According to the Chemical equation 18 mols of water ar produced when 25 mols of Oxigen are consumed. Now we need to know how many mols of Oxigen are consumed in this example. The text give us the required information to solve this, but we need to use the ideal gas law:
[tex]P\cdot V=n\cdot R\cdot T[/tex]Where P is the pressure (1.30 atm), V is the volume (20L), T is the temperature in kelvin (263.15K) and n is the number of mols, what we want to calculate. Therefore we have to solve the ideal gas equation for n and use the provided data:
[tex]n=\frac{P\cdot V}{R\cdot T}=\frac{1.3\text{ atm}\cdot20L}{0.082\frac{atm\cdot L}{K\cdot mol}\cdot263.15K}=1.20\text{ }mols[/tex]so 1.20 mols of O2 are consumed we have to use the conversion factor 18 mols of water/25 mols of O2 to obtain how many mols of water are produced:
[tex]n_{H_2O}=1.20\text{ mols of O}_2\cdot\frac{18\text{ mols of H}_2O}{25\text{ mols of O}_2}=0.867\text{ mols of H}_2O[/tex]And the last step convert those mols into mass. To do that we need the molar mass of water, Mm=18 g/mol:
[tex]m_{H_2O}=n_{H_2O}\cdot Mm_{H_2O}=0.867mols\cdot18\frac{g}{mol}=15.61g[/tex]15.61 grams of water are produced
Calculate the molar concentration of the malonic acid if 38.17 mL of 0.315 M potassium hydroxide was required to neutralize 25.0 mL of malonic acid to a phenolphthalein end point.
The balanced reaction of the neutralization of malonic acid (C3H4O4) with potassium hydroxide (KOH) will be:
[tex]C_3H_4O_4+2KO_{}H\rightarrow K_2C_3H_2O_4+2H_2O[/tex]That is to say, to neutralize a mol of Malonic acid (C3H4O4) two moles of KOH are needed.
Now, we must determine the moles present in the KOH solution, they give us the molarity of the solution. Molarity represents the moles of solute in one liter of solution. So, the moles of KOH will be:
[tex]\begin{gathered} Mo_{}lKOH=GivenVolume\times\frac{0.315molKOH_{}}{1Lsolution} \\ Mo_{}lKOH=38.17mL\times\frac{1L}{1000mL}\times\frac{0.315molKOH_{}}{1Lsolution}=0.012molKOH \end{gathered}[/tex]We will now find the moles of the acid present using the stoichiometry of the reaction. Two moles of KOH neutralize one mole of malonic acid.
[tex]\begin{gathered} Mo_{}lC_3H_4O_4=GivenMolKOH\times\frac{1molC_3H_4O_4}{2molKOH} \\ Mo_{}lC_3H_4O_4=0.012molKOH\times\frac{1molC_3H_4O_4}{2molKOH}=0.006molC_3H_4O_4 \end{gathered}[/tex]That means that in that 25 mL of malonic acid solution there was 0.006 mol of malonic acid.
Now we are going to calculate the molar concentration (molarity) of the initial solution of malonic acid before being neutralized. We have a volume of the solution of 25 mL, therefore the molarity will be:
[tex]\begin{gathered} \text{Molarity}=\frac{\text{Moles of solute}}{L\text{ of solution}} \\ \text{Molarity}C_3H_4O_4=\frac{0.006molC_3H_4O_4}{25mL}\times\frac{1000mL}{1L}=0.24M \end{gathered}[/tex]So, the molar concentration of the malonic acid solution will be 0.24M
Calculate the mass percent of a NaCl solution in which 160 g of NaCl is dissolved in 3.13 L of water at 25 ∘C (density = 0.997 g/mL).Express your answer numerically.
The formula of mass percent is the following:
[tex]\text{mass percent=}\frac{mass\text{ of solute}}{mass\text{ of solution}}[/tex]where mass of solute correspond to the mass of NaCl and mass of solution correspond to the sum of NaCl and water mass.
We need to find the mass of water using its volume and density:
[tex]\text{mass of water=}\frac{\text{density}}{\text{volume}}=\frac{0.997\text{ g/mL}}{3130\text{ mL}}=0.997\text{ g water}[/tex]And we replace it in the first formula like this:
[tex]\text{mass percent=}\frac{160}{160+0.997}=0.9938[/tex]We obtain the percent multiplying the result by 100, so the answer is 99.38%
How many moles of potassium bromide is needed to produced 51.62 moles of bromine?
Answer
Option A = 103.2 mol KBr
Explanation
Given:
[tex]Cl_2+2KBr\rightarrow2KCl+Br_2[/tex]Moles of bromine = 51.62 moles
Required: Moles of KBr
Solution
Use the stoichiometry to calculate the the moles of KBr
The molar ration between Br and KBr is 1:2
Therefore the moles of KBr = 51.62 mol x 2 = 103.24 mol
Which of the following is not an empirical formula:A)N2O4B)PCl3C)N2O5D)CH2
Step 1 - What is an empirical formula?
An empirical formula is, by defnition, a formula showing only the smallest possible proportion between the elements that are part of a substance.
Let's take glucose, C6H12O6, as an example. Note that all numbers (6, 12 and 6 again) are divisible by 6. Therefore, this is not an empirical formula. If we divide all of them by 6, we obtain CH2O, which is indeed the empirical formula for glucose.
Step 2 - How to discover whether a given formula is empirical?
It's actually quite simple: are all the small numbers multiple of some number? If yes, then it is not an empirical formula.
Step 3 - Determining which formula in the exercise is not an empirical one
Note the only formula in which the small numbers can be further divided is N2O4. We can divide them by two, which would result in NO2.
Therefore, N2O4 is not an empirical formula. The correct answer is thus item A) N2O4.
What is missing in the nuclear reaction shown below? 10 is over 5B + 4 is over 2He → ________ + 1 is over 0n
Before balancing the given nuclear reaction, we need to understand what nuclear reaction is.
In simple terms, a nuclear reaction is a process where two nuclei collide to produce one or more new nuclides.
Given the nuclear reaction;
[tex]^{10}_5B+^4_2He\rightarrow^b_aX+^1_0n[/tex]We need to get the unknown element X, the mass number "b" and its atomic number "a"
To get the atomic and the mass number, we will take the sum of the atomic and mass number in the reactant and equate it to the atomic and mass number at the product as shown;
For the atomic number:
[tex]\begin{gathered} 5+2=a+0 \\ 7=a+0 \\ a=7 \end{gathered}[/tex]For the mass number:
[tex]\begin{gathered} 10+4=b+1 \\ 14=b+1 \\ b=14-1 \\ b=13 \end{gathered}[/tex]This shows that the required element needed to balance the chemical reaction is an element with an atomic number of 7 and a mass number of 13. The required element will be Nitrogen-13-atom.
The balanced form of the nuclear reaction will be:
[tex]^{10}_5B+^4_2He\rightarrow^{13}_7N+^1_0n[/tex]Determine whether the following equations are representative of an exothermic or endothermic reaction by checking the boxes below.
Answer:
[tex]\begin{gathered} a)\text{ Endothermic} \\ b)\text{ Exothermic} \end{gathered}[/tex]
Explanation:
Here, we want to know what class does each reaction belong to
An exothermic reaction has heat given off. The temperature of the surroundings is raised as a result of that. In notation terms, the change in enthalpy is negative and heat is added to the product side of the equation of reaction
An endothermic reaction otherwise has heat absorbed. The temperature of the surroundings is reduced as a result. In notation terms, the change in enthalpy is positive and heat is added to the reactant side of the equation of reaction
From all of these, we conclude that:
a) Endothermic
b) Exothermic
Students in ICP are practicing measuring using meter sticks. Two students combine their meter sticks to measure a length of tape on the floor. How many centimeters
is a piece of tape that measures two meter sticks in length?
In this question, we have to determine how many centimeters we have with these two-meter sticks mentioned. If the meter sticks do measure exactly 1 meter, therefore we will have 100 centimeters for each stick, and for the piece of tape, which takes 2-meter sticks to measure, we will then have 200 centimeters, again, considering that each meter stick measures 100 centimeters.
The answer will be 200 centimeters
Balance these chemical equations. (Use the lowest possible whole number coefficients.)CuCl + H2 → Cu + HCl
To balance the the following equation we will take the following steps:
[tex]CuCl+H_2\rightarrow Cu+HCl[/tex]-We will balance the the hydrogen on the product side to be equivalent with that on the reactant side of the equation. The number 2 will go infront of HCl.
-Now that we have two chlorines on the product side, we will balance the chlorines on the reactant side and so the number 2 will go in fron of the CuCl. We also need to add a 2 to to the Cu on the product side. The balanced equation will be:
[tex]2CuCl+H_2\rightarrow2Cu+2HCl[/tex]How does the pressure of a gas over a liquid affect the quantity of gas that dissolves in solution?A.As pressure increases, the quantity of gas that dissolves decreases.B.As pressure increases, the quantity of gas that dissolves increases.C.As pressure increases, the quantity of gas that dissolves decreases to a point and then levels off.D.Pressure has no effect on the quantity of gas that dissolves in a liquid.
• As you increase the pressure of a gas, the collision frequency increases and thus the solubility goes up.
,• This means that as pressure increses, the quantity of the gas that dissolves( solubility) also increases.
,• Option B is correct. [increase in pressure = increase in gas solubility.]
Calculate the number of molecules present in 0.40 moles of H2O.
To find the number of molecules present in 0.40 moles of water (H2O), we have to use Avogadro's number which is 6.02 x 10^(23). Remember that this number represents the number of molecules or atoms per mole of a compound.
The calculation is:
[tex]0.40molesH_2O\cdot\frac{6.02\cdot10^{23}moleculesH_2O}{1moleH_2O}=2.408\cdot10^{23}moleculesH_2O.[/tex]The answer is that 0.40 moles of H2O contain 2.408 x 10^(23) molecules of H2O.
Pentane gas, C5H12 , reacts with oxygen to produce carbon dioxide and water.
C5H12(g)+8O2(g)→5CO2(g)+6H2O(g)
1.) How many grams of CO2 are produced from 32.1 g of O2 ?
2.) How many grams of CO2 are formed if 44.4 g of C5H12 is mixed with 106 g of O2 ?
Pentane gas, C₅H₁₂ , reacts with oxygen to produce carbon dioxide and water.
C₅H₁₂(g)+8O₂(g)→5CO₂(g)+6H₂O(g)
1.) grams of CO₂ are produced from 32.1 g of O₂ is 0.72 gram
2.) grams of CO are formed if 44.4 g of C₅H₁₂ is mixed with 106 g of O₂ is 3.417 gram
Pentane is a volatile and colorless liquid with a characteristic gasoline like odour and the given reaction is
C₅H₁₂(g)+8O₂(g)→5CO₂(g)+6H₂O(g)
And given data is
1. 32.1 g of O₂ and we have to find grams of CO₂ are produced = ?
Gram = mass of solute/formula mass of solute
Gram of CO₂ = 32.1 g/44.01
Gram of CO₂ = 0.72 gram
0.72 gram grams of CO₂ are produced from 32.1 g of O₂
2. given data is 44.4 g of C₅H₁₂ is mixed with 106 g of O₂
Gram = mass of solute/formula mass of solute
Gram = 44.4 g + 106 g /44.01
Gram = 3.417 gram
3.417 gram grams of CO are formed if 44.4 g of C₅H₁₂ is mixed with 106 g of O₂
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1. A gas with a volume of 4.0 L at a pressure of 2 08 atm is allowed to expand to a volume of 12.0 L. What is the pressure in atm in the container if the temperature remains constant. Please show work (including the equation you used). --------- 2. The volume of a gas is 200 mL at 350.0 kPa pressure. What will the volume be when the pressure is reduced to 125.0 kPa, assuming the temperature remains constant. Please show all work for full credit. Please show work (including the equation you used).
Explanation
To solve this, we consider Boyle's Law:
It relates gas pressure to volume while keeping other parameters such as temperature and amount of substance constant.
Equation we use here is p1 x V1 = p2 x V2 (1)
(p for pressure and V for volume)
p1 = 2.08 atm
V1 = 4.0 L
If the gas expands, the pressure will be lower than 2.08 atm.
p2 = unknown
V2 = 12.0 L
From (1) we clear p2:
=> p1 x V1/V2 = p2
=> 2.08 atm x 4.0 L/12.0 L = p2
=>0.69 atm = p2
Answer: p2 = 0.69 atm
How many grams of CaCl2 (MM = 110.98) are needed to make 0.00175× 10^6 mL of 1.50 M235 g100 g200 g150 g291 g
Answer: 291.3 g of CaCl2 would be necessary to prepare the solution given. The best option to answer the question is the last one (letter E, 291 g)
Explanation:
The question reuquires us to determine the mass of CaCl2 necessary to prepare 0.00175× 10^6 mL of a 1.50 M solution.
We can use the following equation, which corresponds to the definition of molarity, to calculate the number of moles of CaCl2 necessary to prepare the solution:
[tex]M=\frac{n}{V}[/tex]Where M is the molarity of the solution (given in mol/L or M), n is the number of moles of solute (in mol) and V is the volume of solution (in L).
Note that we'll need to convert the volume given (0.00175× 10^6 mL) from mL to L:
[tex]V=0.00175\times10^6mL\times\frac{1L}{1000mL}=1.75L[/tex]Next, we can rearrange the equation for molarity in order to calculate the number of moles of CaCl2:
[tex]M=\frac{n}{V}\rightarrow n=M\times V[/tex]And, applying the values given by the question:
[tex]n=(1.50mol/L)\times(1.75L)=2.62mol[/tex]Therefore, 2.62 moles of CaCl2 would be necessary to prepare the solution.
We can use the following equation to convert the number of moles calculated (2.62 mol) to the mass required, considering the molar mass of CaCl2 given (MM = 110.98 g/mol):
[tex]n=\frac{m}{MM}\rightarrow m=n\times MM[/tex]Where n corresponds to the number of moles (in mol), m is the mass of the sample (in grams) and MM corresponds to the molar mass of the compound (in g/mol).
Applying the molarity given by the question and the number of moles calculated, we'll have:
[tex]m=(2.62mol)\times(110.98g/mol)=291.3g[/tex]Therefore, 291.3 g of CaCl2 would be necessary to prepare the solution given. The best option to answer the question is the last one (letter E, 291 g).
40/20 Ca2+ hasa) 2O protons, 20 neutrons, and 18 electronsb) 22 protons, ?O neutrons, and 20 electronsc) 20 protons, 22 neutrons, and 18 electronsd) 22 protons, l8 neutrons, and 18 electrons
The common representation of atomic number, atomic mass and charge of an element is by writing in the following way:
40/20 Ca^2+
On the top, you have the atomic mass, which is the sum of the number of protons and the number of neutrons A = p + n
On the bottom, you have the atomic number, which is equal to the number of protons and number of electrons Z = p = e
The number and the sign indicate the charge of the element, if the element is +2, this means that we have a cation with 2 electrons missing, if the charge was -2, we would have an anion with 2 more electrons
Therefore the best answer for this question is:
20 protons, 20 neutrons, 18 electrons, letter A
True or False? If the concentration of H+ is greater than the concentration of OH- in a solution, the solution is basic.TrueFalse
Answer: the sentence given by the question is false
Explanation:
The question requires us to determine if the sentence given ("If the concentration of H+ is greater than the concentration of OH- in a solution, the solution is basic") is true or false.
The acidity or alkalinity of a solution can be measured by the amount of H+ ions that are present, and it is numerically represented by the pH of a solution. The pH is calculated based on the concentration of H+ ions, as it follows:
[tex]pH=-log\lbrack H^+\rbrack[/tex]where [H+] corresponds to the concentration, in mol/L, of H+ ions in solution.
The equation above tells us that:
- a high concentration of H+ ions leads to a low pH value (acid solution);
- a low concentration of H+ ions leads to a high pH value (alkaline or basic solution).
A similar idea is applied to the value pOH (which corresponds to -log[OH-]), but a high pOH value indicates an alkaline solution, while a low pOH value corresponds to an acidic solution.
Thus, we can say that, if a solution presents more H+ ions than OH-, its pH value should be low while its pOH value should be high, which indicates an acid solution.
Therefore, the sentence given by the question is false.
Calculate the wavelength (in nm) of light with a frequency of 7.66 × 1014 s-1
The relationship between the wavelength and the frequency is given by this equation:
λ = c/f
Where λ is the wavelength, f is the frequency and c is the speed of light.
We are given the frequency. f = 7.66 *10¹⁴ 1/s
C is a constant. c = 3 * 10⁸ m/s
λ is our unknown. So we can calculate the wavelength using the formula that we wrote before.
λ = c/f
λ = (3 * 10⁸ m/s) / (7.66 *10¹⁴ 1/s)
λ = 3.92 *10⁻⁷ m
The wavelength of light with the frequency given is 3.92 *10⁻⁷ m, but we are asked for the answer in nm. We have to convert the answer. There are 1*10⁹ nm in 1 m, then the conversion is:
1 m = 1 * 10⁹ nm
λ = 3.92 *10⁻⁷ m * 1 * 10⁹ nm/(1 m)
λ = 392 nm
Answer: The wavelength is 392 nm.
a gas has a volume of 48.8 Liters at STP. what will it’s volume be at 14.0 atm and 25C
3.81L
Explanations:According to the general gas equation
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]where:
• P1 and P2 are the i,nitial and final pressure
,• T1 and T2 are the ,initial and final temperature
,• V1 and V2 are the ,initial and final volume
At standard temperature and pressure:
[tex]\begin{gathered} P_1=1atm \\ T_1=0^0C=273K \end{gathered}[/tex]Other parameters include
[tex]\begin{gathered} V_1=48.8L \\ P_2=14.0atm \\ T_2=25^0C=25+273 \\ T_2=298K \end{gathered}[/tex]Required
Final volume V2
Substitute the given parameters into the formula to have:
[tex]\begin{gathered} V_2=\frac{P_1V_1T_2}{P_2T_1} \\ V_2=\frac{1\times48.8\times298}{14.0\times273} \\ V_2=\frac{14542.4}{3822} \\ V_2=3.81L \end{gathered}[/tex]Hence the final volume at 14.0atm and 25 degree celsius is 3.81L
Find the percent composition for Ca and Cl in CaCl2
The percent composition for Ca in CaCl2 is 36.1% and the percent composition for Cl in CaCl2 is 63.9%
1st) We need to know the atomic weight of Ca and Cl, and the molecular weight of CaCl2.
We can find the atomic weight in the Periodic Table:
Ca atomic weight: 40.07 g/mol
Cl atomic weight: 35.45 g/mol
So, the molecular weight of CaCl2 is:
40.07 g/mol x (35.45 g/mol x2) = 110.97 g/mol
CaCl2 molecular weight: 110.97 g/mol
2nd) With a mathematical Rule of Three we can calculate the percent composition for Ca and Cl. Here, it is important to rember that the CaCl2 molecular weight (110.97 g/mol) represents the 100%.
- Percent composition for Ca in CaCl2:
[tex]\begin{gathered} 110.97\text{ g/mol-100}\% \\ 40.07\text{ g/mol -x=}\frac{40.07\text{ g/mol}\cdot\text{100}\%}{110.97\text{ g/mol}} \\ x=36.1\% \end{gathered}[/tex]- Percent composition for Cl in CaCl2:
[tex]\begin{gathered} 110.97\text{ g/mol - 100}\% \\ 70.9\text{ g/mol - x=}\frac{70.9\text{ g/mol}\cdot\text{100}\%}{110.97\text{ g/mol}} \\ x=63.9\% \end{gathered}[/tex]So, the percent composition for Ca in CaCl2 is 36.1% and the percent composition for Cl in CaCl2 is 63.9%.
The percent composition of Calcium(Ca) in Calcium chloride CaCl₂ is 36%
To calculate the" Percentage of an element in a compound =Molar mass of element/ Mass of Compound x 100" which is generally Know as Mass Percentage of an element.
Molecular Formula is CaCl₂ Ca+Cl₂→CaCl₂
Which is called as calcium chloride in which Molecular Mass of Calcium (Ca) is 40 and Molecular Mass of chlorine(Cl) is 35.5. In compound we have 1 Calcium and 2 chlorine atoms.
Ca =40x1+ 2cl= 35.5x2 =71 = 40+71= 111 which is Mass of compound.
Percent of Element Ca in CaCl₂ is [tex]\frac{40}{111}[/tex]×100 = 36.06(approx. )36%.
Percent of Cl₂in CaCl₂ is 64%.
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How does a change in heat affect the non-polar molecules compared to polar molecules Which one would boil faster? How do you know when a molecule is in the gas phase?
Answer and Explanation:
Remember that the boiling point shows the temperature of a substance when it reaches the vapor phase.
It is common to see the application of heat to certain substances in boiling, for example, in water, the boiling point at 1 atm would be 100 ° C (373 K).
Remember that water is a polar molecule, so if we compare a polar substance such as water with a non-polar substance (ie, gas nitrogen, N2), you can note that the boiling point of the non-polar substance would be smaller. This is because non-polar molecules contain weaker forces than polar substances and heat "breaks" the bond between these forces to reach the gas phase, so it is more difficult to break strong forces such as the polar molecules.
At ambient temperature, we can identify at a glance if a molecule is in the gas phase by reviewing if it's a polar or nonpolar molecule. Many nonpolar molecules, at ambient temperature, are in the gas phase such as hydrogen gas (H2), oxygen gas (O2), nitrogen gas (N2), methane (CH4), etc.
7. All cells require a continuous supply of ATP to sustain life processes. Cells generate ATP from ADP via
Cells just want sufficient ATP to live to tell the tale, but most cells save lots more than they need. it's miles doubtful why cells hold a lot of ATP, or whether or not this excess ATP has every other cause.
ATP hydrolysis offers the electricity needed for many crucial approaches in organisms and cells. these encompass intracellular signaling, DNA and RNA synthesis, Purinergic signaling, synaptic signaling, energetic transport, and muscle contraction.
One of the first and maximum enduring statistics maximum college students examine in biology class is that every living cells use a small molecule called adenosine triphosphate (ATP) as gas.
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A gas has a volume of 58.4 L and the pressure is 1 atm at a temperature of 2.5 degrees Celsius what is the number of moles of gas?
Step 1
The gas is assumed to be ideal. Therefore, it is applied the next formula:
p x V = n x R x T (1)
p = pressure = 1 atm
T = temperature = 2.5 °C; in K => T = 2.5 °C + 273 = 275.5 K
V = volume = 58.4 L
R = gas constant = 0.082 atm L/mol K (according to the units of the variables)
n = number of moles = unknown
---------------
Step 2
n is cleared from (1):
p x V/R x T = (1 atm x 58.4 L)/(0.082 atm L/mol K x 275.5 K) = 2.6 moles approx.
Answer: n = 2.6 moles
18.When magnesium combines with chlorine, it loses 2 electrons. What is the charge of magnesium?Select one:a. +2b. 0c. -2d. -4
Answer:
[tex]A\text{ : +2}[/tex]Explanation:
Here, we want to get the charge of magnesium
Magnesium is an atom that has 12 protons and 12 electrons
Losing 2 out of the 12 electrons means it would be left with 12 protons and 10 electrons
What that means is that it would be having an excess of 2 protons
This gives it a charge of +2
You have 500 grams of CaCO3. How many moles are present? a 8 b 5 c 2 d 10
Answer
b. 5
Explanation
The moles of CaCO3 present in 500 grams of CCO3 can be calculated using the mole formula.
[tex]Mole=\frac{Mass}{Molar\text{ }mass}[/tex]Molar mass of CaCO3 100 g/mol
Putting mass = 500 g and molar mass = 100 g/mol into the mole formula, we have,
[tex]Mole=\frac{500\text{ }g}{100\text{ }g\text{/}mol}=5\text{ }moles[/tex]The moles of CaCO3 resent in 500 grams of CacCO3 = 5 moles
Copper reacts with sulfur to create copper 1 sulfide. If 10.6g Cu reacts with 3.98g sulfur. How much product will be formed
This is a stoichiometry problem, where we have an initial amount of reactant and we need to find out how much of the product will we end up with, in order to do that we need to:
1. Set up the properly balanced equation, which the question already provided us
2 Cu + S -> Cu2S
2. See how many moles of reactant there are in the given amount of grams, but in this case we have 2 masses given in the question, which will directly lead us to think about limiting and excess reactant, even though the question did not ask us about it, we need to know it in order to find the right amount of product
Another thing important is molar ratio, which is 2:1 between Cu and S, this means that for each 2 moles of Cu we only need 1 mol of S in order to react
10.6g of Cu, and molar mass = 63.55g/mol
3.98g of S, and molar mass = 32.06g/mol
63.55g = 1 mol of Cu
10.6g = x moles of Cu
x = 0.17 moles of Cu
If we have 0.17 moles of Cu, we would need 0.08 moles of S in order to react
but we don't know if 3.98g of S is equal to 0.08 moles, let's check
32.06g = 1 mol of S
x grams = 0.08 moles of S
x = 2.56 grams of S, this means that we have more mass than we actually need to react, since we only need 2.56 grams and we have 3.98 grams, therefore Cu is the limiting reactant and S is in excess
Now we can find out how much product will be formed, using the number of moles of the limiting reactant, the molar ratio between Cu and Cu2S will be 2:1, same explanation as the one given between Cu and S, therefore we will end up with 0.08 moles of Cu2S as product
0.08 moles of Cu2S, molar mass = 159.16g/mol
159.16g = 1 mol of Cu2S
x grams = 0.08 moles of Cu2S
x = 12.73 grams of Cu2S
If gallium and oxygen chemically react to form a compound, what would be thecorrect formula of the resulting compound and would it be classified as molecular orionic?
4Ga + 3O2 = 2Ga2O3
So when gallium reacts with oxygen a compound formed if Ga2O3, and this compound is ionic.
fill out the table shown in the photo (black boxes don’t need to be filled in
Substance Boiling point
Vegetable oil 300 degrees celsius
Maple syrup 103.8 degrees celsius
Substance Melting point
Table salt 801 degrees celsius
Sugar 186 degrees celsius
What is a hydrogen bond?A. a weak bond between a hydrogen atom and anothermoleculeB. a bond where electrons are sharedC. a strong bond between atoms with a chargeD. a bond between atoms in a molecule
Answer:
C. a strong bond between atoms with a charge.
Explanation:
A hydrogen bond is a intermolecular bond, between a hydrogen atom and the electrons of an oxygen, nitrogen or fluorine atom of another molecule :
For example, the molecule of water:
Answer:
Its A
Explanation:
a weak bond between a hydrogen atom and another molecule
The specific heat of a certain type of cooking is 1.75 J(g ⋅ °C). How much heat energy is needed to raise the temperature of 2.44 kg of this oil from 23 °C to 191 °C? q = ___________ J
The heat abvsorved by a substance is calculated according to the following equation:
[tex]q=m\times C\times\Delta T[/tex]Where q is the heat, m is the mass of the substance, C is the specific heat and AT is the temperature increment. Therefore we have all the required data to solve this problem:
[tex]q=2440\text{ g }\times1.75\frac{J}{g\cdot\degree C}\times(191\degree C-23\degree C)=717360\text{ J}[/tex]Just be sure all quantities are in the corresponding units
question is in photo below What is the molarity of a solution that contains 0.75 moles of solute in a 1.5 L of solution?
What is the molarity of a solution that contains 0.75 moles of solute in a 1.5 L of solution?
Molarity is defined as the number of moles of solute per liter of solution. We can calculate it using this formula.
molarity = moles of solute/volume of solution in L
We already know the number of moles of solute and the volume of solution.
moles of solute = 0.75 moles
volume of solution in L = 1.5 L
We can replace those values to get the answer to our problem.
molarity = moles of solute/volume of solution in L
molarity = 0.75 moles/(1.5 L)
molarity = 0.5 M
Answer: the molarity of the solution is 0.50 M
A mixture contains 21.7 g NaCl, 3.74 g KCI,and 9.76 g LICI.How many total moles are in this mixture?Molar Mass NaCl: 58.44 g/molKCI: 74.55 g/mol Licl: 42.39 g/molHint: To find the total number of moles, calculate:(g of NaCl = molar mass of NaCl) +(g of KCI = molar mass of KCl) +(g of LiCl = molar mass of LiCl).
The total number of moles is something relative. If we wanted to be critical, we would have to consider the dissociation of the ions, if in solution, and the number of moles of solvent (if there is one).
However, the hint says that to calculate the number of moles we would have to divide each mass by the corresponding molar mass and add them.
This means that we want the total number of moles of each compound that was added to the mixture, so it is probably just a mixture of the solids.
So, we want the following:
[tex]n=n_{NaCl}+n_{KCl}+n_{LiCl}_{}[/tex]The molar mass, M, is the property that relates the number of moles, n, and the mass, m, of a compound:
[tex]\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}_{} \end{gathered}[/tex]The molar masses are given in the question, so we don't need to calculate them:
[tex]\begin{gathered} M_{NaCl}=58.44g/mol \\ M_{KCl}=74.55g/mol \\ M_{LiCl}=42.39g/mol \end{gathered}[/tex]So, we have:
[tex]\begin{gathered} n_{NaCl}=\frac{m_{NaCl}}{M_{NaCl}} \\ n_{KCl}=\frac{m_{KCl}}{M_{KCl}} \\ n_{LiCl}=\frac{m_{LiCl}}{M_{LiCl}} \end{gathered}[/tex]So, putting them into the equation, we have:
[tex]\begin{gathered} n=n_{NaCl}+n_{KCl}+n_{LiCl} \\ n=\frac{m_{NaCl}}{M_{NaCl}}+\frac{m_{KCl}}{M_{KCl}}+\frac{m_{LiCl}}{M_{LiCl}} \end{gathered}[/tex]Now, we just need to substitute the given values:
[tex]\begin{gathered} n=\frac{21.7g}{58.44g/mol}+\frac{3.74g}{74.55g/mol}+\frac{9.76g}{42.39g/mol} \\ n=0.371321\ldots mol+0.050167\ldots mol+0.230242\ldots mol \\ n=0.651731\ldots mol\approx0.652mol \end{gathered}[/tex]So, there are approximately 0.652 mol in the mixture.