Answer:
a
Explanation:because i did the test
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?
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!
The molecules you use in a chemical reaction are called the:
What is the right answer?
Answer:
equal to zero is the right answer
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.
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
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?
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!
A small coffee cup calorimeter contains 28.0 g of H2O at 19.73 oC. A 2.05 g sample of a metal alloy is heated to 98.88 oC and then placed in the water. The contents of the calorimeter come to a temperature of 21.23 oC. What is the specific heat of lead
Answer:
1.104 J/g°C
Explanation:
Using Q = m × c × ∆T
Where;
m = mass of substance (g)
c = specific hear capacity (J/g°C)
∆T = change in temperature (°C)
For a colorimeter,
Q(water) = - Q(metal)
m. c. ∆T (water) = - m. c. ∆T (metal)
According to the information provided;
For water:
m = 28.0g
c = 4.184 J/g°C
∆T = (21.23 - 19.73°C)
For the metal:
m = 2.05g
c = ?
∆T = (21.23 - 98.88°C)
m. c. ∆T (water) = - m. c. ∆T (metal)
[28 × 4.184 × (21.23 - 19.73°C)] = -[2.05 × c × (21.23 - 98.88°C)]
[117.152 × 1.5] = -[2.05 × c × (-77.65)]
175.728 = -[-159.1825c]
175.728 = 159.1825c
c = 175.728 ÷ 159.1825
c = 1.104
c = 1.104 J/g°C
Excess HCl is added to 234 g of Na2S solution as shown by the equation below:
Na2S(aq) + 2HCl(aq) →2NaCl(aq) + H2S(g)
How many liters of H2S gas are produced at standard temperature and pressure (STP)? The molar mass of Na2S is 78 g.
Answer:
67 L
Explanation:
Step 1: Write the balanced equation
Na₂S(aq) + 2 HCl(aq) → 2 NaCl(aq) + H₂S(g)
Step 2: Calculate the moles corresponding to 234 g of Na₂S
The molar mass of Na₂S is 78 g/mol.
234 g × 1 mol/78 g = 3.0 mol
Step 3: Calculate the moles of H₂S produced from 3.0 moles of Na₂S
The molar ratio of Na₂S to H₂S is 1:1. The moles of H₂S formed are 1/1 × 3.0 mol = 3.0 mol.
Step 4: Calculate the volume occupied by 3.0 moles of H₂S at STP
At STP, 1 mole of H₂S occupies 22.4 L.
3.0 L × 22.4 L/1 mol = 67 L
Name a machine used to reap, thresh and clean the crop.
i will give brainliest who answers first
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 ?
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
Which of the following is an organic compound?
ammonia (NH3)
calcium sulfide (CaS)
octane (C8H18)
sulfur trioxide (SO3)
Organic compround is octane C8H18
The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive.
Answer:
The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity. Rank from most conductive to least conductive.
Explanation:
The given substances are:
barium chloride(BaCl2),
glucose(C6H12O6),
butanoic acid (C3H7COOH) which is a weak acid,
sodium bromide (NaBr).
The conductivity of a solution is proportional to the number of ions present in a particular solution.
1mol. of BaCl2 in water produces a total three mol. of ions.
[tex]BaCl_2 (aq) -> Ba^2^+(aq) + 2Cl^-(aq)[/tex]
Gluocse is a covalent compound and it does not dissociate into ions in water.
So, it does not conduct electricity.
Butanoic acid is a weak acid. But due to the release of H+ ions it can conduct a very less amount of electricity.
NaBr is an ionic compound and in 1mol. of NaBr in water gives two mol. of ions.
NaBr (aq) -> Na+ (aq) + Br- (aq)
Hence, the order of conductivity among the given substances in aqueous solution is:
BaCl2 > NaBr > butanoic acid > glucose
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?
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!
The aldol reaction is catalyzed by acid as well as base. What is the reactive nucleophile in the acid-catalyzed aldol reaction
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
A 10.0 g gold ring with a specific heat 0.129 at 24.00°C is placed in a calorimeter with 118 g of water at 1.00°C.
What will be the final temperature of the system?
Answer:
1.06 °C
Explanation:
From the question given above, the following data were obtained:
Mass of gold (M₉) = 10 g
Specific heat capacity of gold (C₉) = 0.129 J/gºC
Initial temperature of gold (T₉) = 24 °C
Mass of water (Mᵥᵥ) = 118 g
Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Initial temperature of water (Tᵥᵥ) = 1 °C
Equilibrium temperature (Tₑ) =?
The equilibrium temperature of the system can be obtained as follow:
Heat loss by the gold = heat gained by the water
M₉C₉(T₉ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)
10 × 0.129 (24 – Tₑ) = 118 × 4.184 (Tₑ – 1)
1.29(24 – Tₑ) = 493.712 (Tₑ – 1)
Clear bracket
30.96 – 1.29Tₑ = 493.712Tₑ – 493.712
Collect like terms
30.96 + 493.712 = 493.712Tₑ + 1.29Tₑ
524.672 = 495.002Tₑ
Divide both side by 495.002
Tₑ = 524.672 / 495.002
Tₑ = 1.06 °C
Therefore, the temperature of the system is 1.06 °C
The amount of heat of the system is measured by a device called a calorimeter. The final temperature of the system will be 1.06 degrees celsius.
What is equilibrium temperature?The equilibrium temperature is the temperature that follows the law of thermodynamics and is said to be the system that has alike temperatures.
Given,
Mass of Ag [tex]\rm (M_{g})[/tex] = 10g
Specific heat capacity of Ag [tex](\rm C_{g})[/tex] = [tex]\rm 0.129 J/g^{\circ}C[/tex]
The initial temperature of Ag [tex](\rm T_{g})[/tex] = [tex]24 ^{\circ}\;\rm C[/tex]
Mass of water [tex](\rm M_{w})[/tex] = 118 g
Specific heat capacity of water [tex](\rm C_{w})[/tex] = [tex]4.184 \rm \;J/g^{\circ}\;\rm C[/tex]
The initial temperature of water [tex](\rm T_{w})[/tex] = [tex]1 ^{\circ}\;\rm C[/tex]
Equilibrium temperature = [tex](\rm T_{e})[/tex]
The equilibrium temperature can be shown as, heat loss by the gold = heat gained by the water:
[tex]\rm \rm M_{g}C_{g}(T_{g} - T_{e}) = M_{w}C_{w}(T_{e}-C_{w})[/tex]
Substituting values in the equation:
[tex]\begin{aligned} 10 \times 0.129 (24 - \rm T_{e}) &= 118 \times 4.184 (\rm T_{e} - 1)\\\\\rm 1.29(24 - T_{e}) &= 493.712 (\rm T_{e} - 1)\\\\524.672 &= 495.002 \;\rm T_{e}\end{aligned}[/tex]
Now divide both the sides by 495.002:
[tex]\begin{aligned} \rm T_{e} &= \dfrac{524.672 }{495.002}\\\\\rm T_{e} &= 1.06 \;^{\circ}\rm C\end{aligned}[/tex]
Therefore, the final temperature of the system is 1.06 degrees celsius.
Learn more about equilibrium temperature here:
https://brainly.com/question/16207236
Help please with this question
Picture above
Answer:
3:372-88U:771-772828
When 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48°C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat (in GJ) produced by combustion of a ton of coal.
Answer:
32.0 kJ
General Formulas and Concepts:
Thermochemistry
Specific Heat Formula: q = mcΔT
q is heat (in J) m is mass (in g) c is specific heat (in J/g °C) ΔT is change in temperature (in °C)Explanation:
Step 1: Define
Identify variables
[Given] m = 1.00 g
[Given] ΔT = 1.48 °C
[Given] c = 21.6 kJ/g °C
[Solve] q
Step 2: Find Heat
Substitute in variables [Specific Heat Formula]: q = (1.00 g)(21.6 kJ/g °C)(1.48 °C)Multiply [Cancel out units]: q = (21.6 kJ/°C)(1.48 °C)Multiply [Cancel out units]: q = 31.968 kJStep 3: Check
Follow sig fig rules and round. We are given 3 sig figs.
31.968 kJ ≈ 32.0 kJ
how to test for the presence of water in a substance
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
Acetylide ions react with aldehydes and ketones to give alcohol addition products.
a. True
b. False
Answer:
a
Explanation:
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
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.
An alkyl groyp has a formula of CxH6, determine the value of x
Answer:
i didnt understand
Explanation:
Answer:
x = 2 (C₂H₆)
Explanation:
The general formula for alkanes is CₓH₂ₓ₊₂
2x + 2 = 6
Simply solve for n:
2x = 4
x = 2
Which of the following material is the weakest thermal conducters
please help me asap!!
Answer:
Al2o3 is 101
(nh4)2O IS 52
S8 is 256.56
Ba(oh)2 is 171.35
Cacl2 is 110.98
H2O is 18.01
Explanation:
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
The correct answer is mixture
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
Answer:
Explanation:
578kj/mol
A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make 500 mL (three significant figures) of solution. Calculate the pH of this buffer.
Answer:
A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make 500 mL of solution.
Calculate the pH of this buffer.
Explanation:
The pH of a buffer solution can be calculated by using the Henderson-Hesselbalch equation:
[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]
The pH of the given buffer solution can be calculated as shown below:
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?
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
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
.
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.
Learn more about polar molecules at: https://brainly.com/question/1433127
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Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coat the metal preventing it from further oxidation. Which of the following is the correct balanced equation for the reaction?
a. Al(s) + 302(g) → 2903(s).
b. 2Al(s) + O2(g) → 2A10(s).
c. AI(s) + O2(g) → AIO (s).
d. 4Al(s) + 3O2(g) →2Al2O3.
e. 3Al(s) + O2 → Al3O2.
Answer:
d. 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
Explanation:
Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coats the metal preventing it from further oxidation: aluminum oxide. Aluminum is a cation with charge 3+ (Al³⁻) and oxide is an anion with charge 2- (O²⁻). Thus, the neutral compound aluminum oxide has the chemical formula Al₂O₃. The unbalanced chemical equation is:
Al(s) + O₂(g) → Al₂O₃(s)
We can balance using the trial and error method. First, we will balance O atoms by multiplying Al₂O₃ by 2 and O₂ by 3.
Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
Finally, we get the balanced equation by multiplying Al by 4.
4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
Element compound2. Select all the compounds from the following elements
HE
Fe2O3
O2
P4
C2H4O2
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.
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*
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]