Answer:
dimond is stronger
Explanation:
Answer:
Graphite and Diamond are different because they have different structures. ... However each carbon atom in Diamond has 4 covalent bonds with other Carbons, making it extremely strong and hard. On the other hand, each carbon in graphite is bonded to three carbons, and therefore graphite is formed in layer
Also:
Each carbon atom in a diamond is linked to four other carbon atoms. Each carbon atom in graphite is linked to three other carbon atoms. Diamond is poor conductor of electricity due to the absence of free electrons. Graphite is good conductor of electricity due to the presence of free electrons in its structure.
Explanation:
Hope this helps
How are solutions and compounds similar?
Answer:
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A compound is a pure substance that is composed of elements chemically bonded in definite proportions. A compound can be broken down into simpler substances only by chemical reactions, such as electrolysis.
A solution is a homogeneous mixture, meaning that it is the same throughout. A solution is composed of one or more solutes dissolved in a solvent. The proportions of the solute(s) can vary, as the components of a solution are not chemically bonded. The components of a mixture can be separated by physical means, such as filtration and distillation
How are tadpoles and larvae similer
Answer: Tadpole, also called polliwog, aquatic larval stage of frogs and toads. Compared with the larvae of salamanders, tadpoles have short, oval bodies, with broad tails, small mouths, and no external gills. The internal gills are concealed by a covering known as an operculum.
Explanation:
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:
Which of the following is true for a gas under conditions of very high pressure? (5
points)
1) PV > nRT, because the real volume of the gas would be more than the ideal
volume.
2) PV = nRT, because intermolecular forces are considerable at very high
pressures.
3) PV = nRT, because all gases behave as ideal gases at very high pressures.
04) PV = nRT, because the volume of the gas would become negligible.
Answer:
1) PV > nRT, because the real volume of the gas would be more than the ideal
volume.
Explanation:
According to the ideal gas equation; PV = nRT. Let us recall that this equation only holds under ideal conditions.
Gases exhibit ideal behavior under high temperature and low pressure. At higher pressure, the real volume of the gas is larger than the ideal volume of the gas.
Thus, at high pressure, PV > nRT, because the real volume of the gas would be more than the ideal volume.
Answer:
1) PV > nRT, because the real volume of the gas would be more than the ideal volume.
Explanation:
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An experimental measurement was taken of 10.4mL and the actual measurement was 9.7mL. What is the percent error?
Answer:
13%
Explanation:
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!
If 3.13 mol of an ideal gas has a pressure of 2.33 atm and a volume of 72.31 L, what is the temperature of the sample in degrees Celsius?
Answer:
382.49 C degree Celsius
Explanation:
Hello,
This problem deals with understanding the ideal gas law which hopes to predict how ideal gases might behave in any given condition. I listed the formula below and we are basically just going to solve for temperature by rearranging the equation as seen on the picture (there's also other rearranged ones in case you need to solve for those).
Universal gas constant R has a value of 0.0821 L * atm/(mole * K) when working with these given units so it will be part of this equation. R value changes based on what units you have.
T = PV/nR
= (2.33) (72.31) / (3.13)(0.0821)
= 655.64 K
Question is asking temperature in celsius so we employ the formula attached below:
C = K - 273.15
= 655.64-273.15
= 382.49 degree Celsius
382.49 degree Celsius is the answer!
Determine the hydroxide ion concentration in
a solution that is 0.00034 M Ca(OH)2.
Answer in units of M.
Answer:
[tex]0.00068M[/tex]
Explanation:
Hello there!
In this case, according to the ionization of calcium hydroxide, a strong base:
[tex]Ca(OH)_2\rightarrow Ca^{2+}+2OH^-[/tex]
Thus, since there is a 1:2 mole ratio of calcium hydroxide to hydroxide ions, we apply the following proportional factor to obtain:
[tex]0.00034\frac{molCa(OH)_2}{L}*\frac{2molOH^-}{1molCa(OH)_2} \\\\=0.00068\frac{OH^-}{L}\\\\=0.00068M[/tex]
Regards!
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.
How can heat energy transform from mechanical energy?
A)Burning
B)Friction
C)Light
D)Flames
Answer:
A
Explanation:
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PLEASE HELP!! ORGANIC CHEMISTRY
A sample of a diatonic gas is loaded into an evacuated bottle at STP. The 0.25 L bottle contains 1.76 grams of the unidentified gas. Calculate the molar mass of the gas. What is the identity of the diatomic gas?
Answer:
(a) 157.7 g
(b) 7.04 g/dm³
Explanation:
(a) From the question,
According to Avogadro's Law,
1 mole of every gas at STP occupies a volume of 22.4 dm³
But mass of 1 mole of the diatomic gas = molar mass of the gas.
This Implies that,
The molar mass of the gas at STP occupies a volume of 22.4 dm³
From the question,
If,
0.25 L bottle contain 1.76 g of the gas,
Therefore,
Molar mass of the gas = (1.76×22.4)/0.25
Molar mass of the gas = 157.7 g.
(b) Density of the gas = mass/volume
D = m/v
Given: m = 1.76 g, v = 0.25 L = 0.25 dm³
Therefore,
D = 1.76/0,25
D = 7.04 g/dm³
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.
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.
What is the limiting reactant in the following equation? How much Fe2O3 will be produced if 2.1 g of Fe reacts with 2.1 g of O2?
4 Fe + 3O2 —> 2Fe2O3
Answer:
Fe is limiting reactant and 3.00g of Fe2O3 will be produced
Explanation:
To solve this question we must convert the mass of each reactant to moles and, using the reaction we can find limiting reactant. With moles of limiting reactant we can find moles of Fe2O3 and its mass as follows:
Moles Fe -Molar mass: 55.845g/mol-
2.1g * (1mol / 55.845g) = 0.0376 moles
Moles O2 -Molar mass: 32g/mol-
2.1g * (1mol / 32g) = 0.0656 moles
For a complete reaction of 0.0656 moles of O2 are needed:
0.0656moles O2 * (4mol Fe / 3 mol O2) = 0.0875 moles Fe
As there are just 0.0376 moles,
Fe is limiting reactant
The mass of Fe2O3 is:
Moles:
0.0376 moles Fe* (2mol Fe2O3 / 4mol Fe) = 0.0188 moles Fe2O3
Mass:
0.0188 moles Fe2O3 * (159.69g / mol) =
3.00g of Fe2O3 will be produced
What is one movement that liquid water CANNOT do while on or at the Earth's surface? (GIVE RIGHT ANSWER OR I DELETE 100 POINTS)
Answer:
One movement that i can't do is float in mid air
Explanation:
8
What happens to solid waste in the circulatory system?
A it's expelled through the lungs
B
It pumps into the coronary circulation
C
It's dropped off in the kidneys
D
It's deposited in the aorta
Answer:c it’s dropped off in the kidneys
Explanation:
I took the quiz
The solid wastes are from the circulatory system is expelled to the kidney where, the nitrogenous wastes like urea and uric acid is excreted as urine from the body.
What is circulatory system ?A circulatory system is an organ system, where the blood is purified and oxygenation of blood takes place. Through circulatory system, the blood reaches throughout the body pumped from the heart through veins.
The organs included in circulatory system are lungs, heart, aorta, veins, blood vessels etc. There are various kinds of blood vessels each having specific functions.
There is a network of blood vessels including arteries and large veins, capillaries that join the venules and other veins. All the nutrients and ions are circulated throughout the body through blood and solid wastes are then expelled to kidney.
Kidney function as a sieve to clean the good fluid from waste products. Uric acid and urea along with water excreated as urine then. Thus, option C is correct.
To find more about circulatory system, refer the link below:
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What compound(s) does pure water contain?
1. Which individuals are most likely to die before reproducing, those with adaptive traits or
nonadaptive traits? Why? (Hint: You may use the newt population as an example in your
explanation.)
At 27.0°C, the volume of a gas is 630 L. At the same pressure, its volume is 92,0 mL at a temperature of
Answer:
–272.96 °C
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 27.0 °C
Initial volume (V₁) = 630 L.
Final volume (V₂) = 92.0 mL
Final temperature (T₂) =?
Next, we shall convert 27.0 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 27.0 °C
Initial temperature (T₁) = 27.0 °C + 273
Initial temperature (T₁) = 300 K
Next, we shall convert 92.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
92 mL = 92 mL × 1 L / 1000 mL
92 mL = 0.092 L
Next, we shall determine the final temperature.
Initial temperature (T₁) = 300 K
Initial volume (V₁) = 630 L.
Final volume (V₂) = 0.092 L
Final temperature (T₂) =?
V₁ / T₁ = V₂ / T₂
630 / 300 = 0.092 / T₂
2.1 = 0.092 / T₂
Cross multiply
2.1 × T₂ = 0.092
Divide both side by 2.1
T₂ = 0.092 / 2.1
T₂ = 0.04 K
Finally, we shall convert 0.04 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
Final temperature (T₂) = 0.04 K
Final temperature (T₂) = 0.04 – 273
Final temperature (T₂) = –272.96 °C
12. An electrolysis reaction is
A. hydrophobic.
B. spontaneous.
C. exothermic.
D. non-spontaneous.
Answer: D.) non-spontaneous.
Explanation:
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 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)
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
Every cell is surrounded by a thin membrane. What is the main function of this cell membrane?
A.
to protect the cell from invasion by bacteria and viruses
B.
to allow each cell to form connections with other cells
C.
to limit the size of the cell and keep the shape of the cell the same
D.
to separate the inside of the cell from the outside environment
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:
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].
Learn more about Solid Potassium at
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200.0g of a 3.0% NaF solution, how much distilled water do we weigh out?
197g of distilled water
194g of distilled water
140g of distilled water
170g of distilled water
Answer:
194g of distilled water.
Explanation:
Hello there!
In this case, according to the given information for this problem, it turns out possible for us to use the given mass of the solution and the percent by mass of NaF to firstly calculate the grams of this solute as shown below:
[tex]\%m=\frac{m_{solute}}{m_{solution}} *100\%\\\\m_{solute}=\frac{\%m*m_{solution}}{100\%} \\\\m_{solute}=\frac{3.0\%*200.0g}{100\%} \\\\m_{solute}=6g[/tex]
And finally, since the mass of solution is calculated by adding mass of solute and mass of solvent we obtain the mass of water (solvent) as follows:
[tex]m_w=200g-6g=194g[/tex]
Therefore, the answer is 194g of distilled water
Regards!
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is contained in a 0.200-kg aluminum calorimeter cup. The final temperature of the system is 30.5 Celsius. What is the specific heat of the metal alloy in J/Kg.Celsius
Answer:
[tex]C_{alloy}=0.497\frac{J}{g\°C}[/tex]
Explanation:
Hello there!
In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:
[tex]Q_{allow}=-(Q_{water}+Q_{Al})[/tex]
Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:
[tex]m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})[/tex]
Then, we solve for specific heat of the metallic alloy to obtain:
[tex]C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}[/tex]
Thereby, we plug in the given data to obtain:
[tex]C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}[/tex]
Regards!
Do u believe that you are beautiful/handsome?
Yes or No?
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:
Hl Weakly dissociates in water according to the chemical equation below. H20+ Hl <-> H3O^+ + l- What is a conjugate acid-base pair in this reaction?
Answer:
https://www.clutchprep.com/chemistry/practice-problems/70217/hi-aq-h2o-l-h3o-aq-i-aq-identify-each-as-either-a-bronsted-lowry-acid-bronsted-l
Explanation:
https://www.clutchprep.com/chemistry/practice-problems/70217/hi-aq-h2o-l-h3o-aq-i-aq-identify-each-as-either-a-bronsted-lowry-acid-bronsted-l