Answer:
A. Number of energy levels
A gas has a pressure of 699.0 mm Hg at 40.0 0C. This temperature is
K.
What is the temperature at 760.0 mm Hg? K (round to 1 decimal place) or
degrees Celsius (round to one decimal place.
Answer:
initial temp: 313.2 Kfinal temp: 340.5 Kin Celsius: 67.3 °CExplanation:
Given a volume and quantity of gas at 699.0 mm Hg pressure has a temperature of 40.0 °C, you want to know the temperature at 760.0 mm Hg pressure.
Pressure and TemperaturePressure and absolute temperature of a gas are proportional for a given quantity and volume.
Initial temperatureThe temperature of 40.0 °C is an absolute temperature of ...
40.0 K + 273.15 K = 313.15 K ≈ 313.2 K
New temperatureThe temperature after increasing the pressure can be found from ...
T/313.15 = (760.0 mm)/(699.0 mm)
T = 313.15 · 760/699 ≈ 340.5 . . . . Kelvin
In Celsius, that is ...
340.5 K -273.2 = 67.3 °C
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?CH4 → ?C6H6 + ?H2
?C2H6 + ?O2 → ?CO2 + ?H2O
I need to know the number that goes in front of the equations
Answer:
You basically want to balance these equations.
(6)CH4 → (1)C6H6 + (9)H2
(2)C2H6 + (7)O2 → (4)CO2 + (6)H2O
NOTE: DO NOT PUT BRACKETS IN YOUR ANSWER, I HAVE PUT HERE FOR THE ANSWERS INSTEAD OF QUESTION MARKS.
Calculate the number of molecules found in 35 g of Sodium Hydroxide?
Answer:
5.27*10^23 (rounded to 3 significant figures)
Explanation:
The amount of molecules in one mole of anything is equal to Avogadro's number: 6.022×10^23
To find the number of moles of NaOH in 35 grams of it, do 35 divided by the molar mass (39.997): 35/39.997=0.87506562 moles of NaOH
To find the number of molecules, multiply the moles of NaOH by Avogadro's number: 0.87506562×(6.022×10^23)=5.26964522*10^23
Answer:
5.27x10²³ molecules
Explanation:
In order to solve this problem we first convert 35 g of Sodium Hydroxide (NaOH) into moles, using its molar mass:
35 g ÷ 40 g/mol = 0.875 molFinally we convert 0.875 moles into number of molecules, using Avogadro's number:
0.875 mol * 6.023x10²³ molecules/mol = 5.27x10²³ moleculesHow many molecules of CO2 are there in 3.55 g of CO2?
Answer:
4.9∗1022 molecules of CO2 in a 3.6 gram sample.
Explanation:
The gram molecular weight of co2 is 44 gms.
So the number of molecules in 44 gms of co2 is 6.023×10^23 .
Then the number of molecules in 3.6 gms of co2 is (6.023×10^23×3.6)÷44 = 4.92×10^22
Please help meeee, I will brainliest. and
Put it into your own words, if u use from internet.
Answer:
energy is used to break bonds in reactants and energy is released when new bonds form in products.The law of conservation of energy states that matter cannot be created or destroyed. Whether a chemical reaction absorbs or releases energy there is no overall change in the amount of energy during the reaction.
Explanation:
Sorry if im wrong
How many polyatomic ions (s) below contain a total of four atoms?
hydroxide ion
nitrate ion
ammonium ion
sulfite ion
Group of answer choices
two
four
three
one
Answer:
I think the answer is two
Atoms with 5 or more valence electrons tend to form __________________ ions, called _________.
Answer:
positively-charged cations
Explanation:
hope this helps! :)
At 40°C how many grams of sugar would dissolve in 100 grams of water
Answer:
I think is C 240 yup yup
Explanation:
At 40°C how many grams of sugar would dissolve in 100 grams of water?
arrange the following group of atoms in order of increasing atomic size:B,Al,Ga
Answer:
Al,Ga,B
Explanation:
Now since i helped you can you help me with this plz
Matteo took 5 math quizzes. The mean of the 5 quizzes was 8.2. Here are four of his quiz scores 7, 7, 8, 10. What is the 5th quiz score? Show work.
Boron ( B )
Aluminium ( Al )
Gallium ( Ga )
What do the symbols in the parentheses indicate
Answer:
(C) the physical state of each reactant and product
Explanation:
Hope this helps
what is an example of an electrolyte solution?
Answer:
nacl with water
they are capable of conducting electricity
All chemical reactions use reactants in a specific proportion or stoichiometry to regulate the amount of products produced. The __________ reactant is consumed completely during the reaction but some of the __________ reactant is left over.
a. limiting, excess
b. excess, limiting
c. proportional, ideal
d. chemical, stoichiometric
Answer: The answer is a. limiting, excess. I took the test.
Explanation:
Gallium (, ) is a metalloid obtained from its salts during the smelting of ores of other elements, like Zinc. has broad applicability in the electronics industry. It is also used as a safe replacement for mercury in thermometers as it melts at and has a heat of fusion of . What is the entropy change of of gallium in as it melts when placed on a surface at
The given question is incomplete. The complete question is:
Gallium (Ga, 69.723 g/mol) is a metalloid obtained from its salts during the smelting of ores of other elements, like Zinc. has broad applicability in the electronics industry. It is also used as a safe replacement for mercury in thermometers as it melts at 29.8 °C and has a heat of fusion of 5.59 kJ/mol. What is the entropy change of 22 g of gallium in J/K as it melts when placed on a surface at 29.8°C?
Answer: The entropy change of of gallium in as it melts when placed on a surface at is 5.81 J/K
Explanation:
Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.
Amount of heat required to melt 1 mole of Ga = 5.59 kJ
Mass of Ga given = 22 gram
Heat required to melt 69.723 g of Ga = 5.59 kJ
Thus Heat required to melt 22 g of Ga = [tex]\frac{5.59kJ}{69.723}\times 22=1.76kJ[/tex]
Temperature = [tex]29.8^0C=(273+29.8)kJ=302.8kJ[/tex]
Now entropy change = [tex]\frac{\text {heat of fusion}}{\text {temperature in K}}=\frac{1.76kJ}{302.8K}=0.00581kJ/mol=5.81J/K[/tex]
Thus the entropy change of of gallium in as it melts when placed on a surface at [tex]29.8^0C[/tex] is 5.81 J/K
Which factor contributes the most to the rates of diffusion and effusion between two gases in a mixture?
a)The size of the particles.
b)The molar mass of the particles.
c)The interactions between the particles.
d)The relative volume of the particles to each other.
Answer:
b)The molar mass of the particles.
Explanation:
The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham's law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases.
The reaction CO2+H2O=H2OCO3 is classified as
g At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H 2 (g) Br 2 (g) 2HBr (g) A mixture of 0.682 mol of H 2 and 0.440 mol of Br 2 is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.516 mol of H 2 present. At equilibrium, there are ________ mol of Br 2 present in the reaction vessel.
Answer: At equilibrium , there are 0.274 moles of [tex]Br_2[/tex]
Explanation:
Moles of [tex]H_2[/tex] = 0.682 mole
Moles of [tex]Br_2[/tex] = 0.440 mole
Volume of solution = 2.00 L
Initial concentration of [tex]H_2[/tex] = [tex]\frac{0.682}{2.00}=0.341 M[/tex]
Initial concentration of [tex]Br_2[/tex] = [tex]\frac{0.440}{2.00}=0.220 M[/tex]
Equilibrium concentration of [tex]H_2[/tex] = [tex]\frac{0.516}{2.00}=0.258 M[/tex]
The given balanced equilibrium reaction is,
[tex]H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)[/tex]
Initial conc. 0.341 M 0.220 M 0 M
At eqm. conc. (0.341-x) M (0.220-x) M (2x) M
Given : (0.341-x) M = 0.258 M
x= 0.083 M
Thus equilibrium concentartion of [tex]Br_2[/tex] = (0.220-0.083) M = 0.137 M
Thus moles of [tex]Br_2[/tex] at equilibrium = [tex]0.137M\times 2.00L=0.274mol[/tex]
At equilibrium , there are 0.274 moles of [tex]Br_2[/tex]
Bases are typically found in
Answer:
they are typically found in
1)soap
2) detergent
3) baking soda etc..
Help it’s due right now I will give you 15 points or more
Answer:
1. Temperature, the average kinetic energy of particles, indicates how warm something is. Thermal energy, or the overall kinetic energy of the particles, indicates how a substance or material will transmit heat or chill something else.
2. A thermal expansion is a matter to change in volume in response to a change in temperature.
3. Heat is the form of energy that is transferred between systems or objects with different temperatures.
4. Air temperature near or over bodies of water is much different from that over land due to differences in the way water and land heat and cool. Properties that affect water temperature are transparency, ability to circulate, and specific heat.
5. Radiation is the transfer of heat energy through space by electromagnetic radiation.
6. Radiation is natural and found everywhere, it comes from outer space, the air we breathe, and the earth we tread.
7. When a fluid, such as air or a liquid, is heated and then travels away from the source, it carries the thermal energy along. The fluid above a hot surface expands, becomes less dense, and rises.
8. Convection currents in the Earth occur in the mantle
9. The fire's heat causes molecules in the pan to vibrate faster, making it hotter. These vibrating molecules collide with their neighboring molecules, making them also vibrate faster.
10. Since air is a poor conductor, most energy transfer by conduction occurs right near Earth's surface. Conduction directly affects air temperature only a few centimeters into the atmosphere.
- Hope this helps!
What is factor that is covered up by another trait Dominant or recessive
Answer: Recessive
Explanation: The recessive trait is often faded away by the dominant/other
Also looking at the word dominant it means powerful, most important, influential meaning it would most likely overpower
Goodluck :)
Question List
Question 4 of 8
Total Points: 0 out of 80
Calculate the volume that will be occupied by
240 mL of hydrogen, measured at 360 mm Hg, when
the pressure is changed to 1080 mm Hg.
Answer:
80 mL
Explanation:
Step 1: Given data
Initial pressure of hydrogen (P₁): 360 mmHgInitial volume of hydrogen (V₁): 240 mLFinal pressure of hydrogen (P₂): 1080 mmHgFinal volume of hydrogen (V₂): ?Step 2: Calculate the final volume of hydrogen
If we assume ideal behavior and constant temperature, we can calculate the final volume occupied by hydrogen using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 360 mmHg × 240 mL/1080 mmHg = 80 mL
prop-1-yne + 2HBr/H2O2 = A;
A + 2H2O = B;
B + K2CO3(aq) = C;
C + heat = D;
D + HBr = E.
find the compounds A, B, C, D and E
Based on the given reactions, the compounds are as follows:
A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.
B: The product formed when compound A reacts with 2H2O.
C: The product formed when compound B reacts with K2CO3(aq).
D: The product formed from the heat-induced reaction of compound C.
E: The product formed when compound D reacts with HBr.
Based on the given reactions, let's analyze the compounds involved:
Reaction 1: prop-1-yne + 2HBr/H2O2 = A
The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.
Reaction 2: A + 2H2O = B
Compound A reacts with 2H2O (water) to form compound B.
Reaction 3: B + K2CO3(aq) = C
Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.
Reaction 4: C + heat = D
Compound C undergoes a heat-induced reaction to form compound D.
Reaction 5: D + HBr = E
Compound D reacts with HBr (hydrobromic acid) to form compound E.
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Write a balanced nuclear equation for the beta decay of 234/90Th
The balanced nuclear equation for the beta decay of 234/90Th can be represented as follows: ^234/90Th --> ^234/91Pa + e^0/-1β
In this equation, the nucleus of thorium-234 (234/90Th) undergoes beta decay. During beta decay, a neutron within the nucleus is converted into a proton, resulting in the emission of an electron (beta particle). As a result, the thorium-234 nucleus is transformed into protactinium-234 (234/91Pa) by gaining one proton.
The beta particle emitted during the decay process is represented as e^0/-1β, where the superscript 0 denotes that the electron has no charge (neutral), and the subscript -1 indicates that the electron carries a negative charge of -1.
It is important to note that in a nuclear equation, the total atomic mass and atomic number on both sides of the equation must be equal to maintain a balanced equation and conserve mass and charge.
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Good Morning I have a question I need help and can not find the answer o it maybe someone help me? The question is _______ Most Of the Energy that drives water cycle comes from__________? (this is from Science)
Answer:
hydro, water
Explanation:
Helllllp me please
Part D ends with neon Express your answer as an integer. 20
Answer:
attach file please
Which statement is accurate about variations?
Organisms with variations that give an advantage are more likely to survive
Organisms with variations are less likely to pass these variations to their offspring
In all cases vanations become more common in a population
In all cases variations affect how likely it is that an organism reproduces
Answer:
A,C,D
Explanation:
Variations effect the likelyhood of survival for a organism, it can give off an advantage or disadvantage.
Brainliest? pls
34 g of O2 are reacted with excess Cs, causing a production of 199 g of Cs2O. What is the percent yield of this
reaction?
2 Cs + O2 ---> Cs2O
Answer:
33.23 %
Explanation:
4 Cs + O₂ → 2Cs₂OFirst we convert 34 g of O₂ into moles, using its molar mass:
34 g O₂ ÷ 32 g/mol = 1.0625 mol O₂Then we convert O₂ moles into Cs₂O moles, using the stoichiometric coefficients of the balanced reaction:
1.0625 mol O₂ * [tex]\frac{2molCs_2O}{1molO_2}[/tex] = 2.125 mol Cs₂ONow we convert 2.125 moles of Cs₂O into grams, using its molar mass:
2.125 mol Cs₂O * 281.81 g/mol = 598.85 g Cs₂O598.85 g is the theoretical yield. Finally we proceed to calculate the percent yield:
199 / 598.85 * 100% = 33.23 %What is a special characteristic of Jupiter, Saturn, Neptune and Uranus?
Answer:
The gas and ice giants — Jupiter, Saturn, Uranus and Neptune — are outliers. They are much larger than the terrestrial planets, but their cores are small and icy. Most of their size is formed by a combination of gases that become denser and hotter as you get closer to the core.
T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round___ Intermetallic compounds are usually hard and brittle.___ For effective dispersion strengthening, the dispersed phase should be continuous.___ Stoichiometric intermetallic compounds exist overa range of compositions.___ Faster solidification results in smaller interlamellar spacing
Answer:
TRUEFALSETRUEFALSEFALSETRUEExplanation:
At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack. Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hardFor the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous Stoichiometric intermetallic compounds exist over a range of compositions : FALSEFaster solidification results in smaller interlamellar spacing : TRUEA sample of O2 occupies 75 L at 1 atm. If the volume of the
sample doubles, what is the new pressure of Oz?
Answer:
the new pressure of Oz is 0.5 atm
Explanation:
The computation of the new pressure of Oz is shown below:
Given that
Original pressure of O2 (P1) = 1 atm
And, the Original volume of oxygen O2 (V1) = 75L
Now
New pressure of O2 (P2) = ?
And, the New volume of O2 (V2) = 2V1 i.e
= (2 × 75L)
= 150L
Here we applied the Boyle's law formula
P1V1 = P2V2
1 atm × 75L = P2 × 150L
75 atm × L = 150L ×P2
Now Divide both sides by 150L to get P2
75 atm × L ÷ 150L = 150L × P2 ÷ 150L
0.5 atm = P2
Hence, the new pressure of Oz is 0.5 atm
Write a structural formula for the principal organic product formed in the reaction of methyl bromide with each of the following compounds:
1. Sodium hydroxide
2. Potassium ethoxide
3. Sodium benzoate
4. Lithium azide
Answer:
1. CH3Br + NaOH --------> H3C---OH + Na---Br
Explanation: