Answer:
they're all fake
Explanation:
12. An electrolysis reaction is
A. hydrophobic.
B. spontaneous.
C. exothermic.
D. non-spontaneous.
Answer: D.) non-spontaneous.
Explanation:
How are solutions and compounds similar?
Answer:
hope you liked it!!!!!!
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
Which of the following would result in being able to dissolve a greater amount of gas in a solution?
Answer:
Lower the temperature of the solution
An experimental measurement was taken of 10.4mL and the actual measurement was 9.7mL. What is the percent error?
Answer:
13%
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.
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!
Use the equations below to calculate the enthalpy of formation for propane gas, C3H8, from its elements, hydrogen gas and solid carbon. Please hurry!!
Answer: I got -4542.9kg/mol
Explanation:
Using the enthalpy relation, the enthalpy of formation of propane gas in the given equation is 4542.9 kJ/mol
[tex] △H_{f} = Product [/tex]Using the enthalpy value of [tex] CO_{2} [/tex] and[tex] H_{2}0[/tex] given :
Product = [tex] 3CO_{2} + 4H_{2}0[/tex] Product = 3(-393.5) + 4(-285.8) = - 2323.70Reactant = [tex] C_{3}H_{8} + 5H_{2}0[/tex] Water, H20 has △H = 0Reactant = [tex] C_{3}H_{8} + 0[/tex]
Enthalpy of formation = product - Reactant
2219.2 = -2323.70 - (propane + 0)
2219.2 = - 2323.70 - propane
Propane = - 2323.70 - 2219.2
[tex] C_{3}H_{8} = - 4542.9 [/tex]
Therefore, the entalphy of formation of propane gas is - 4542.9 kJ/mol.
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The molar mass of water is 18.02 g/mol. A mass of 160.0 grams of water is equivalent to how many moles?
please show work!
Answer:
8.879 moles.
Explanation:
From the question given above, the following data were obtained:
Molar mass of water = 18.02 g/mol
Mass of water = 160.0 g
Mole of water =?
Mole is defined by the following equation:
Mole = mass / molar mass
With the above formula, we can obtain the number of mole present in 160 g of water. This can be obtained as follow:
Molar mass of water = 18.02 g/mol
Mass of water = 160.0 g
Mole of water =?
Mole = mass / molar mass
Mole of water = 160 / 18.02
Mole of water = 8.879 moles.
Therefore, 160 g of water contains 8.879 moles.
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.)
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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10 ml of a 0.25M solution is diluted to make exactly 250 ml of solution. What's the concentration of the diluted solution?
Answer:
0.01 M
Explanation:
As this problem deals with a dilution process, we can solve it by using the following formula:
C₁V₁=C₂V₂Where subscript 1 stands for the initial concentration and volume, while 2 stands for the final conditions.
That means that in this case:
C₁ = 0.25 MV₁ = 10 mLC₂ = ?V₂ = 250 mLWe input the given data:
0.25 M * 10 mL = C₂ * 250 mLC₂ = 0.01 Mhow 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:
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!
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:
Which organic compound listed below does not have a carbonyl group ?
A. Aldehyde
B. Ketone
C. Ethers
D. Protein
E. Ester
F. Amides
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!
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
What would the products be for the reaction between Na3PO4 + MgSO4?
MgSO4 + Na3PO4 = Na2SO4 + Mg3(PO4)2
Answer: The products of Na3PO4 + MgSO4 are Na2SO4 + Mg3(PO4)2
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
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:
How many grams of oxygen are required to burn 3.01 x 10^23 propane molecules?
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.
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:
Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted
between S and q is the force exerted between S and p.
O 1/2
O 2 times
O 1/4
O 4 times
Answer:1/4
Explanation:
differences between diamond and graphite
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 can heat energy transform from mechanical energy?
A)Burning
B)Friction
C)Light
D)Flames
Answer:
A
Explanation:
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What [H3O+] in a solution of a PH of 0.60 is
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.
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