Answer:
Because of the artificial lights influencing the dark night, turning it into a grey.
In a given compound, the mass of sulfur (S) is 40.05 g and the mass of oxygen (O) is 59.95 g. What is
the empirical formula of the compound
Answer:
[tex]SO_3[/tex]
Explanation:
Hello there!
In this case, according to the definition of the empirical formula as the smallest representation of the molecular formula of a chemical compound; for us to determine it, we first need to calculate the moles of sulfur in 40.05 g and those of oxygen in 59.95 g as shown below:
[tex]n_S=\frac{40.05g}{32.07g/mol}= 1.25molS\\\\n_O=\frac{59.95g}{16.00g/mol}=3.75molO[/tex]
Now, we perform the following mole ratios to figure out the subscripts in the empirical formula, by dividing over the fewest number of moles:
[tex]S=\frac{1.25}{1.25}=1 \\\\O=\frac{3.75}{1.25}=3[/tex]
Therefore, the empirical formula turns out:
[tex]SO_3[/tex]
Best regards!
arrange the following group of atoms in order of decreasing atomic size:B,Al,Ga
Answer:
B<Ga<Al
Explanation:
Hope this helps
For any molecule at the surface the resulting surrounding forces equal zero
True
False
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A voltaic cell is constructed in which the anode is a Fe|Fe2 half cell and the cathode is a Cd|Cd2 half cell. The half-cell compartments are connected by a salt bridge. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) The anode reaction is:
Answer: The Anode reaction is [tex]Fe(s)\rightarrow Fe^{2+}(aq)+2e^-[/tex]
Explanation:
Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.
Oxidation reaction is defined as the reaction in which a substance looses its electrons.Oxidation occurs at anode.
Anode : [tex]Fe(s)\rightarrow Fe^{2+}(aq)+2e^-[/tex]
Reduction reaction is defined as the reaction in which a substance gains electrons. Reduction occurs at cathode.
Cathode : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]
A compound containing nitrogen and oxygen is decomposed in the laboratory. It produces 24.5 g nitrogen and 70.0 g oxygen. Calculate the empirical formula of the compound.
Answer:
N2O5
Explanation:
1. Convert to moles
24.5g N * 1mol/14g = 1.75
70.0g * 1mol/16g = 4.375
2. Divide each value by the smallest
1.75/1.75 = 1
4.375/1.75 = 2.5
3. Multiply each by a whole number so that they are both whole numbers
1*2 = 2
2.5*2 = 5
4. These are moles of elements present in the compound
Answer: N2O5
Question: You decompose a compound containing nitrogen and oxygen in the laboratory and produce 24.5 g of nitrogen and 70.0 g of oxygen. Calculate the empirical formula of the compound.
The solubility of a solid in a liquid generally increases with increase temperature.
True
Or
False
Answer:
False, Solubility increases with temperature for most solids.
Explanation:
A buffered solution _______. Select the correct answer below: fails to keep hydronium and hydroxide ion concentrations nearly constant when strong acids or bases are added. maintains a constant or nearly constant pH when small amounts of strong acids or bases are added. acts to keep the hydroxide ion concentration nearly constant. acts to keep the hydronium ion concentration nearly constant.
The correct option for the given question about Buffer Solution is Option B) which is maintains a constant or nearly constant pH when small amounts of strong acids or bases are added.
What is a Buffer Solution?When a little quantity of acid or base is diluted or added, the buffer solution undergoes very slight variations in its hydrogen ion concentration (pH). pH may be maintained in buffer solutions, which are mixtures of a weak acid and its conjugate base or a weak base and its conjugate acid.Acidic and alkaline buffers are the two main groups into which buffer solutions are commonly categorized.A weak acid and its salt are combined with a strong base to create an acidic buffer, which has an acidic pH.A weak base, its salt, and a strong acid are combined to create an alkaline buffer, which has a basic pH.
Thus we conclude that when weak acids or bases are supplied in small amounts, the pH remains steady or almost constant.
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15. Kinetic and potential energy both relate to
a. friction
a. heat
b. light
d. motion
Answer:
All forms of energy are either potential or kinetic energy. Potential refers to stored energy while kinetic is energy in motion.
Explanation:
hope help you pls thanks...
D is correct option
Nitrous oxide (N2O), more commonly known as laughing gas, is used as a mild sedatitive during various dental procedures.As a gas, it has a densityof 1.977 x 10-3g/mL.Wheniron is exposed to oxygen it forms rust (Fe2O3), which is a solid and has a density value of 5.25 g/mL.Why are the density values so different among these substances?
a)The metal atoms weigh more than the atoms of the gas.
b)The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.
c)The metal is a solid, and solids weigh more based on the principles of their states of matter.
d)There are fewer gas particles than solid particles in the same volume.
Answer:
B.
Explanation:
The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.
How many moles of magnesium nitrate are produced when he reacts 0.34 moles of nitric acid with excess magnesium?
Bases are typically found in
Answer:
they are typically found in
1)soap
2) detergent
3) baking soda etc..
A 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
what volume of 1.75 h2so4 would be needed to neutralize 350 ml of 3.33 m sodium hydroxide?
Answer:
1.332 L
Explanation:
The reaction that takes place is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂OFirst we calculate the number of NaOH moles, using the given volume and concentration:
350 mL * 3.33 M = 1165.5 mmol NaOHThen we convert NaOH moles into H₂SO₄ moles, using the stoichiometric coefficients:
1165.5 mmol NaOH * [tex]\frac{2mmolH_2SO_4}{1mmolNaOH}[/tex] = 2331 mmol H₂SO₄Finally we calculate the required volume of a 1.75 M solution of H₂SO₄, using the concentration and number of moles:
2331 mmol / 1.75 M = 1332 mLWe can convert mL into L:
1332 mL / 1000 = 1.332 LReaction of Nitrile with Grignard Grignard reagents react with nitriles to give an intermediate imine anion that is hydrolyzed by the addition of water to yield a ketone. The mechanism is similar to the reduction of a nitrile to an amine except that only one nucleophilic addition occurs instead of two, and the nucleophile is a carbanion rather than a hydride ion.
Required:
Draw curved arrows to show the movement of electrons. in this step of the mechanism. Arrow-pushing Instructions
Answer:
its a.
Explanation:
if a gas sample has a pressure of 30.7 kPa at 273 K, what does the temperature have to decrease to lower the pressure to 28.4 kPa?
253K
268K
262K
Which of the following sentences describe how latitude affects a biome? (There may be more than one answer).
Question 4 options:
In the far north, the summers are too short and cold for trees.
As you travel east, the climate gets colder.
The closer the area is near to the equator, the hotter the weather.
The higher in elevation, the colder the environment.
Answer:
A and C
Explanation:
The other answers didn't seem reasonable,
I took the quiz and got it right.
If I am wrong, please correct me
The biome can be given as the collection of flora and fauna in the region. The latitude affects the biome, with the short summers in the north and hot weather at the equator.
What is latitude?The latitude can be given as the horizontal lines imaginary drawn on the earth. The equator is the center of latitude passing from the middle of the earth.
The sun's rays fall directly towards the equator, while the poles in the north and the south experiences the transverse rays of the sun.
The direct rays falling at the equator termed the area closer to the equator warmer, while the falling rays to the north resulted in the short summers and cold for trees. Thus, options A and C are correct.
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You are given three liquids, pentane (C5H12), water (H2O), and tetrachloromethane (CCl4), and are told to make solutions.
a. Predict whether you will be able to make a solution with pentane and tetrachloromethane
b. Predict whether you will be able to make a solution with pentane and water, and give your reasoning.
Answer:
A. predict whether you will be able to make a solution with pentane and tetrachloromethane
Explanation:
Sana nakatulong
Several elements must be transmuted to sea the real implications of this event
a.surpassed
b.estimated
c.changed
d.summed
thanks guys :)
How much heat is transferred when 7.19 grams of H2 reacts with excess nitrogen, according to the following equation: N2(g) + 3 H2 (g) --> 2 NH3 (g) \DeltaΔH = +46.2 kJ
Answer:
[tex]Q=54.8kJ[/tex]
Explanation:
Hello there!
In this case, according to the given chemical reaction, it is possible for us to realize that the 46.2 kJ of energy are given per mole of reaction, which are related to 3 moles of hydrogen; Thus, we can calculate the energy per mole of hydrogen as shown below:
[tex]\Delta H=\frac{46.2kJ}{mol} *\frac{1mol}{3molH_2}\\\\ \Delta H=15.4\frac{kJ}{molH_2}[/tex]
Now, to calculate the total energy, we convert the grams to moles of hydrogen as shown below:
[tex]Q=7.19gH_2*\frac{1molH_2}{2.02gmolH_2}*15.4\frac{kJ}{molH_2} \\\\Q=54.8kJ[/tex]
Best regards!
How 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
A scientist wants to display the numbers of gallons of fuel that were used up
in a one-hour period by three different types of motors. What type of visual
display should the scientist use?
Help please
Answer:
bar graph
Explanation:
3. What is the mass % of a solution that contains 36g KCl in 475g of water?
Answer:
%KCl = 7.05%
%Water = 92.95%
Explanation:
Step 1: Given data
Mass of KCl (solute): 36 gMass of water (solvent): 475 gStep 2: Calculate the mass of the solution
The mass of the solution is equal to the sum of the masses of the solute and the solvent.
m = 36 g + 475 g = 511 g
Step 3: Calculate the mass percentage of the solution
We will use the following expression.
%Component = mComponent/mSolution × 100%
%KCl = 36 g/511 g × 100% = 7.05%
%Water = 475 g/511 g × 100% = 92.95%
If 5.32 mols N2 and 15.8 mols H2 react together, what mass NH3 can be
produced? Which is the limiting reactant?
Answer:
2.87 gram
N2 is the limiting agent
Explanation:
We will find out if there is sufficient N2 and h2 to produce NH3
a) For 2.36 grams of N2
Molar mass of N2 = 28.02
Number of moles of N2 in 2.36 grams = 2.36/28.02
Mass of NH3 = 17.034 g
Now NH3 produced form 2.36 grams of N2 =
2.36/28.02 * 2 * 17.034 = 2.87 g NH3
b) For 1.52 g of H2
NH3 produced = 1.52/2.016 * (2/3) * 17.034 = 8.56
N2 Is not enough to produce 2.87 g of NH3 and also H2 is not enough to make 8.56 g of NH3.
N2 is the limiting agent as it has smaller product mass
A 0.10M NH3 solution is 1.3% ionized, calculate the hydrogen ion concentration
Answer:NH3 + H2O <==> NH4+ + OH-
0.10 x 0.013 = 0.0013 M (this is the hydroxide concentration)
Kw = [H+] [H-]
1.00 x 10^-14 = (x) (0.0013)
x = 7.7 x 10^-12 M
Explanation:
At some temperature for the equilibrium PX3(g) + X2(g) # PX5(g) the equilibrium constant is 0.74. At the same temperature the equilibrium constant for PX5(9) + PX3(g) X2(g) is
Answer: The equilibrium constant for [tex]PX_5(g)\rightarrow PX_3(g)+X_2(g)[/tex] is 1.35
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_c[/tex]
For the given chemical reaction:
[tex]PX_3(g)+X_2(g)\rightarrow PX_5(g)[/tex]
The expression for [tex]K_c[/tex] is written as:
[tex]K_c=\frac{[PX_5]^1}{[PX_3]^1[X_2]^1}[/tex]
[tex]0.74=\frac{[PX_5]^1}{[PX_3]^1[X_2]^1}[/tex]
For the reverse chemical reaction:
[tex]PX_5(g)\rightarrow PX_3(g)+X_2(g)[/tex]
The expression for [tex]K_c'[/tex] is written as:
[tex]K_c'=\frac{[PX_3]^1[X_2]^1}{[PX_5]^1}[/tex]
[tex]K_c'=\frac{1}{K_c}=\frac{1}{0.74}=1.35[/tex]
The equilibrium constant for [tex]PX_5(g)\rightarrow PX_3(g)+X_2(g)[/tex] is 1.35
arrange
these liquids in order of increasing surface tension: water,
hexane, ethanol.
The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension:
water, alcohol (C2H5OH) and hexane [CH3(CH2)4CH3)].
Answer:
Hexane, Methanol, Water
Explanation:
I did the assessment
2. Calculate the concentration (in molarity) of a NaOH solution if 25.0 mL of the solution are needed to neutralize 17.4 mL of a 0.312 M HCl solution.
Answer:
0.217 M NaOH
Explanation:
M1V1 = M2V2
M1 = 0.312 M HCl
V1 = 17.4 mL HCl
M2 = ?
V2 = 25.0 mL NaOH
Solve for M2 --> M2 = M1V1/V2
M2 = (0.312 M)(17.4 mL) / (25.0 mL) = 0.217 M NaOH
Titrations provide a method of quantitatively measuring the concentration of an unknown solution.
a. Trueb. False
Answer: True
Explanation:
Titration is a quantitative technique where a solution of known concentration is used to determine the concentration of an unknown solution. The titrant (the known solution) is added from a burette to a known quantity of the the unknown solution until the reaction is complete.
By knowing the volume of titrant, we can determine the concentration of the unknown.
Thus the statement Titrations provide a method of quantitatively measuring the concentration of an unknown solution is True .
how many mg are in 125ml of
phosphoric acid?
Answer:
125000mg
Explanation:
1ml = 1000mg
125(1000) = 125000mg
2C 2 H 2 (g)+5O 2 (g) CO 2 (g) + 2H 2 O (l)
Answer:
2c2gh2+5cg2o4+2h2lo
Explanation: