chemical were to be polar and mix)
Based on the "Reactivity in Substitution Reactions" experiment, which molecule would be expected to react the fastest using AgNO3 in water-ethanol ?
Answer:
C) EtOH 1% AgNO3
Somebody help me!!
Calculate the mass of 2.046L of NO2
1. Why is it necessary to equalize the pressure(i.e, have the water level the same in each tube) before taking a volume reading?
2. Why is it important to use water that has been pre-saturated with CO2 in the gas burettes?
3.If your antacid sample had been contaminated by moisture, what effect(if any )would you expect this to have on your result
4.Explain why an'antacid is called as such,what is the role of the NAHCO3 or CACO3 in reactions?
Answer:
If you contact water with a gas at a certain temperature and (partial) pressure, the concentration of the gas in the water will reach an equilibrium ('saturation') according to Henry's law.
Explanation:
This means: if you increase the pressure (e.g. by keeping the vial closed), the CO2 concentration will increase. So it simply depends what concentration you need for your assay: 'CO2-saturated' water at low pressure or 'CO2-saturated' water at high pressure.
A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
Explanation:
Ore contains ---- 65.2% [tex]TiO_2[/tex]
Mass% of titanium in TiO2 can be calculated as shown below:
[tex]mass percentage of Ti in TiO2=\frac{mass of Ti}{mass of TiO_2} *100\\=(47.86g/79.866g)* 100\\=59.9[/tex]
Given 10.0 metric tons of titanium is required.
The mass of ore that should be processed can be calculated as shown below:
Mass of Ti = ore x TiO2 % x Ti mass %
10.0 x 1000 kg = M (mass of ore) x (65.2/100) x (59.9/100) Ti
=>M=(10.0 metric tons) / (0.652 x 0.599)
=>M=25.6 metric tons
Hence, the mass of ore required is 25.6 metric tons.
The minimum amount of rutile ore to be processed is 25.6 metric tons.
Explanation:
Given :
The sample of rutile, 65.2% [tex]TiO_2[/tex] (s) by mass.
To find:
The minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
Solution:
The atomic mass of titanium = 47.867 g/mol
The atomic mass of oxygen = 15.999 g/mol
The molar mass of [tex]TiO_2[/tex] [tex]= 47.867 g/mol+2\times 15.999 g/mol=79.865 g/mol[/tex]
Percentage of the titanium in [tex]TiO_2[/tex] :
[tex]Ti(\%)=\frac{1\times 47.867 g/mol}{79.865 g/mol}\times 100\\=59.935\%[/tex]
Quantity of titanium required = 10.0 metric ton
Quantity of [tex]TiO_2[/tex] from rutile ore = x
[tex]59.935\%=\frac{10.0 \text{metric ton}}{x}\times 100\\x=\frac{10.0 \text{metric ton}}{59.935}\times 100=16.7 \text{metric ton}[/tex]
Mass of [tex]TiO_2[/tex] form rutile ore= 16.7 metric ton
The percentage of [tex]TiO_2[/tex] in rutile = 65.2 %
The quantity of rutile ore to be processed = M
[tex]65.2\%=\frac{16.7 \text{metric ton}}{M}\times 100\\M=\frac{16.7 \text{metric ton}}{65.2}\times 100=25.6 \text{metric ton}[/tex]
The minimum amount of rutile ore to be processed is 25.6 metric tons.
Learn more about the percent of an element in the compound here:
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Classify each of the reactions listed below as a single-displacement, double-displacement, synthesis,
decomposition, oxidation reduction or combustion reaction.
Reaction Type
: 2Na + Cl2 → 2NaCl
: C2H4 + 3O2 → 2CO2 + 2H2O
: 2Ag2O-> 4Ag + O2
: BaCl2 + Na2SO4->BaSO4 +2NaCl
: 2AI + Fe2O3-> 2Fe + Al2O3
