Many types of fungi are responsible for the breakdown and
recycling of organic material in an ecosystem. Fungi are an
example of a(n)
Answer:
mushroom are a type of fungi
Explanation:
What is the mass of 4.50mol Ca
Any time a reaction procedure involves separation steps during the workup, it is a good idea to outline a separation scheme.
a. True
b. False
Answer:
True
Explanation:
Benzene is formed as a result when benzoic acid is generated by Grignard reaction. We must therefore isolate it from the desired benzoic acid compound, as the rest is not dismissed after dissolution in aqueous NAOH solution. Therefore, it a good idea to outline a separation scheme for such cases.
This implies that the given statement is true.
This equation shows the combustion of methanol.
2CH2OH(1) +302(g) → 2C02(g) + 4H2O(g)
How many liters of carbon dioxide gas at STP are produced by the combustion of 64.0 g of methanol
(CH3OH)
A. 44.8L
B. 89.6 L
C. 11.2L. D. 22.4L
Answer: 44.8 L
Explination:
44.8 liters of carbon dioxide gas at STP are produced by the combustion of 64.0 g of methanol. Therefore, option A is correct.
What is combustion ?Combustion is a chemical reaction between two or more substances, usually involving oxygen, that produces heat and light in the form of a flame.
Combustion is a high-temperature exothermic redox chemical reaction that occurs between a fuel and an oxidant, frequently atmospheric oxygen, to produce oxidized, often gaseous products in a mixture known as smoke.
The flash point of methanol is extremely low. This means that even trace amounts of ignition material can start a fire. 44.8 liters of carbon dioxide gas at STP are produced by the combustion of 64.0 g of methanol.
Thus, option A is correct.
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if a compound formula mass was experimentally determined to be 58, the chemical formula could be?
Answer:
NaCl
Explanation:
If you follow traditional rounding, Na atomic mass is 23 and Cl atomic mass 35, 23 +35 = 58.
I was taught that Cl is one of the elements you round to 35.5, but this works if you’re not in an advanced chem class.
how proved that carbon is present in sugar?
In all living organisms, in substances like carbohydrate, starch, urea etc. To prove sugar contains carbon, take a little amount of sugar in spoon and heat it gently till it melts and turned into yellowish color. On further heating it changes into brown colored mass and then black, which is charcoal.
Answer: sugar treated with strong sulphur acid turns to black
Explanation: sulphur acid removes water from sugars and carbon is left in residue.
This should not to try at home. Water boils when high temperature is achieved.
Also smell of burned sugar occurs
TRUST PLZ HELP MATCH THE BOXES
Answer:
Transform = the third one
convergent = the first one
divergent = the second one
Explanation:
what is molecular attraction? Am I right?
Answer:
tendency of molecules with unlike electrical charges to attract each other.
Explanation:
Ethanol in the body is oxidized to acetaldehyde by liver alcohol dehydrogenase (LADH). Other alcohols are also oxidized by LADH. For example, methanol, which is mildly intoxicating, is oxidized by LADH to the quite toxic product of formaldehyde. The toxic effects of ingesting methanol (a component of many commercial solvents) can be reduced by administering ethanol. The ethanol acts as a competitive inhibitor of methanol by displacing it from LADH. This provides sufficient time for the methanol to be harmlessly excreted by the kidneys. If an individual has ingested 30 mL of methanol (a lethal dose), how much 100 proof whiskey (50% ethanol by volume) must be imbibed to reduce the activity of her LADH towards methanol to 5% of its original value? The adult human body contains ~40L of aqueous fluids throughout which ingested alcohols are rapidly and uniformly mixed. The densities of ethanol and methanol are both 0.79 g/cm3. Assume the KM values of LADH for ethanol and methanol to be 10-3 M and 10-2 M, respectively, and that Ki = KM for ethanol
Answer: The changes are the mL of methanol ingestion (30mL) and the proof of the whiskey (40% ethanol; 80 proof), and the percentage of reduction (1%). I applied the formulas from the last post in reference to this question but am completely lost. The formulas used last time were: 2. Formulas Applied alpha=1 + ([etOH]/KetOH) (V[meOH]/V[etOH])= (Vmax*[meOH]/KmeOH+[meOH])/(Vmax*[meOH]/alpha*KmeOH+[meOH]) which reduces to (V[meOH]/V[etOH])=(alpha*KmeOH+[meOH])/(KmeOH+[meOH]) 3. My attempt Molarity of methanol: 30mL; which equates to 23.7g of methanol; in 40L that is equal to 0.5925 g/L Dividing the molecular weight by 32.04g/mol I get 0.0184925 which is approximately 0.02M; Km is 0.01M Since the molar mass of methanol and ethanol are two fold, I can multiply the g/l by 4. However, unlike the previous problem, I cannot multiply by 2 because I do not have 50% EtOH, so because 40 is less than 50 I assume to multiply by 2.5 yielding: (30mL)(4)(2.5)=300mL 300mL of EtOH to effectively reduce the Methanol to 1%.
1. a) Define the term electricity ?
Answer:
a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current.
Explanation:
Which of the different components of smog depicted in the graph are most likely released from automobile exhaust?
A
A and B
B
B and C
с
A, B, and C
D
B, C, and D
Answer:
B, B and C
Explanation:
The two components (AA, hydrocarbons and BB, nitrogen oxide) are most likely released from automobile exhaust.
What is Smog?This is referred to a type of air pollution which could be from carbon emission etc and reduces the visibility.
(AA, hydrocarbons and BB, nitrogen oxide peaked midmorning when traffic is highest which means they were most likely released from automobile exhaust?.
Read more about Smog here https://brainly.com/question/14029972
A liquid is placed in a closed container and time passes until
the level of the liquid is constant
The general situation described by the rate of evaporation is
equal to the rate of condensation called a:
a
state balance
b static equilibrium
chemical balance
d dynamic equilibrium
Answer: Dynamic equilibrium
Explanation:
What substance is oxidized in the following reaction? 4HCl + MnO2 → Cl2 + 2H2O + MnCl2
Answer:
Cl⁻ was oxidized.
Explanation:
4HCl + MnO₂ → Cl₂ + 2H₂O + MnCl₂Oxidation can be defined as the process in which the oxidation number of a substance increases.
On the left side of the equation, Cl has a charge of -1 (in HCl); while on the right side of the equation Cl has a charge of 0 in Cl₂.
Thus, Cl⁻ was oxidized.
If energy cannot be created or destroyed, what happens to the kinetic and potential energy?
Answer:
According to the law of conservation of energy, energy cannot be created or destroyed, although it can be changed from one form to another. KE + PE = constant. ... As the car coasts down the hill, it moves faster and so it's kinetic energy increases and it's potential energy decreases. hope this helps
Explanation:
Answer:
They cannot be destroyed or created however,
Explanation:
it can only be transformed or transferred from one form to another.
A solution has a [OH-] of 1.43x10-9. What is the pH of this solution?
a. 5.2
b. 8.8
C. 9.1
d. 4.9
Answer:
a. 5.2
Explanation:
pOH = -log(1.43 x 10^-9) = 8.845
pH = 14 - pOH = 14 - 8.845 = 5.155 ≈ 5.2
Heat flows from hotter objects to colder objects until it reaches a state where heat is no longer flowing.What have the objects reached when heat is not longer flowing?
QUESTION 6 Consider the following reaction between the diatomic and monatomic forms of iodine: I2 (g) <-> 2I (g) When 0.095 M I2 is initially placed in a previously empty container and sealed, the system slowly reaches equilibrium. When equilibrium is reached, it is found that there is an equilibrium concentration of 0.0055 M of the monatomic form of iodine. Calculate the (unitless) equilibrium constant Kc. Round your answer to two sig figs, and express it in scientific notation.
Answer: The equilibrium constant is [tex]3.3\times 10^{-4}[/tex]
Explanation:
Initial concentration of [tex]I_2[/tex] = 0.095 M
The given balanced equilibrium reaction is,
[tex]I_2(g)\rightleftharpoons 2I(g)[/tex]
Initial conc. 0.095 M 0 M
At eqm. conc. (0.095-x) M (2x) M
Given : 2x = 0.0055
x = 0.00275
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[l]^2}{[I_2]}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{(0.0055)^2}{(0.095-0.00275)}[/tex]
[tex]K_c=\frac{(0.0055)^2}{0.09225}=0.00033[/tex]
Thus the equilibrium constant is [tex]3.3\times 10^{-4}[/tex]
g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thin film gate of 500 nm in 50 minutes. What is the original field oxide thickness d0 (in nm)?
Answer:
Explanation:
From the information given:
oxidation of oxidized solution takes place at 950° C
For wet oxidation:
The linear and parabolic coefficient can be computed as:
[tex]\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big][/tex]
Using [tex]D_o[/tex] and [tex]E_a[/tex] values obtained from the graph:
Thus;
[tex]\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr[/tex]
[tex]B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr[/tex]
So, the initial time required to grow oxidation is expressed as:
[tex]t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)[/tex]
[tex]where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719[/tex]
∴
[tex]2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)[/tex]
[tex]2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2 \\ \\ t_o(initial) = 1.4267 \ hr[/tex]
NOW;
[tex]1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\ 1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453[/tex]
[tex]d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}[/tex]
[tex]d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \ \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}[/tex]
[tex]d_o =0.02609 \ OR \ -0.0939[/tex]
Thus; since we will consider the positive sign, the initial thickness [tex]d_o[/tex] is ;
≅ 0.261 μm
Simone is helping her mom cook. She takes a glass casserole dish directly from the refrigerator and put it in the oven. Which best explains the result of putting the cold dish in the hot oven?
Group of answer choices
The dish stays cool and will not warm.
The dish quickly changes color.
The dish cracks or breaks.
The dish contracts and shrinks.
Answer:
I think the dish contracts and shrinks
Answer:
The dish cracks or breaks.
Explanation:
When glass changes temperature rapidly, it may undergo thermal shock. This is stress that is produced by a sudden change in temperature.
A mixture of nitrogen and xenon gases contains nitrogen at a partial pressure of 251 mm Hg and xenon at a partial pressure of 472 mm Hg. What is the mole fraction of each gas in the mixture?
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
ANSWERIT AND YOU WILL BE MARKED THE BRAINLIEST
Answer:
The answer is definitely D
Explanation:
What is the concentration of a 500 mL solution with 25 mol of HF? Write your answer with TWO decimal places and round accordingly.
Answer:
oop I'm thinking hmmmmm
A certain metal M forms a soluble nitrate salt M(NO3), Suppose the left half cell ofa galvanic cell apparatus is filled with a 4.00 M solution of M (NO,), and the right half cell with a 20.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 35.0 °C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show?
Answer:
Explanation:
The left electrode will be positive on the grounds that focus on the concentration that the cell electron moves from a lower concentration fixation to a higher concentration. Thus right electrode will go about and act as an anode and will be negative. Also, the left electrode will be the cathode and will be positive.
The concentration cell [tex]E_{Cell} = \dfrac{-0.0591}{1} log \dfrac{0.02}{4}[/tex]
[tex]E_{Cell} =0.136 V[/tex]
= 136 mV
Show your reasoning in describing what would be the effect on the calculated equilibrium constant for each of the following mistakes.
1. Your buret was not rinsed with NaNCS solution before use and still contained traces of distilled water.
2. The spectrophotometer was set well-past infinite absorbance upon calibration with the sample holder empty.
Answer:
The readings of equilibrium constant is incorrect.
Explanation:
There is a mistake in the value of equilibrium constant if the burette contains traces of distilled water because the water will be present in the solution from which equilibrium constant is calculated. The spectrophotometer did not give the correct reading because its calibration is incorrect. If we want to gain correct readings from the spectrophotometer, we must have to calibrate the instrument correctly.
Incorrect value of equilibrium constant is obtained.
What is equilibrium constant?Equilibrium constant of any reaction tells about the relative ration of the concentration of products to the reactants.
Value of equilibrium constant can be calculated in the aqueous medium where concentration of water also present. If we don't rinse the burette with the reaction solution then the already present traces of water affect the concentration of water during the calculation of equilibrium constant.
As the absorbance of any solution is also directly proportional to the concentration, so if the spectrometer is not calibrated accurately then incorrect value of absorbance also affects the value of equilibrium constant.
Hence, we get incorrect value of equilibrium constant for each of the given mistakes.
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2)Explain: How can the shapes of a boat and an airplane help them move faster?
Answer:
Their sleek shape makes them more aerodynamic which reduces the contact force.
Answer: The are formed in a way so they create less friction against the air and water. Helping the jet stream create less traction so it moves faster. Like how a bullet is designed. The pointed front and narrow body helps it move faster through the air. Less things sticking of these object so friction is avoided, and they can easily move speedily. The same thing goes for cars, the more narrow they are as well as pointed makes it easier to push through the air making the wind resistance extremely low.
Hope this helps :)
The ratio of the measured molar volume of a real gas to the measured molar volume if the gas
were to behave ideally is known as
Answer: compressibility
Explanation:
The ratio of the measured molar volume of a real gas to the measured molar volume of an ideal gas is known as compressibility.
Compressibility is defined by the symbol Z and is calculated by the formula :
[tex]Z=\frac{PV}{nRT}[/tex]
For a gas with ideal behavior, molar volume of an ideal gas is equal to the molar volume of a real gas so Z= 1.
Describe two functions of blood.
Answer:
Transports Dissolved gases and Controls pH The pH of blood must remain in the range 6.8 to 7.4, otherwise it begins to damage cells.
Explanation:
main functions tho Transportation. Nutrients from the foods you eat are absorbed in the digestive tract. ...
Defense. Many types of WBCs protect the body from external threats, such as disease-causing bacteria that have entered the bloodstream in a wound.
Maintenance of Homeostasis. Recall that body temperature is regulated via a negative-feedback loop.
Mg(s) + 2HCl (aq) -----> MgCl2 (aq) + H2
You have 6.00 g HCl and 4.8 g Mg.
What is the limiting reaction
Answer:
HCl is the limiting reactant
Explanation:
To find the limiting reactant of a reaction we must find the moles of each one as follows:
Moles Mg (Molar mass: 24.305g/mol)
4.8g Mg * (qmol / 24.305g) = 0.20 moles Mg
Moles HCl (Molar mass: 36.46g/mol)
6.00g HCl* (1mol / 36.46g) = 0.165 moles HCl
Based on the chemical reaction, for a complete reaction of 0.20 moles of Mg are necessaries:
0.20 moles Mg * (2mol HCl / 1mol Mg) = 0.40 moles HCl
As there are just 0.165 moles of HCl:
HCl is the limiting reactantCalculate the Rf value of compounds X and Y based on information (X travels 0.5.0 cm, Y travels 1.0 cm from the origin line; the solvent front is 3.0 cm). Comment on the polarity and separation of X and Y on the TLC plate.
Answer:
Retention factor or [tex]R_f[/tex] value is defined as the ratio of distance traveled by the unknown to the distance traveled by the solvent front.
[tex]R_f=\frac{\text{ distance by unknown}}{\text {distance by solvent}}[/tex]
Putting in the values we get:
[tex]R_f_{X}=\frac{0.50cm}{3.0cm}=0.16[/tex]
[tex]R_f_{Y}=\frac{1.0cm}{3.0cm}=0.33[/tex]
As the [tex]R_f[/tex] value of Y is more than X , it follows that Y compound will be retained less strongly than the compound X. Thus Y is more polar than X.
Uranium, an important component of both nuclear weapons and nuclearreactors, has two major isotopes, U-238, which has a half-life of approximately billion years, and U-235, which has a half-life of approximately million years. Both were present in equal amounts at the time of thecreation of the Earth, billion years ago. How many years after the creationof the Earth had the amount of radiation from uranium decayed to half theamount present at the time of the creation of the Earth
Answer:
Hello your question is poorly written below is the well written question
Uranium, an important component of both nuclear weapons and nuclear reactors, has two major isotopes, U-238, which has a half-life of approximately 4.5 billion years, and U-235, which has a half-life of approximately 700 million years. Both were present in equal amounts at the time of the creation of the Earth, 4.5 billion years ago. How many years after the creation of the Earth had the amount of radiation from uranium decayed to half the amount present at the time of the creation of the Earth
Answer : 140 billion years
Explanation:
Given that :
U-238 h1/2 = 4.5 billion years
U-235 h1/2 = 700 million years
At the beginning both Isotopes where present in equal amount
Determine the T years before the amount of Uranium decays to Half
T = ? N'2 = N1 / 2
we know that N = No ( 1/2 )^h where h = time / half-life time
attached below is the detailed solution of the given problem