Answer:
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
Explanation:
kh uh
A substance that contains two or more kinds of atoms is
1. (An example of non-point-source pollution is
A. Wastewater from a leaking pipe.
B. Fertilizer runoff from lawns.
C. Oil from a leaking underground tank.
D. Chemicals released from a paper mill
Answer:
i think the answer is c.
Explanation:
True or false. Law of action/reaction: for every action there is an equal and the same reaction. Pls don’t give me a link and tell me the answe
Convert 0.0338 moles of K3PO4 to grams.
Which fossil fuel was formed from the bodies of prehistoric animals and plants?
Answer:
coal
Explanation:
What is the source of Earth's magnetic field according to what you have read in chapter 222
Answer:
Scientists know that today the Earth's magnetic field is powered by the solidification of the planet's liquid iron core. The cooling and crystallization of the core stirs up the surrounding liquid iron, creating powerful electric currents that generate a magnetic field stretching far out into space.
Two gas particles collide together in a sealed container. What can be said about the kinetic energy of the two gas particles?
a)During the collision, each gas particle transfersall of its kinetic energy to the other particle.
b)The total kinetic energy of both gas particles will remain the same, but they can transfer any amount to each other.
c)The new kinetic energy of each gas particle is the sum of the kinetic energy that each particle is traveling with.
d)The kinetic energy of each gas particle remains the same as kinetic energy cannot be transferred between particles with zero volume.
Explanation:
c)The new kinetic energy of each gas particle is the sum of the kinetic energy that each particle is traveling with.
pls help me with this question thank you
A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by
mass.
Question 1: The empirical formula for this compound is :
Question 2: The molar mass for this compound is 82.98 g/mol.
The Molecular formula for this compound is:
Answer: 1. The empirical formula is [tex]PNF_2[/tex]
2. The molecular formula is [tex]PNF_2[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of P = 37.32 g
Mass of N = 16.88 g
Mass of F = 45.79 g
Step 1 : convert given masses into moles.
Moles of P =[tex]\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles[/tex]
Moles of F =[tex]\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For P = [tex]\frac{1.20}{1.20}=1[/tex]
For N = [tex]\frac{1.20}{1.20}=1[/tex]
For F =[tex]\frac{2.41}{1.20}=2[/tex]
The ratio of P: N: F= 1: 1: 2
Hence the empirical formula is [tex]PNF_2[/tex]
The empirical weight of [tex]PNF_2[/tex]= 1(31)+1(14)+2(19)= 82.98 g.
The molecular weight = 82.98 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1[/tex]
The molecular formula will be=[tex]1\times PNF_2=PNF_2[/tex]
Is chemistry required to get into speech language pathology?
Answer:
As of right now (4-7-2021), the physical science requirement for the SLP certificate must be met by completing coursework in the areas of either chemistry or physics.
PLEASE HURRY! What is the name of this alkane?
A skeletal model has a hexagon ring with points at the top and bottom. C H 3 is bonded to the top point, and a carbon and another C H 3 are bonded to the bottom right point.
1-methyl,3-ethylnonane
1-methyl,5-ethylisohexane
1-ethyl,3-methylcyclohexane
1-ethyl,5-methylcyclohexane
Answer:
C. 1-ethyl, 3-methylcyclohexane
(Photo for proof at the bottom.)
Explanation:
The 1-ethyl is because you start numbering from the longest branch, towards the next closest branch. Prefix "eth-" means two, there are 2 carbons in the longest branch. 3-methyl is because the next branch is at number 3, and prefix "meth-" means 1, there is 1 carbon in that chain. "Cyclo" in cyclohexane means the skeletal model is shaped like a ring, and the "hexane" means there are 6 carbons in the ring. Prefix "hex" means 6.
Here's a photo of the unit review on Edge. Refer to the 2nd attachment for a visualization.
Please click the heart if this helped.
Answer:
C. 1-ethyl,3-methylcyclohexane
Explanation:
Nitric oxide (NO) reacts with oxygen gas to produce nitrogen dioxide. A gaseous mixture contains 0.66 g of nitric oxide and 0.58 g of oxygen gas. After the reaction is complete, what mass of nitrogen dioxide is formed? Which reactant is in excess? How do you know? Suppose you actually recovered 0.91 g of nitrogen dioxide. What is your percent yield?
Answer:
NO is the limiting reagent.
In this reaction 0.886 mole of NO2 is produced
Explanation:
The chemical equation for this reaction is
2NO(g) + O2(g) → 2NO2(g)
In this limiting reagent reaction, 2 moles of NO reacts with one mole of O2 to produce 2 mole of 2NO2
0.886 mole of NO * (2 mole of NO2/2 mole of NO) = 0.886 mole of NO2
0.503 mole of O2 * (1 mole of NO2/1 mole of O2) = 1.01 mole of NO2
Hence, NO is the limiting reagent.
In this reaction 0.886 mole of NO2 is produced
A car generates 2552 N and weighs 2250 kg. What is its rate of acceleration
2 m/s ^2
0.88m/s^2
5,742,000m/s^2
1.13m/s^2
Answer:
[tex]a=1.134\frac{m}{s^2}[/tex]
Explanation:
Hello there!
In this case, by considering the physical definition of force in terms of mass and acceleration:
[tex]F=m*a[/tex]
Given the generated force and the involved mass, we can compute the required acceleration as shown below:
[tex]a=\frac{F}{m}\\\\a=\frac{2552N}{114kg}[/tex]
Yet it is necessary to break out Newtons to:
[tex]a=\frac{2552\frac{kg*m}{s^2} }{114kg}\\\\a=1.134\frac{m}{s^2}[/tex]
Best regards!
What would you call the bright blue liquid?
What is the change in enthalpy when 180 g of water vapor condenses at 100°C? (AH, = 40.67 kJ/mol)
a. 565 kJ
b. -565 kJ c.
-407 kJ d.
407 kJ
Answer:
[tex]Q=-407kJ[/tex]
Explanation:
Hello there!
In this case, considering that the heat has two forms, sensible (variable temperature) and latent (constant temperature), we can notice that phase changes account for latent heat as the temperature remains the same. In such a way, given the enthalpy of vaporization of water, 40.67 kJ/mol, the enthalpy of condensation (reverse process) is the negative value, -40.67 kJ/mol; therefore, the associated latent heat would be:
[tex]Q=180g*\frac{1mol}{18.02g} *-40.67\frac{kJ}{mol} \\\\Q=-407kJ[/tex]
Best regards!
The answer is
C) -407 kJ d
Which of the following best describes the type of particle found in the cloud around an atom’s nucleus?
It has a negative charge and much less mass than a neutron
It has a negative charge and about the same mass as a neutron.
It has a negative charge and about the same mass as a neutron.
It has a positive charge and about the same mass as a neutron.
1. What volume of a 2.50M Kl(aq) is needed to make 200 ml of a 1.OOM KI)aq)?
Answer:
80 ml
Explanation:
From the question,
Applying Dilution formular
MV = mv................... Equation 1
Where M = Molarity of Kl before dilution, V = Volume of Kl before dilution, m = molarity of Kl after dilution, v = volume of Kl after dilution.
make V the subject of the equation
V = mv/M............. Equation 2
Given: m = 1.00 M, v = 200 ml, M = 2.50 M
Substitute these values into equation 2
V = (1.00×200)/2.50
V = 80 ml
When 0.20 mol of hydrogen gas and 0.15 mol of iodine gas are heated at 450 °C until equilibrium is established, the equilibrium mixture is found to contain 0.26 mol of hydrogen iodide. The equation for the reaction is as follows
H2((g) + I2(g) ⇋ 2HI(g)
What is the correct expression for the equilibrium constant, Kc?
Answer:
[tex]Kc=\frac{[HI]^2}{[I_2][H_2]}[/tex]
Explanation:
Hello there!
In this case, for these equilibrium problems it is firstly necessary to know the balanced reaction at equilibrium:
H2((g) + I2(g) ⇋ 2HI(g)
Next, by means of the law of mass action, it turns out possible for us to write the required and correct expression for the equilibrium constant by considering the concentrations and the coefficients in the aforementioned reaction:
[tex]Kc=\frac{[HI]^2}{[I_2][H_2]}[/tex]
Best regards!
Which nitrogenous base is NOT found in DNA? *
A. cytosine
B. guanine
C. uracil
D. adenine
Answer:
c. uracil
Explanation:
uracil is not found in DNA. the missing base would be thymine
Argon occupies a volume of 28.3 L at a pressure of 0.36 atm. Find the volume of argon when the pressure is increased to 0.85 atm.
Answer:
12
Explanation:
Boyles law
P1 (.36)
V1(28.3)
P2(.85)
V2(?)
P1*V1=P2*V2
plug it in, and you get 12
Why aren't gas molecules attracted to each other? Explain.
Answer:
As gas is compressed, the individual molecules begin to move in each others way creating a very repulsive force. It acts to oppose any further volume decrease. At very close distances, all molecules repel each other as their electron clouds emerge.
Explanation:
PLEASE HELP!!!!!!!
Answer:
2nd option
Explanation:
How many moles of magnesium nitrate are produced when he reacts 0.34 moles of nitric acid with excess magnesium?
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:
What keeps the Earth orbiting on its current path
in our solar system?
Step 2: Now that you have counted the atoms on each side, add coefficients to balance the equation. Check your work by updating the number of atoms on each side in the table below (Remember to multiply coefficients by subscripts): NaF + NaBr + CaF2 CaBr2 Reactants Products Na- Na- Br- Br- Ca- Ca- IF-
Answer:
The answers to your questions are given below.
Explanation:
__ NaBr + CaF₂ —> __ NaF + CaBr₂
The above equation can be balance as follow:
NaBr + CaF₂ —> NaF + CaBr₂
There are 2 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 2 before NaF as shown below:
NaBr + CaF₂ —> 2NaF + CaBr₂
There are 2 atoms of Na on the right side and 1 atom on the left side. It can be balance by writing 2 before NaBr as shown below:
2NaBr + CaF₂ —> 2NaF + CaBr₂
Now, the equation is balanced.
Atom >>>> Reactant >>>> Product
Na >>>>>> 2 >>>>>>>>>>> 2
Br >>>>>>> 2 >>>>>>>>>>> 2
Ca >>>>>> 1 >>>>>>>>>>>> 1
F >>>>>>> 2 >>>>>>>>>>>> 2
32.14 mL of a 0.05 M sulfuric acid (H2SO4) solution are required to neutralise 25 mL of a sodium hydroxide (NaOH) solution.
What is the concentration (in mol L-1) of SO42- ions in the final solution?
The concentration of SO42- ions in the final solution is 0.01 mol L-1.
The neutralisation reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) requires the use of a chemical equation that is balanced:
H2SO4 + NaOH = Na2SO4 + 2H2O
We may deduce from the balanced equation that one mole of H2SO4 reacts with one mole of Na2SO4, meaning that the amount of SO42- ions produced is equal to the amount of H2SO4 utilised.
We may determine how many moles of H2SO4 were used by calculating the amount of 0.05 M H2SO4 solution needed to neutralise 25 mL of NaOH solution:
H2SO4 moles are equal to the product of the volume of H2SO4 (in L) and the concentration of H2SO4 (in mol L-1): 0.03214 L 0.05 mol L-1 = 0.001607 mol
The formation of SO42- ions results in the same amount of moles as the
The resulting solution contains the following amounts of SO42- ions per mole of H2SO4 used:
Moles of SO42- ions/volume of solution (in L) = 0.001607 mol/(0.025 L + 0.03214 L) = 0.01 mol L-1; concentration of SO42- ions (in mol L-1)
As a result, the final solution contains 0.01 mol L-1 of SO42- ions.
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The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine: The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate: In the analysis of a 1.54-g sample, 0.0596 g of barium iodate was recovered. Express the results of this analysis as percent potassium iodide.
Answer: The percentage of potassium iodide in the sample is 2.63 %.
Explanation:
The chemical equation for the reaction of iodide ions with bromine gas follows:
[tex]I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+[/tex] (i)
The chemical equation for the reaction of iodate ions with barium ions follows:
[tex]Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2[/tex] ......(ii)
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of barium iodate = 0.0596 g
Molar mass of barium iodate = 487.13 g/mol
Using equation 1:
[tex]\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles[/tex]
By Stoichiometry of the reaction (ii):
1 mole of barium iodate is produced by 2 moles of iodate ions
So, [tex]1.22\times 10^{-4} moles[/tex] of barium iodate will be produced by [tex]\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodate ions
By the stoichiometry of the reaction (i):
1 mole of iodate ions are produced by 1 moles of iodine ions
So, [tex]2.44\times 10^{-4} moles[/tex] of iodate ions will be produced by [tex]\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodine ions
Moles of potassium iodide = Moles of iodide ions = [tex]2.44\times 10^{-4}[/tex]
Since, the molar mass of potassium iodide = 166 g/mol
Using equation 1:
[tex]\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g[/tex]
To calculate the percentage by mass of a substance, the equation used is:
[tex]\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100[/tex]
Mass of a solution = 1.54 g
Mass of potassium iodide = 0.0405 g
Using above equation:
[tex]\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%[/tex]
Hence, the percentage of potassium iodide in the sample is 2.63 %.
Expressing the results of potassium iodide in percentage = 2.63%
The chemical reaction of iodine ions with Bromine gas can be expressed as :
I⁻ + 3Br₂ + 3H₂O -- > 6Br⁻ + IO₃ + 6H⁺ ----- ( 1 )
Chemical reaction between iodate ions with barium ions can be expressed as : Ba²⁺ + 2IO⁻₃ ------> Ba ( IO₃ )₂ --------- ( 2 )
Step 1 : Calculate the number of Barium iodate moles
mass of Barium iodate = 0.0596 g
molar mass of Barium iodate = 487.13 g/mol
from equation ( 1 )
moles of Barium iodate = ( 0.0596 ) / ( 487.13 ) = 1.22 * 10⁻⁴ moles
also from equation ( 1 ) the moles of potassium iodide = moles of iodide ions
= 2.44 * 10⁻⁴
molar mass of potassium iodide = 166 g/mol
Next step : Determine the mass of potassium iodide
moles of potassium * molar mass
= 2.44 * 10⁻⁴ * 166 g/mol = 0.0405 g
Final step : Determine the percentage of potassium iodide in the solution
Percentage = ( mass of potassium iodide / mass of solution ) * 100
= ( 0.0405 / 1.54 ) * 100
= 2.63%
Hence we can conclude that potassium iodide in percentage = 2.63%
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When NH3(g) reacts with O2(g) to form N2O(g) and H2O(l), 342 kJ of energy are evolved for each mole of NH3(g) that reacts. Write a balanced thermochemical equation for the reaction with an energy term in kJ as part of the equation.
Answer: [tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]
Explanation:
The skeletal thermochemical equation for the reaction is:
[tex]NH_3(g)+O_2(g)\rightarrow N_2O(g)+H_2O(l)+342kJ[/tex]
The balanced thermochemical equation for the reaction is:
[tex]2NH_3(g)+2O_2(g)\rightarrow N_2O(g)+3H_2O(l)+684kJ[/tex]
When 1 mole of [tex]NH_3[/tex] reacts with oxygen , heat released = 342 kJ
Thus when 2 moles of [tex]NH_3[/tex] reacts with oxygen , heat released = [tex]\frac{2}{1}\times 342 kJ=684kJ[/tex]
How many grams of Co, are produced when 88 g of o, is reacted with an excess of
butane?
Answer:
[tex]74.5gCO_2[/tex]
Explanation:
Hello there!
In this case, for the described reaction, it is possible to realize there is a 13:8 mole ratio between oxygen (O2) and carbon dioxide (CO2); moreover, since the molar mass of the former is 32.00 g/mol and that of latter is 44.01 g/mol, the produced mass of the required product turns out to be:
[tex]88gO_2*\frac{1molO_2}{32.00gO_2}*\frac{8molCO_2}{13molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\=74.5gCO_2[/tex]
Best regards!