Answer:
high melting point
Explanation:
The filament of a bulb is often heated to very high temperatures as the bulb is in operation.
Many times, electric bulbs may have to be on for a whole day and they may reach temperatures that are outrageously high in the process.
The material of the filament must have a very high melting point so that it doesn't melt while the bulb is still in operation.
Answer:
Actually the answer is High Vapor pressure
Explanation:
For an ideal gas, classify the pairs of properties as directly or inversely proportional. You are currently in a sorting module. a. P and n b. V and n c. P and T d. T and V e. P and V1. directly proportional2. inversely proportional
Answer:
the result for the following are (a) P is directly proportional to n
(b) V is directly proportional to T (c) P is directly proportional to T (d) T is inversly proportional to V (e) P is inversely proportional to V
How many moles of carbon dioxide at STP will fit in a 50 liter container?
Answer:
n = 2.23 moles
Explanation:
Given the following data;
Standard temperature = 273 K
Standard pressure = 101.325 kPa
Volume = 50 liter
R = 8.314 J/mol·K
To find the number of moles, we would use the ideal gas law formula;
PV = nRT
Where;
P is the pressure.V is the volume.n is the number of moles of substance.R is the ideal gas constant.T is the temperature.Making n the subject of formula, we have;
[tex] n = \frac {PV}{RT} [/tex]
Substituting into the formula, we have;
[tex] n = \frac {101.325*50}{8.314*273} [/tex]
[tex] n = \frac {5066.25}{2269.722} [/tex]
n = 2.23 moles
Therefore, 2.23 moles of carbon dioxide at STP will fit in a 50 liter container.
A piece of solid tin is submerged in silver nitrate solution a reaction occurs producing tin(IV) nitrate solution and solid silver
Write a word equation write a skeleton equation write a balanced chemical equation
Answer:
Tin + silver trioxonitrate V -------->Tin IV nitrate + silver
Explanation:
The term word equation refers to an equation in which the reactants and products are written in words rather than chemical symbols.
Note than tin is above silver in the electrochemical series hence silver will be displaced by tin as follows;
Tin + silver trioxonitrate V -------->Tin IV nitrate + silver
Senario: 2 years ago, a fruit was smuggled into California on a plane from an exotic, far away land. The homeowner saw that the fruit had maggots and tossed it into the backyard, hoping the seed would grow. The larvae hatched out and moved throughout the area. This fictitious insect will destroy fruit and has the possibility of spreading disease killing the trees. The insect consumes plants in the Prunus species of stone fruits? Look up the plant genus Prunus.
Discussion: The insect has spread over a large area of Southern California, discovered at UC Riverside. What steps would you do to control or eradicate the destructive exotic insect?
PLZ HELP THX WITH COLLEGE LEVEL EXPERICENCE
Answer:
The best existing methods of control—artemisinin-based drug treatment and insect control with chemical sprays and treated bednets—can reduce the burden of disease substantially, and can even eliminate the disease in some regions,
hope this will help you more
Predict the products of below reaction, and whether the solution at equilibrium will be acidic, basic, or neutral.
N2O5 + 3H2O → __________
Answer: The product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.
Explanation:
When two or more chemical substances react together then it forms new substances and these new substances are called products.
For example, [tex]3N_{2}O_{5} + 3H_{2}O \rightarrow 6HNO_{3}[/tex]
This shows that nitric acid [tex](HNO_{3})[/tex] is the product formed and it is an acidic substance.
Hence, the solution at equilibrium will be acidic in nature.
Thus, we can conclude that the product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.
Please help me ASAP I’ll mark Brainly
Answer:
cell
chloroplast and cell wall
nucleus
life processes
cell membrane
shape and size
vacuole
Hope it helps
Question 1 Points 3 23 and Louis immerses his left hand in a beaker containing cold water and immerses his right hand in a beaker containing warm water. Then, he immerses both his hands on a beaker containing water at room temperature. Which of the following statements are true? 1. The hand that was in hot water would feel cold. 2. The hand that was in cold water would feel hot. 3. His two hands will feel the same hotness. Que O2 and 3 0 1 and 2 o 1 and 3 1.2, and 3
Answer:look down below
Explanation:
The statements that are true about hands that are immersed in the water are:
1. The hand that was in hot water would feel cold.
2. The hand that was in cold water would feel hot.
The correct option is B 1. and 2.
What is temperature?
Temperature is the measurement of the hotness or coldness of any object. It is measured in Celsius or kelvin. Our body has nerves that feel the different temperatures of any object. The high temperature is called hot and the low temperature is called cold.
When Louis put his hand in the warm water and one hand in the cold water. He feels the temperature of both glasses of water. Then he put both hands in the normal water.
So the hand that is warm would feel the water as cold and the hand with cold water would feel the water as hot.
Thus, the correct option is B. 1. and 2.
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Based on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --> Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) + Ag(s) --> No reaction Cu(
Answer:
Ni
Explanation:
An active metal is a highly reactive metal. Active metals are found high up in the activity series.
Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.
Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.
Which substrate is used in the last step of glycolysis
Pyruvate Kinase
Pyruvate Kinase performs a substrate level phosphorylation on ADP to generate an ATP and pyruvate, the final product of glycolysis.
PK dificiency is transmitted in an autosomal recessive disorder in which both alleles must contain the mutated gene, PK-LR.
Hope it helps you! ＼(^ᴥ^)／
define saturated and unsaturated fats
Explanation:
Saturated fats are defined as the fat where fatty acid chains contain only single bonds.
For example, stearic acid, palmitic acid etc.
Unsaturated fats are defined as the fat where fatty acids contain one or more number of double bonds on the carbon atoms.
For example, oleic acid, palmitoleic acid etc.
A
(c) 2 C(s) + MnO2(s)
Mn(s) + 2 CO(g)
O combination reaction
O decomposition reaction
O combustion reaction
O single-displacement reaction
Answer: The reaction, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
Explanation:
A chemical reaction in which one element of a compound is replaced by another element participating in the reaction.
For example, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex]
Here, the element manganese is replaced by carbon atom. As only one element gets replaced so, it is a single-displacement reaction.
Thus, we can conclude that [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Answer:
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Explanation:
The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:
The answer is (R)-2-chlorobutane.
The reaction take splace through [tex]S_{N} _2[/tex] mechansim and inversion in configuration happens.
In the reactionpyruvate lactatethat is catalyzed by the enzyme lactate dehydrogenase, the compound on which the enzyme works, pyruvate, is called the _______.
Answer:
Substrate
Explanation:
In biochemical sciences, a substrate is a substance that is acted upon by an enzyme to yield a product. Enzymes are known for catalyzing biochemical reactions. The substances that are usually worked with during this catalytic process are termed as SUBSTRATES.
Substrates, which are usually changed during the process, binds to the active site on the enzyme and form an enzyme-substrate complex.
According to this question, pyruvate is converted to lactate in a reaction that is catalyzed by the enzyme lactate dehydrogenase. This means that the compound on which the enzyme works, pyruvate, is called the SUBSTRATE.
howtocalculatethevolumeofcarbondioxideproducedwhen400gofmarblewereats.t.p
Answer:
so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.
viagnesiumi anu
If I have 100g of Magnesium, how much Magnesium Nitride will I theoretically create?
O 24.3g Mg3 N2
O 138.4g Mg3 N2
415.2g Mg3 N2
O 200g Mg3 N2
Answer:
Theoretical yield is 138.4 g
Explanation:
In the first step we determine the reaction:
3Mg + N₂ → Mg₃N₂
Mass of reactant is 100 g. We assume the nitrogen is in excess, so we work with Mg. We convert mass to moles:
100 g . 1mol/ 24.3g = 4.11 moles of Mg.
Ratio is 3:1. 3 moles of Mg can produce 1 mol of nitride
Our 4.11 moles, may produce (4.11 . 1)/3 = 1.37 moles of Mg₃N₂
We convert mass to moles, to find the theoretical yield:
1.37 mol . 100.9 g/mol = 138.2 g
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Answer:
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Explanation:
a. A red blood cell is a microscopic particle.
It can be viewed under a microscope.
b. A sugar molecule is also a microscopic substance.
It can be viewed under a microscope.
c. Baking powder is macroscopic substance.
an emerald can be described as...
Answer:
green gemstone
Explanation:
hope this helps someone
I need help solving this!
For the reaction C + 2H2 → CH4, how many moles of hydrogen are needed to make 146.6 grams of methane, CH4 ?
Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0
Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:
Element
Molar Mass
Hydrogen
1
Carbon
12
Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol[/tex]
The given reaction equation is as follows.
[tex]C + 2H_{2} \rightarrow CH_{4}[/tex]
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
[tex]Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol[/tex]
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].
The doctor has ordered Claforan 1 g in 100 ml D5W to run IV piggyback for 30 minutes twice daily. The pharmacy sends Claforn 2 g in a powdered form, which when reconstituted has a concentration of 180 mg Claforan per ml. How much Claforn will you add to the bag of D5W
Answer:
0.111 g
Explanation:
1 g = 1000 mg
Doctor ordered the following concentration of Claforan:
C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL
If we add 2 g iof Claforan, we obtain:
2 g Claforn ---- 180 mg/mL Claforan
To reach a concentration equal to C (10 mg/mL), we need:
10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn
Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.
If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).
Answer:
[tex]{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}[/tex]
the ability of organism to sense changes in its body is an example of
Answer:
the ability of organism to sense changes in its body is an example of responsiveness.Hope it is helpful to you
What is the percent yield of the reaction below if 84.0 grams of Al2O3(s) is recovered from a reaction whose theoretical yield of Al2O3(s) is 104 grams?
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
Answer:
80.8%
Explanation:
Let's consider the following balanced equation.
4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
The mass obtained of Al₂O₃ (experimental yield) is 84.0 g. The theoretical yield of Al₂O₃ is 104 g. We can calculate the percent yield of Al₂O₃ using the following expression.
%yield = (experimental yield / theoretical yield) × 100%
%yield = (84.0 g / 104 g) × 100% = 80.8%
Answer:
Percent Yield = 80.8%
Explanation:
We can find the percent yield of a reaction using the equation:
Percent yield = Actual yield (g) / Theoretical Yield (g) * 100
Where Actual yield is the amount of product produced (84.0g)
And theoretical yield is the mass produced assuming a 100% of product (104.0g)
Replacing the computed values:
Percent yield = 84.0g / 104.0g * 100
Percent Yield = 80.8%
what class of organic compound is formed when cyclopentanone reacts with ethylamine in the presence of trace acid
The question is incomplete, the complete question is;
What functional group results when cyclopentanone reacts with ethylamine in the presence of trace acid? A) cyanohydrin B) semicarbazone C) imine D) enamine E) oxime
Answer:
imine
Explanation:
An imine is an unsaturated amine. An imine contains the carbon- nitrogen double bond.
Imines are obtained when a carbonyl compound is condensed with NH3 or an amine. The reaction involves several steps in its mechanism.
Since cyclopentanone is a ketone (carbonyl compound) and ethylamine is an amine,in the presence of trace acid, condensation of the two compounds occur to yield an imine
determine the number of moles of Cl in 1.9 mol of CaCl2
Explanation:
In the formula of [tex]CaCl_2[/tex] we can see that there are 2mol Cl in 1 molecule of CaCl2. So, now we can do some math to find out how many mol of Cl there are in 1.9mol CaCl2!
[tex]1.9molCaCl_2*\frac{2molCl}{1molCaCl_2}[/tex]
So, there are 3.8mol Cl
The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?
Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
Explanation:
Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]
Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g
Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.
[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]
Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
Explanation:
Given:
Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]
To find:
The number of atoms of tantalum in 37.1 milligrams.
Solution:
Mass of tantalum = 37.1 mg
[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]
The number of atoms in 0.0371 grams of tantalum = N
Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]
Then a mass of N atoms of tantalum will be:
[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
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What is bond energy
c) Solar energy is the source of all forms of energy.give reasons
Answer:
All energy is made by the sun because without the sun there would be no humans to produce other energy
Explanation:
We use many different forms of energy here on earth, but here’s the thing: almost all of them originate with the sun, not just light and heat (thermal) energy! The law of conservation of energy says that energy can’t be created or destroyed, but can change its form. And that’s what happens with energy from the sun—it changes into lots of different forms:
Plants convert light energy from the sun into chemical energy (food) by the process of photosynthesis. Animals eat plants and use that same chemical energy for all their activities.
Heat energy from the sun causes changing weather patterns that produce wind. Wind turbines then convert wind power into electrical energy.
Hydroelectricity is electrical energy produced from moving water, and water flows because heat energy from the sun causes evaporation that keeps water moving through the water cycle.
Right now, much human activity uses energy from fossil fuels such as coal, oil, and natural gas. These energy sources are created over very long periods of time from decayed and fossilized living matter (animals and plants), and the energy in that living matter originally came from the sun through photosynthesis.
solar panel shows what is the ultimate source of energy
4019K Express your answer as a nuclear equation.
Answer:
[tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v
Explanation:
[tex]_{19} ^{40} K[/tex] is an unstable potassium isotope which has the capacity to undergo beta decay. During the decay process, it transforms to calcium atom with a release of beta particle and an antineutrino.
The required nuclear equation is:
[tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v
The calcium formed due to this process is a stable atom.
A 1.5-m 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 150 kPa. Determine the entropy change of carbon dioxide during this process. Assume constant specific heats
Answer:
The entropy change of carbon dioxide = 0.719 kJ/k
Explanation:
Given:
1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa
The objective is to determine the entropy change of carbon dioxide
Formula used:
ΔS=
Solution:
On considering,
[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]
ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]
On substituting the values,
ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]
ΔS=0.719 kJ/k
The entropy change is "0.719 kJ/K".
Given values are:
Mass of tank,
m = 2.7 kgPressure,
P₁ = 100 kPaRised pressure,
P₂ = 150 kPaAssumption of constant specific heat is,
[tex]C_v = 0.657 \ kJ/kgK[/tex]As we know the formula,
→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]
[tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]
[tex]= 1.7739\times 0.4055[/tex]
[tex]= 0.7193 \ kJ/K[/tex]
Thus above answer is right.
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Calculate the moment of inertia of a CH³⁵CL₃ molecule around a rotational axis that contains the C-H bond. The C-Cl bond length is 177pm and the HCCl angle is 107⁰f
Answer:
The correct answer is "[tex]4.991\times 10^{-45} \ kg.m^2[/tex]".
Explanation:
According to the question,
[tex]R_{C-Cl} = 177 \ pm[/tex]
or,
[tex]=1.77\times 10^{-10} \ m[/tex]
[tex]\alpha = 107^{\circ}[/tex]
[tex]m_{Cl}=34.97 \ m.u[/tex]
or,
[tex]=34.97\times 1.66\times 10^{-27}[/tex]
[tex]=5.807\times 10^{-26} \ kg[/tex]
The moment of inertia around the rotational axis will be:
⇒ [tex]I=3\times m_{Cl}\times (R_{C-Cl})^2 \ Sin^2 \alpha[/tex]
By putting the values, we get
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2 \ Sin^2 (107)[/tex]
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2\times 0.91452[/tex]
[tex]=4.991\times 10^{-45} \ kg.m^2[/tex]
