Answers
Answer:
–0.13 Pa.m²
Explanation:
From the question given above, the following data were obtained:
Measurement (Pa.mm²) = –1.3×10⁵ Pa.mm²
Measurement (Pa.m²) =?
We can convert from Pa.mm² to Pa.m² by doing the following:
1 Pa.mm² = 1×10¯⁶ Pa.m²
Therefore,
–1.3×10⁵ Pa.mm² = –1.3×10⁵ Pa.mm² × 1×10¯⁶ Pa.m² / 1 Pa.mm²
–1.3×10⁵ Pa.mm² = –0.13 Pa.m²
Thus, –1.3×10⁵ Pa.mm² is equivalent to –0.13 Pa.m².
The complete equation will be:
[tex](-1.3\times 10^5 Pa.mm^2)\times 10^{-6}=(-0.13) Pa.m^2[/tex]
Explanation:
Given:
The equation to convert a measurement:
[tex](-1.3\times 10^5 Pa.mm^2)\times ? = ? Pa.m^2[/tex]
To find:
The missing part of the equation.
Solution:
[tex](-1.3\times 10^5 Pa.mm^2)\times ? = ? Pa.m^2[/tex]
On LHS the unit is in [tex]Pa. mm^2[/tex] and RHS the unit is in [tex]Pa.m^2[/tex] which means that we have to convert [tex]mm^2[/tex] to [tex]m^2[/tex]
In 1 millimeter there are 0.001 meters.
[tex]1 mm = 0.001 m\\1 mm^2=0.000001 m^2=10^{-6} m^2[/tex]
So, the complete equation will be:
[tex](-1.3\times 10^5 Pa.mm^2)\times 10^{-6}=(-0.13) Pa.m^2[/tex]
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Related Questions
The aldol reaction is catalyzed by acid as well as base. What is the reactive nucleophile in the acid-catalyzed aldol reaction
Answers
Answer:
The reactive nucleophile is Ketone.
Explanation:
In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.
The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.
Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles
Choose true or false
1. Acetone cannot be used as a solvent because the Grignard reagent will react with its carbonyl, instead of reacting with the planned synthesis carbonyl.
2. Tetrahydrofuran is not a suitable solvent for the Grignard reaction because his oxygen may form complexes with the Mg, deactivating the Grignard.
3. Phenol can be used as a solvent in Grignard reaction as long as is anhydrous and moisture is kept out of the system with a drying tube with drierite.
4. Syringes are used in the Grignard experiment to avoid mixing all the reagents.
5. We use syringes to inject the reagents through a septum preventing moisture to enter the system.
Answers
Explanation:
Grignard reagent reacts with ketones. Upon chemical reaction of acetone and Grignard reagent there will be formation of tertiary alcohol.
1). So the statement, acetone cannot be used as a solvent because the Grignard reagent will react with its carbonyl, instead of reacting with the planned synthesis carbonyl is true.
2). Tetrahydrofuran is not a suitable solvent for the Grignard reaction because his oxygen may form complexes with the Mg, deactivating the Grignard is a false statement.
3). Phenol can be used as a solvent in Grignard reaction as long as is anhydrous and moisture is kept out of the system with a drying tube with drierite is false statement.
4). Syringes are used in the Grignard experiment to avoid mixing all the reagents is false statement.
5). We use syringes to inject the reagents through a septum preventing moisture to enter the system is true statement.
A straight chain hydrocarbon with the formula C5H8_____
Answers
Answer:
I has 2 double carbon carbon bonds
Which of the following is an organic compound?
ammonia (NH3)
calcium sulfide (CaS)
octane (C8H18)
sulfur trioxide (SO3)
Answers
Organic compround is octane C8H18
8. An experiment requires a solution that is 80%
methyl alcohol by volume. What volume of
methyl alcohol should be added to 200 mL of
water to make this solution?
Answers
Answer:
[tex]v_{solute}=160mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the volume of methyl alcohol solute by using the definition of by-volume percentage:
[tex]\%v=\frac{v_{solute}}{v_{solution}} *100\%[/tex]
Whereas we solve for the volume of the solute as shown below:
[tex]v_{solute}=\frac{\%v*v_{solution}}{100\%} \\\\v_{solute}=\frac{80\%*200mL}{100\%}\\\\ v_{solute}=160mL[/tex]
Regards!
A 24.803 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 79.733 g of water. A 10.560 g aliquot of this solution is then titrated with 0.1077 M HCl . It required 32.37 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
Answers
Answer:
2.37 (w/w)% of NH3 in the fertilizer
Explanation:
The HCl reacts with NH3 as follows:
HCl + NH3 ⇄ NH4Cl
To solve this question we must find the moles of HCl used in the titration = Moles NH3. With its molar mass we can find mass of NH3 and using the dilutions we can find weight percent as follows:
Moles HCl = Moles NH3
32.37mL = 0.03237L * (0.1077mol/L) =
Mass NH3 in the dilution -Molar mass: 17.031g/mol-
0.003486moles NH3 * (17.031g/mol) = 0.05937g NH3
Mass NH3 in the sample:
0.05937g NH3 * (79.733g + 24.803g) / 10.560g =
0.588g NH3
Weight percent:
0.588g NH3 / 24.803g * 100 =
2.37 (w/w)% of NH3 in the fertilizer
Acetylide ions react with aldehydes and ketones to give alcohol addition products.
a. True
b. False
Answers
Answer:
a
Explanation:
Electrode A has a standard reduction potential of -0.21 volts and electrode B has a standard reduction potential of -0.15 volts. What electrode is the anode?
will give brainiest to person with the best answer ♡
Answers
So, we know the reduction potentials of the electrodes :
Electrode A = -0.21 V
Electrode B = -0.15 V
We want our cell to be spontaneous - this means that the voltage has to be positive. In order to do so, we need to turn the more negative reduction half reaction to be an oxidation half reaction.
-0.21 is more negative than -0.15, so electrode A will be the oxidation half reaction.
Anode is where the oxidation half reaction takes place, so electrode A is the correct answer . A fun way to remember this is An(ode)Ox(idation) = AnOx and Red(uction)Cat(hode) = RedCat
The E° cell will be 0.21 -0.15 V = 0.6 V
Help please with this question
Picture above
Answers
Answer:
3:372-88U:771-772828
What is the right answer?
Answers
Answer:
equal to zero is the right answer
Which statement best describes the formula equation Cl2(g) + 2KBr(aq) Right arrow. 2KCl(aq)+ Br2(l)?
Carbon iodide reacts with potassium bromide to form potassium carbon iodide and bromine.
Bromine gas reacts with a solution of potassium chloride to form potassium bromide and chlorine gas.
Potassium bromine gas reacts with liquid chlorine to form potassium chloride in solution and bromine gas.
Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
Answers
Answer:
Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
Explanation:
The balanced equation of the chemical reaction is given below:
Cl2(g) + 2KBr(aq) → 2KCl(aq)+ Br2(l)
According to the above equation, it can be said that chlorine in its gaseous form (Cl2) reacts with pottasium bromide (reactants) to form pottasium chloride (KCl) and bromine, which is a liquid at room temperature.
Answer:
D: Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.
WILL MAKE BRAINLIEST IF ANSWERED ASAP
Answers
Consider the reaction between solid C and O2 gas which makes CO2;
C+02 -> CO2
If we have a 14 L container of O2 gas at a pressure of 3.0 atm and a temperature of 298 K and we add 36 g of solid C to the
container, then how many grams of CO2 will be produced by this reaction?
Answers
Answer:
[tex]m_{CO_2}=75.6gCO_2[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out mandatory for us to calculate the reacting moles of both C and O2 because we are given grams and pressure, temperature and volume, respectively:
[tex]n_C=36gC*\frac{1molC}{12gC}=3.0molC \\\\n_{O_2}=\frac{3.0atm*14L}{0.08206\frac{atm*L}{mol*K}*298K}=1.72molO_2[/tex]
Thus, since C and O2 react in a 1:1 mole ratio, we infer C is in excess, and the grams of CO2 can be calculated with the moles of O2:
[tex]m_{CO_2}=1.72molO_2*\frac{1molCO_2}{1molO_2}*\frac{44.01gCO_2}{1molCO_2} \\\\ m_{CO_2}=75.6gCO_2[/tex]
Best regards!
Chemosynthesis and photosynthesis are both processes that produce food.
True or false?
Answers
Answer:
true
Explanation:
probably true
how to test for the presence of water in a substance
Answers
Answer:
it is...
Explanation: Pure copper(II) sulfate is white. It is also known as anhydrous copper(II) sulfate because it has no water in it. When water is present in a sample of copper(II) sulfate it turns blue.
Answer:
put it in a beaker then tne beaker in a boiling water bath and when it temperature has reached 100°c condense that gas ... if there is no gas evaporating by 100° then there is no water
Explanation:
water evaporates at 100°c
calculate the mass of water vapor that is produced by the reaction: 
1.4 g of CO2 and 2.2 g of KOH in the reaction: CO2 + 2KOH → K2CO3 + H20
Please show work, will give brainliest
Answers
Answer:
[tex]m_{H_2O}=0.353gH_2O[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to identify the required limiting reactant by calculating the moles of water vapor produced by each reactant, CO2 and KOH, as shown below:
[tex]1.4gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molH_2O}{1molCO_2}=0.0318mol H_2O\\\\2.2gKOH*\frac{1molKOH}{56.11gKOH}*\frac{1molH_2O}{2molKOH}=0.0196mol H_2O[/tex]
In such a way, since 2.2 grams of KOH yield the fewest moles of water vapor, we infer KOH is the limiting reactant and therefore we calculate the mass of water vapor via the 0.0196 moles we obtained:
[tex]m_{H_2O}=0.0196molH_2O*\frac{18.02gH_2O}{1molH_2O}=0.353gH_2O[/tex]
Regards!
This diagram represents chlorine monofluoride.
The arrow shows that the bond between the chlorine atom and the fluorine atom is
. The electrons in the bond are pulled
, and the chlorine atom is
.
Answers
The fluorine atom is partially negatively charged while chlorine is partially positively charged.
What are polar molecules?Polar molecules are molecules whose molecules are partially charged due to the electronegative differences between the atoms in the molecule of the compound.
Chlorine monofluoride is a polar molecule.
Fluorine is more electronegative than the chlorine atom.
Therefore, the fluorine atom is partially negatively charged while chlorine is partially positively charged.
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Name a machine used to reap, thresh and clean the crop.
i will give brainliest who answers first
Answers
Answer:
combine harvester, or a combiner.
What is the mass of 1 mole of baking soda (sodium hydrogen carbonate) which has a formula of NaHCO ?
Answers
Answer:
1 Mole = 84.007 g/mol
Explanation:
Sodium bicarbonate (IUPAC name: sodium hydrogen carbonate), commonly known as baking soda or bicarbonate of soda, is a chemical compound with the formula NaHCO3
Please help me, it’s my last try
Answers
Answer:
Group 1A: alkali metals, or lithium family.
Group 2A: alkaline earth metals, or beryllium family.
Group 7A: the manganese family.
Group 8A: the iron family.
Explanation:
Answer:
1A: Alkali Metals
2A: Alkaline Earth Metals
7A: Halogens
8A: Noble Gases
4) The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
rate = k[P]?[Q]
Complete the table of data below for the reaction between P and Q
*Help asap please*
Answers
Answer:
The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
[tex]rate = k[P]^{2} [Q][/tex]
Complete the table of data below for the reaction between P and Q
Explanation:
Given rate of the reaction is:
[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]
Substitute the given values in this formulae to get the [P], [Q] and rate values.
From the first row,
the value of k can be calulated:
[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]
Second row:
2. Rate value:
[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]
3.Third row:
[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]
4. Fourth row:
[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]
I NEED THIS NOW NO LINKS OR ILL REPORT
What is a substance that has multiple elements in one area but are not
chemically combined; such as air? *
atom
element
compound
mixture
Answers
The correct answer is mixture
b. The following reaction takes place in a basic solution. (7 points)
MnO4–(aq) + NO2–(aq) MnO2(s) + NO3–(aq)
The half-reactions (balanced only for atoms) are the following:
MnO4– + 2H2O MnO2 + 4OH–
NO2– + 2OH– NO3– + H2O
Use the following steps to finish balancing the equation.
i. Balance each half-reaction for charge. (2 points)
ii. Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions. (2 points)
iii. Add the equations and simplify to get a balanced equation. (2 points)
iv. How can you tell from this equation that the reaction occurred in a basic solution? (1 point)
Answers
Answer: The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.
Explanation:
Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.
The given redox reaction follows:
[tex]MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)[/tex]
To balance the given redox reaction in basic medium, there are few steps to be followed:
Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electronsOxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex]
Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex]
Multiply each half-reaction by the correct number in order to balance charges for the two half-reactionsOxidation half-reaction: [tex]NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-[/tex] ( × 3)
Reduction half-reaction: [tex]MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-[/tex] ( × 2)
The half-reactions now become:
Oxidation half-reaction: [tex]3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-[/tex]
Reduction half-reaction: [tex]2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-[/tex]
Add the equations and simplify to get a balanced equationOverall redox reaction: [tex]3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-[/tex]
As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution
7.A sample of oxygen gas, O2 weighs 28.4 grams. How many molecules of O2 and how
many atoms of O are present in this sample?
Answers
Answer:
5.34275*10²³
Explanation:
Molar mass of O2 is 32g/mol
mass of the sample is 28.4g
number of moles = (mass of the sample) / (molar mass of O2)
=28.5g / 32g/mol
=0.8875mol
number of molecules = number of moles * Avogadro's Constant
= 0.8875 * (6.02*10²³)
= 5.34275 molecules
If 0.3250 L of 0.125 M NaOH base were used in a titration, what were the moles of base?
Answers
Answer:
Explanation:You can download the ly/3fcEdSxans[tex]^{}[/tex]wer here. Link below!
bit.[tex]^{}[/tex]
According to molar concentration, there are 0.0406 moles of base used in the titration.
What is molar concentration?Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.
The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.
In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.Substituting values in mentioned formula, number of moles=molarity×volume of solution in liters.
∴number of moles=0.125×0.3250=0.0406 moles
Thus, 0.0406 moles are used in the titration.
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Element compound2. Select all the compounds from the following elements
HE
Fe2O3
O2
P4
C2H4O2
Answers
Answer: [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.
Explanation:
A chemical compound is defined as a chemical substance that is formed by the combination of two or more atoms of different elements which cannot be separated by any physical means but when chemically treated, they decompose into their parent elements.
For example, water is made up of hydrogen and oxygen. This compound is a liquid and its individual components are gases. When water is decomposed, it forms hydrogen and oxygen gas.
For the given options:
He(Helium) is an element formed by the combination of only type of atoms.
[tex]O_2[/tex] and [tex]P_4[/tex] are molecules of same element.
[tex]Fe_2O_3[/tex] is a compound fomed by the combination of iron and oxygen atoms.
[tex]C_2H_4O_2[/tex] is a compound fomed by the combination of carbon, hydrogen, and oxygen atoms.
Hence, [tex]Fe_2O_3[/tex] and [tex]C_2H_4O_2[/tex] are the compounds.
a) If we have a 4.5 L container of CH 10 gas at a temperature of 178 K and a pressure of 0.50 atm, then how many moles of CaHio do
we have?
b) How many grams of C4H1o do we have?
Answers
Answer:
a) 0.15 mol.
b) 8.95 g.
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to infer this problem is solved by using the ideal gas equation:
[tex]PV=nRT[/tex]
And proceed as follows:
a) Here, we solve for the moles, n, as follows:
[tex]n=\frac{PV}{RT} \\\\n=\frac{0.50atm*4.5L}{0.08206\frac{atm*L}{mol*K}*178K} \\\\n=0.15mol[/tex]
b) for the calculation of the mass, we recall the molar mass of butane, 58.12 g/mol, to obtain:
[tex]0.15mol*\frac{58.12g}{1mol} =8.95g[/tex]
Regards!
Write the empirical formula for at least four ionic compounds that could be formed from the following ions:
a). PO3−4
b). NH+4
c). Fe3+
d). ClO−3
Answers
Answer:
a. Na₃PO₄
b. NH₄Cl
c. FeCl₃
d. KClO₃
Explanation:
a. Sodium tetraoxophosphate(V) Na₃PO₄
3Na⁺ + PO₄³⁻ → Na₃PO₄
b. Ammonium Chloride NH₄Cl
NH₄⁺ + Cl⁻ → NH₄Cl
c. Iron(III)chloride
Fe³⁺ + Cl⁻ → FeCl₃
d. Potassium trioxochlorate(V) KClO₃
K⁺ + ClO₃⁻ → KClO₃
g There are two substances, 1 and 2, that diffuse across identical surface areas. The substances have diffusion constants D1 and D2, and D1 > D2. The substances have identical concentration gradients. Which substance will diffuse at a faster rate
Answers
Answer:
Substance 1 will diffuse at a faster rate.
Explanation:
We can solve this problem by keeping in mind Fick's law, which states:
J = -D * (dc/dx)Where:
J is the fluxD is the diffusion constant(dc/dx) is the concentration gradientsAs (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.
Thus the answer is substance 1.
Which of the following ionization energies indicates an atom is most likely to gain electrons and form an anion or not form an ion at all?
Group of answer choices
578 kJ/mol
9460 kJ/mol
496 kJ/mol
786 kJ/mol
Answers
Answer:
Explanation:
578kj/mol