Answer:
Explain some of the uses of metals based on their properties.
Explanation:
Metals have a shiny or metallic luster and are good conductors of heat and electricity, they can be bent and pounded in various shapes, so they can be used on cars, coins, some pipes, keys, and and a flag.
Washing machines use a large amount of water. A student suggested that old pairs of stained jeans which have to be washed more frequently should be replaced by new pairs of jeans to conserve water. Which of these statements best describes the suggestion made by the student? Question 9 options: It is not practical because an old pair of jeans needs less water to be washed than a new pair of jeans. It is practical because a huge amount of water can be conserved by this method. It is practical because it is easy to implement. It is not practical because it takes a huge amount of water to produce a new pair of jeans.
Answer:
C |||| It is not practical because it takes a huge amount of water to produce a new pair of jeans
Explanation:
If you're doing flvs then it's C.
Answer:
C!
Explanation:
i got it right on the test UwU
Look at the picture below. Name the isotope.
Answer:
carbon-13
Explanation:
In the reaction of aluminum metal and oxygen to make aluminum oxide, how many grams of oxygen gas will react with 2.2 moles aluminum metal?
Answer:
52.8 g of O2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4Al + 3O2 —> 2Al2O3
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3
Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O2.
Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.
Thus, 1.65 moles of O2 is needed for the reaction.
Finally, we shall determine the mass of O2 needed as shown below:
Mole of O2 = 1.65 moles
Molar mass of O2 = 2 × 16= 32 g/mol
Mass of O2 =?
Mole = mass/Molar mass
1.65 = mass of O2 /32
Cross multiply
Mass of O2 = 1.65 × 32
Mass of O2 = 52.8 g
Therefore, 52.8 g of O2 is needed for the reaction.
Which of the two forces is greater the force on the firefly or the force on the bus explain your answer please I really need help please
It’s due today please
Answer: The force on the firefly
The unfortunate firefly hitting the bus does not change the velocity of the bus very much. Technically there is a change, but it's so very small and miniscule that it barely registers. To any casual observer not paying very close attention, they don't notice anything at all. So effectively the force on the firefly is a lot greater since the firefly got the worst end of the deal.
So in short, we look at the velocity of each object and see which velocity changed the most. In this case, the firefly's velocity changed from whatever speed it was flying to 0 when it stops flying all together. That's why the force is greater on the bug.
Using the van der Waals equation, calculate the pressure for a 1.25 mol sample of xenon contained in a volume of 1.000L at 75°C; a = 4.194 L2 atm/mol2 and b = 0.05105 L/mol for Xe. Compare these results to that predicted by the ideal gas law.
Answer:
ABC
Explanation:
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What does this image represent?
Amine group
Carbonyl group
Ether group
Hydroxyl group
Answer:
ether group
Explanation: I looked it up
in the reaction of magnesium metal with hydrochloric acid, how do you determine when the magnesium metal has reacted completely
Answer:
When there is no spark from the reaction
Explanation:
The reaction between the acid and the magnesium metal goes into completion when there is no more spark from the magnesium. Also, the magnesium would have totally disappeared into solution.
These physical observation marks the end of the reaction.
When Mg reacts with HCl hydrogen bubbles are released. The completion of the given reaction can be determined by the presence of bubbles above the HCl.
The reaction between HCl and Mg:[tex]\bold {Mg+ 2HCl\rightarrow MgCl_2 + H_2}[/tex]
In this reaction, Mg reacts with 2 moles of HCl to form 1 mole of Magnesium chloride and release hydrogen gas.
During reaction,
Magnesium slowly disappears as it reacts with HCl.The hydrogen gas is released during the reaction, bubbles can be observed.Therefore, the completion of the given reaction can be determined by the presence of bubbles above the HCl.
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The temperature of a sample of CH4 gas (10.34 g) in a 50.0 L vessel at 1.33 atm is ________ °C.
a.
984
b.
-195
c.
-1260
d.
-195
Answer:
option C is correct
Explanation:
Considering the ideal gas law and the definition of Avogadro's Number, the correct option is option a. The temperature of a sample of CH₄ gas (10.34 g) in a 50.0 L vessel at 1.33 atm is 984 °C.
In first place, you have to know that ideal gases are a simplification of real gases that is done to study them more easily.
It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
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, being the molar mass of CH₄ being 16 [tex]\frac{g}{mole}[/tex], that is, the mass present in one mole of an element or compound, the number of moles that 10.34 grams contains is calculated as:
[tex]10.34 g*\frac{1 mole}{16.04 g} = 0.645 moles[/tex]
So, you know:
P= 1.33 atmV= 50 Ln= 0.645 molesR=0.082 (atm×L)/ (mol×K)T= ?Replacing:
1.33 atm × 50 L= 0.645 moles× 0.082 (atm×L)/ (mol×K) ×T
Solving:
T= [1.33 atm × 50 L] ÷ [0.645 moles× 0.082 (atm×L)/ (mol×K) ]
T≅ 1257 K
Being 273 K equivalent to 0 C, then:
T= 1257 K= 984 C
In summary, the correct option is option a. The temperature of a sample of CH₄ gas (10.34 g) in a 50.0 L vessel at 1.33 atm is 984 °C.
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The initial concentration of NOCl in the second-order reaction 2NOClâ2NO+Cl2 is 0.878M. After 763,200 seconds, the concentration of NOCl is 0.83M. What is the rate constant k for the reaction? Report your answer in scientific notation rounded to two significant figures. Use the multiplication symbol when reporting your answer rather than the letter x. Provide your answer below: $$ 1/M s
Answer:
[tex]k=8.63x10^{-8}\frac{1}{M*s}[/tex]
Explanation:
Hello!
In this case, since the differential rate law of a second-order reaction is:
[tex]\frac{dC_A}{dt}=-kC_A^2[/tex]
Whereas A stands for NOCl and the corresponding integrated rate law is:
[tex]\frac{1}{C_A} =kt+\frac{1}{C_A_0}[/tex]
Thus, since we know the concentrations and the elapsed time, we compute the rate constant as shown below:
[tex]k=( \frac{1}{C_A}-\frac{1}{C_A_0} )/t\\\\k=( \frac{1}{0.83M}-\frac{1}{0.878M} )/763,200s\\\\k=8.63x10^{-8}\frac{1}{M*s}[/tex]
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Determine the correct hybridization (from left to right) about each central atom in
CH3CH2OH.
A) 1st C sp3 ; 2nd C sp2 ; O sp B) 1st C sp3 ; 2nd C sp3 ; O sp3
C) 1st C sp ; 2nd C sp2 ; O sp3 D) 1st C sp3 ; 2nd C sp2 ; O sp2
Answer:
B) 1st C sp3 ; 2nd C sp3 ; O sp3
Explanation:
Both carbons in ethanol have an sp³ configuration of their orbitals, given that each carbon atom binds to four other atoms, as can be seen on the extended formula in the attachment.
In these carbons (1st and 2nd) one s orbital and three p orbitals are hybridized to form four hybrid orbitals sp³. These orbitals form four bonds and their bond angles are close to the tetrahedral value of 109.5º,as can be seen for methanol which also has an sp³ configuration.
The oxygen atom can also be described as having sp³ hybrid orbitals. Two of the four hybrid sp³ orbitals in oxygen are occupied by lone electron pairs, and two are used to form bonds with other atoms (in this case the oxygen is bonded to a carbon atom and a hydrogen atom).
Therefore, the three central atoms have an sp³ configuration.
Identify the titration curve for a monoprotic weak acid titrated with a strong base.
Answer:
Titration is a technique to determine the concentration of an unknown solution. As illustrated in the titration setup above, a solution of known concentration (titrant) is used to determine the concentration of an unknown solution (titrand or analyte).
Typically, the titrant (the solution of known concentration) is added through a burette to a known volume of the analyte (the solution of unknown concentration) until the reaction is complete. Knowing the volume of titrant added allows us to determine the concentration of the unknown analyte. Often, an indicator is used to signal the end of the reaction, the endpoint. Titrant and analyte is a pair of acid and base. Acid-base titrations are monitored by the change of pH as titration progresses.
Let us be clear about some terminologies before we get into the discussion of titration curves.
Would you want to build the wire for a cell phone charger out of nitrogen (N)? Why or why not? Be sure to explain how the properties and structure of nitrogen influence your decision.
Answer:
No
Explanation:
Nitrogen cannot be used to build the wire for a cell phone charger.
What are the elements used to build charger ?
Inside phone chargers is aluminum. Aluminum is used in phone chargers because it does a better job of conducting electricity than copper does. Bauxite is used to make aluminum. The crust of the earth contains aluminum.A phone charger is also made of tungsten. The wire's resistance is provided by tungsten. Wolframite make up tungsten. The crust of the Earth contains tungsten.Additionally, phone chargers contain gold. Because it is a conductor, gold is used in phone chargers. However, gold is also resistant to oxidation. Copper ores are used to make gold. Gold can be found everywhere on Earth and in its oceans.Silver is within your mobile charger. Silver conducts electricity. Silver is comprised of various minerals. Silver can be found free in nature at times, but it can also be found in copper, gold, and nickel.Phone chargers contain copper. Copper is also used to carry electric current.Chalcopyrite make up copper. Copper ore deposits can be found all over the world.Copper windings in a transformer are used to convert a 220 V input voltage into a 5 V operating output voltage in mobile phone chargers. The output voltage is greater than the input voltage if the secondary coil has more turns than the primary coil.Nitrogen (N), a nonmetallic element in the periodic table's Group 15 [Va] group. It is the most abundant element in Earth's atmosphere and a component of all living matter. It is a colorless, tasteless, and odorless gas. Nitrogen is used to shield motors, prevent oxidation and other chemical reactions, pressurize cable jackets, and more in the electrical industry.To know more about Nitrogen check this:
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Which numerical setup can be used to calculate the
atomic mass of the element bromine?
Answer:
From the numerical steps highlighted under explanation, the average atomic mass of bromine is 79.91 u
Explanation:
The steps to be taken will involve;
1) Find the number of isotopes of bromine.
2) Identify the atomic mass and relative abundance of each of the isotopes.
3) Multiply the atomic mass of each of the isotopes by their corresponding values relative abundance value.
4) Add the value in step 3 above to get the average atomic mass of bromine.
Now;
Bromine has 2 isotopes namely;
Isotope 1: Atomic mass = 78.92amu and a relative abundance of 50.69%.
Isotope 2: Atomic mass = 80.92amu and a relative abundance of 49.31%.
Using step 3 above, we have;
(78.92 × 50.69%)
And (80.92 × 49.31%)
Using step 4 above, we have;
(78.92 × 50.69%) + (80.92 × 49.31%) ≈ 79.91 u
The atomic mass of bromine has been calculated as 79.9062 amu.
The relative atomic mass of the element has been given as the mass of each isotope with respect to their abundance.
The atomic mass has been given as:
[tex]amu=\sum mass\;\times\;\% Abundance[/tex]
The available isotopes of Bromine have been:
Isotope 1 = 78.92 amu, and 0.5069 % abundance
Isotope 2= 80.92 amu, and 0.4931 % abundance
Submitting the values for the atomic mass unit of bromine ([tex]amu_{\text{Br}}[/tex]):
[tex]amu_{\text {Br}}= (78.92\;\times\;0.5069)\;+\;(80.92\;\times\;0.4931)\\amu_{\text{Br}}=40.0045\;+\;39.9016\\amu_{\text{Br}}=79.9062[/tex]
The atomic mass of bromine has been calculated as 79.9062 amu.
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