Answer:
The main function of the cell membrane is to protect the cell from the outer environment.
Answer: The answer is D.) to separate the inside of the cell from the outside environment
Explanation:
1. How does a virus differ from a common cell?
A. It has no nucleus, cell wall, or organelles.
B. It has two nuclei and no cell wall or organelles.
C. A virus has no cell well, no nucleus, and only organelles for
movement.
D. A virus differs from a cell only in shape.
How much water, in grams, can be made from 1.84 × 1024 hydrogen molecules?
Answer:
55.0g water can be made
Explanation:
To solve this question, we must convert the molecules of H2 to moles using Avogadro's constant. With the moles, and the reaction:
H2 + 1/2O2 → H2O
We can find the moles of H2O = Moles H2 and its mass of using molar mass of water -H2O = 18.01g/mol-
Moles H2 = Moles H2O:
1.84x10²⁴ molecules * (1mol / 6.022x10²³ molecules) = 3.055 moles H2O
Mass:
3.055 moles H2O * (18.01g / mol) = 55.0g water can be made
explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity
PLEASE HELP I HAVE 19 MINUTES LEFT I WILL MARK BRAINLIEST
How much more acidic is a pH of 4 as compared to a pH of 6.5?
Answer:
316.227766
Explanation:
There are four stages to the classical demographic transition model Pre-transitional Europe was characterized by high and
fluctuating mortality and a high birth rate. The transition model began to progress into and through stage 2 in the late 18th and early
19th century. All BUT ONE contributed to the decline in mortality.
S- -1]))
A)
Enacting measures to provide clean water supplies.
B)
Public health advances including quarantine of settlements undergoing
epidemics
The development of vaccines to prevent disease and antibiotics to treat
infection.
D)
Widespread acceptance of germ theory resulting in more hygienic
practices, including hand washing and sterilizing medical equipment and
infants' bottles.
0.850 moles of N2 originally at 85°C is cooled such that it now occupies 17.55L at 1.25 atm. What is the final temperature of the gas?
The final temperature of the gas is 269.9 K, or -3.25 °C, which was calculated with the help of ideal gas equation.
What is the ideal gas equation?The ideal gas equation gives the relation between pressure, volume and temperature.
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
To find the initial volume of the gas. We can use the fact that the number of moles of gas does not change during the cooling process:
n = 0.850 moles
We can also use the ideal gas law to find the initial volume of the gas:
PV = nRT
V = nRT/P
where R = 0.08206 L atm/K mol is the gas constant.
Convert the initial temperature from Celsius to Kelvin:
T1 = 85 °C + 273.15 = 358.15 K
Substitute the given values into the equation:
V₁ = (0.850 mol)(0.08206 L atm/K mol)(358.15 K)/(1 atm) = 24.03 L
Now we can use the combined gas law to find the final temperature:
(P₁V₁/T₁) = (P₂V₂/T₂)
Substitute the given values into the equation:
(1.25 atm)(24.03 L)/(358.15 K) = (P₂) (17.55 L)/(T2)
Solve for T₂:
T₂ = (P₂)(17.55 L)(358.15 K)/(1.25 atm)(24.03 L)
T₂ = 269.9 K
Therefore, the final temperature of the gas is 269.9 K, or -3.25 °C.
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Calculate how many grams of methane (CH4) are in a sealed 800. mL flask at room temperature (22 °C) and 780. mm of pressure. Show work pls.
"0.0340" mol of CH₄ are in sealed flask.
Methane (CH₄)Methane would also be a greenhouse gas, therefore its existence tends to affect humanity's surface temp as well as weather patterns framework; it is released into the atmosphere from such a wide assortment of life forms as well as biogenic.
According to the question,
Volume, V = 800 mL or, 0.800 L
Temperature, T = 22°C or, 295
Pressure, P = [tex]\frac{780}{760}[/tex] = 1.03 atm
As we know the relation,
The gram of moles will be will be:
→ n = [tex]\frac{PV}{RT}[/tex]
By substituting the values, we get
= [tex]\frac{1.03\times 0.800}{0.08206\times 295}[/tex]
= [tex]\frac{0.824}{242.077}[/tex]
= 0.0340
Thus the response above is correct.
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What are the missing coefficients for C3H8 + o2 = Co2 +H2O
Answer: C3H8 + 5O2 = 3Co2 +4H2O
Explanation: Equations must be balanced
You must have the same amount of C
H and O on both sides of the equation
Explain what matter is, and all of the states it can have.
Answer:
matter is anything that occupies space
states of matter : solid,liquid, gas,plasma
Answer:
matter can be anything, tables chairs, literally anything. it has volume and takes up space.
Explanation:
Solids, liquids, gases, plasmas, and Bose-Einstein condensates (BEC)
130 cm of a gas at 20°C exerts a pressure of
750 mm Hg. Calculate its pressure if its volume
is increased to 150 cm3 at 35 °C.
Answer: The pressure is 1137.5 mm Hg its pressure if its volume is increased to 150 [tex]cm^{3}[/tex] at 35 °C
Explanation:
Given: [tex]P_{1}[/tex] = 750 mm Hg, [tex]V_{1} = 130 cm^{3}[/tex], [tex]T_{1} = 20^{o}C[/tex]
[tex]P_{2}[/tex] = ?, [tex]V_{2} = 150 cm^{3}[/tex], [tex]T_{2} = 35^{o}C[/tex]
Formula used to calculate the new pressure is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{750 mm Hg \times 130 cm^{3}}{20^{o}C} = \frac{P_{2} \times 150 cm^{3}}{35^{o}C}\\P_{2} = 1137.5 mm Hg[/tex]
Thus, we can conclude that the pressure is 1137.5 mm Hg its pressure if its volume is increased to 150 [tex]cm^{3}[/tex] at 35 °C.
write half-reactions that show how H2O2 can act as either an oxidizing agent or a reducing agent, and describe where each of these situations occurred in your testing.
Answer:
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
Explanation:
When
H2O2 acts as an oxidizing agent
H2O2 + 2e- 2H+---> 2H2O
Reducing agent
H2O2 --> O2 + 2e + 2H+
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
Robert Hooke observed a slice of cork under a microscope. He saw what looked like “pores” or “cells” in it. Hooke can assume that the cork is most likely - Answer F A man made material G A type of living organism H A type of mineral J A petroleum based product
The cork oak tree from which cork is extracted is native to southwest europe and northwest africa. Cork is extracted from cork oak trees without harming the tree. So cork is a type of mineral. The correct option is C.
What is a cork cell?The dead cells without having intercellular spaces are defined as the cork cells. They appear at the periphery of roots and stems when they grow older and increase in girth. They also have a chemical called suberin in their walls.
It is the suberin which makes them impervious to gases and water. The outer protective coat of a tree is called the cork. It is one of the components of bark of the tree. The tissues of bark become old and the secondary mersitem replaces them.
Cork is made up of multiple thick layers and it protects the tree from bacterial or fungal infection.
Thus the correct option is C.
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A balloon is inflated to a volume of 8.0 L on a day when the atmospheric pressure is 1.013 bar . The next day, a storm front arrives, and the atmospheric pressure drops to 0.968 bar . Assuming the temperature remains constant, what is the new volume of the balloon, in liters
Answer:
[tex]V_2=8.4L[/tex]
Explanation:
Hello there!
In this case, according to the definition of the Boyle's law, which describes de pressure-volume behavior as an inversely proportional relationship, it is possible for us to write:
[tex]P_1V_1=P_2V_2[/tex]
Thus, since we are given the initial pressure and temperature, and the final pressure, we are able to calculate the final volume as shown below:
[tex]baV_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{8.0L*1.013bar}{ 0.968bar}\\\\V_2=8.4L[/tex]
Regards!
when rolling a number cube 500 times, how many times you expect to get a 3?
Answer:
[tex]\frac{250}{3}[/tex]
Explanation:
you can expect to get a 3 (theoretically) 1 time every 6 times you roll. A 1/6 chance.
Here's the equation:
[tex]\frac{1}{6} =\frac{x}{500}[/tex]
cross multiply (i think that's what it is called)
500=6x
divide by 6 on both sides:
x=[tex]\frac{250}{3}[/tex] or approx 83 times.
Hope this helps! Lmk if u have more questions <3
What is the relative humidity of the air when the dry-bulb temperature is 4°C and the dewpoint is -4°C?
1.
42%
2.
46%
3
51%
4
56%
Answer:56%
Explanation:
In the dewpoint chart when you line it up it ends up at 56%
The relative humidity of the air when the dry temperature is about 4 degrees and the dew point is about -4 degrees C is 56%. Hence option 4 is correct.
What is dewpoint?The dew point is the temperature at which the air needs to be colled at a ta constant pressure in order to have an RH of 100%. The dew point is minus four degrees and the RH of the air is 4 degrees.
For the forces to initiate the air must get cooler and a constant addition of water pressure should be there. Thus substrating the temperature of the wet bulb, putting the thermometer on the dry bulb, and making use of the RH chart tells us that RH is 56%
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Calculate the volume of solvent present in a 55.5%
by volume of 10.5 mL alcohol solution.
Answer:
I dont know
Explanation:
good luck
The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts
Answer:
E: the point where the acid and base have been added in proper stoichiometric amounts
Explanation:
Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.
The Equivalence point occurs before the endpoint.
Thus, option E is correct.
What does the cell theory state? Answer F All organisms are composed of a nucleus G All prokaryotes are composed of multiple cells H All prokaryotes are single celled organisms J All organisms are composed of cells
Answer:
(J) All organisms are composed of cells
A student weighs 0.347 g of KHP on a laboratory balance. The KHP was titrated with NaOH and the concentration of the NaOH determined to be 0.110 M. For the second titration, the student correctly diluted 6 M HCl from the reagent shelf using a graduated cylinder to obtain approximately 0.6 M HCl. This solution was titrated with the original NaOH solution. The student calculated the concentration of NaOH from the experiment to be 0.099 M. In which experiment should the student be more confident of the concentration of the NaOH solution
Answer:
Following are the solution to the given question:
Explanation:
Each method through KHP is somewhat more precise since we have weighed that requisite quantity, we exactly know the KHP intensity appropriately. Its initial 6 M HCl concentration was never considered mandatory. They have probably prepared 6 M HCl solution although long ago and could have changed its concentration over even a period.
When 12.00 moles of potassium chlorate decomposes, how many dm3 of oxygen are produced at 325K and 188 kPa?
2KClO3 →2KCl + 3O2
show work pls
Answer:
258.71 dm³
Explanation:
We'll begin by calculating the number of mole of O₂ produced by the decomposition of 12 moles of KClO₃. This can be obtained as follow:
The balanced equation for the reaction is given below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Therefore, 12 moles of KClO₃ will decompose to produce = (12 × 3)/2 = 18 moles of O₂.
Finally, we shall determine the volume of the O₂. This can be obtained as follow:
Temperature (T) = 325 K
Pressure (P) = 188 KPa
Number of mole (n) = 18 moles
Gas constant (R) = 8.314 KPa.dm³/Kmol
Volume (V) =?
PV = nRT
188 × V = 18 × 8.314 × 325
188 × V = 48636.9
Divide both side by 188
V = 48636.9 / 188
V = 258.71 dm³
Thus, 258.71 dm³ of oxygen were obtained from the reaction.
In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?
Answer:
Refraction
Explanation:
A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5 °C. Find the
mass of the water.
Answer:
1.5 × 10³ g
Explanation:
Step 1: Given and required data
Transferred heat (Q): 41,840 JInitial temperature: 22.0 °CFinal temperature: 28.5 °CSpecific heat capacity of water (c): 4.184 J/g.°CStep 2: Calculate the temperature change
ΔT = 28.5°C - 22.0 °C = 6.5 °C
Step 3: Calculate the mass (m) of water
We will use the following expression.
Q = c × m × ΔT
m = Q / c × ΔT
m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g
Cell membranes are selectively permeable. This means that A. only water can move freely across the cell membrane. B. any substance can move across the cell membrane, but chemical energy will always be required. C. some substances can move freely across the cell membrane, while others must be transported. D. no substances can move freely across the cell membrane.
Answer:
C. some substances can move freely across the cell membrane, while others must be transported.
Explanation:
Convert 7.8 moles of carbon tetrafluoride into grams.
Answer:
686.43363984 is the answer when 7.8 moles is converted.
Which of the following would be considered an Exothermic Reaction?*
Ice Melting
Burning wood
Boiling water
Baking a cake
Answer:
Burning wood
Explanation:
the fire releases heat into the air from the burning wood
Explain the term orbital.
Answer: a mathematical expression describing the probability of finding an electron at various locations; usually represented by the region of space around the nucleus where there is a high probability of finding an electron
Explanation:
how many molecules in 400g of acetic acid
Answer:chemical formula of acetic acid is or
so, molecular mass of acetic acid = 2 × atomic mass of C + 4 × atomic mass of H + 2 × atomic mass of O
= 2 × 12 + 4 × 1 + 2 × 16
= 24 + 4 + 32
= 60g/mol
given mass of acetic acid = 22g
so, no of moles of acetic acid = given mass/molecular mass
= 22/60 ≈ 0.367
so, number of moles of acetic acid is 0.367mol
number of molecules in 0.367 mol of acetic acid = 6.022 × 10²³ × 0.367
= 2.21 × 10²³
Explanation:
A student pours 10.0 g of salt into a container of water and observes the amount of time it takes for the salt to dissolve. She then repeats the process using the same amounts of salt and water but this time she slowly stirs the mixture while it is dissolving. The student performs the experiment one more time but this time she stirs the mixture rapidly.
Answer:
It will go faster each time because she is stirring therefore the water can get to the salt faster than it just sitting at the top
Explanation:
Inquiry Extension Consider a reaction that occurs between solid potassium and chlorine gas. If you start with an initial mass of 15.20 g K, and an initial mass of 2.830 g Cl2, calculate which reactant is limiting. Explain how to determine how much more of the limiting reactant would be needed to completely consume the excess reactant. Verify your explanation with an example
The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].
To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.
The balanced chemical equation for the reaction between solid potassium and chlorine gas is:
2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)
From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.
First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:
moles of K = 15.20 g / 39.10 g/mol = 0.388 mol
moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol
Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:
Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl
Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl
We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.
To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.
We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:
moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])
= 0.080 mol K
This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:
mass of K needed = 0.080 mol K x 39.10 g/mol
= 3.13 g K
Therefore, The 3.13 g of K would be needed to completely react with the remaining.
Example verification:
Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?
Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]
Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol
Moles of additional [tex]Cl_2[/tex] = 0.0070 mol
The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:
0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl
Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:
5.95 g + 1.04 g = 6.99 g
The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].
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