Answer:
Explanation:
we will do ratio method
Aluminum chloride : Aluminum
2 : 2
0.48 : x
(cross multiply)
0.48 x 2 / 2 = 0.48 moles of aluminum
mass = 1 mole of aluminum chloride x moles
mass = 133.33 x 0.48
mass = 63.9984g (round off) = 64g
I hope this helps.
Lookup, verify and write down the correct Lewis dot structure for tartaric acid. How many total lone pairs are there in one molecule of tartaric acid
A radioactive sample has a half life of 1 hour. If you start with 1.000 gram of it at noon, how much of it remains at 4pm
The amount of the sample remaining at 4pm is 0.0625 g
We'll begin by calculating the number of half-lives that has elapsed
Half-life (t½) = 1 hour
Time (t) = 4 hour
Number of half-lives (n) =?n = t / t½
n = 4 / 1
n = 4Finally, we shall determine the amount the sample remaining at 4pmNumber of half-lives (n) = 4
Initial amount (N₀) = 1 g
Amount remaining (N) =?[tex]N = \frac{N_0}{ {2}^{n}} \\ \\ N = \frac{1}{ {2}^{4}} \\ \\ N = \frac{1}{16} \\ \\ N = 0.0625 \: g[/tex]
Thus, the amount remaining at 4pm is 0.0625 g
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please help asap!!!!!!!
Answer:
c) single bonds, double bonds, carbon atoms and hydrogen atoms
what is the best way to make a supersaturated solution?
A: Heat the solution
B: Stir the Solution
C: Evaporate the solution
D: Cool the solution
Answer:
heat the solution
Explanation:
i think
Answer:
The way to make a supersaturated solution is to add heat, but just a little heat won't do the job. You have to heat the water close to the boiling point. When the water gets this hot, the water molecules have more freedom to move around, and there is more space for solute molecules between them.
2. A solution NaF is add dropwise to a solution that is .0122 M in Ba . When the concentration of F exceeds ______M, BaF2 will precipitate. Neglect volume changes. BaF2 K
Explanation:
BaF2(s) <------> Ba2+(aq) + 2F-(aq)
Ksp BaF2 = 1.0 x 10^-6.
Ksp BaF2 = [Ba2+(aq)]×[F-(aq)]^2 at equilibrium
When Qsp >Ksp, BaF2 will precipitate
Qsp = [Ba2+(aq)]×[F-(aq)]^2
[Ba2+(aq)]×[F-(aq)]^2 > 1.0 x 10^-6.
0.0122 moldm-3 × [F-(aq)]^2 > 1.0 x 10^-6
[F-(aq)]^2 > 1×10^-6 / 0.0122 mol2dm-6
[F-(aq)]^2 > 81.96 × 10^-6 mol2dm-6
[F- (aq)] > 9.05 × 10^-3 moldm-3
So F- concentration should be more than 9.05 × 10^-3 moldm-3
3.833 kJ of heat is required to convert a 36.8 g sample of ethyl
alcohol from the solid to liquid phase. What is the heat of
fusion of ethyl alcohol in J/g?
The heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.
The given parameters:
heat required to convert 36.8 g sample of ethyl alcohol, Q = 3.3833 kJmass of the ethyl alcohol, m = 36.8 gThe heat of fusion of the given sample of the ethyl alcohol converted from solid to liquid phase is calculated as follows;
[tex]H_f = \frac{Q}{m} \\\\H_f = \frac{3.833 \times 10^3\ J}{36.8 \ g} \\\\H_f = 104.16 \ J/g[/tex]
Thus, the heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.
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Page No. Date Chemistry traignment Scientific Notation Covert 0.00000046 in to Scientific Notation
Answer:
answer is 4.6×10 with exponent 7
Answer:
[tex]4.6\times10^{-7}[/tex]
Explanation:
Remember that if the decimal point is to the right then the exponent is a negative.
[tex]4.6\times10^{-7}[/tex]
Move the decimal point to the left:
[tex].0.0.0.0.0.0.4.6[/tex]
[tex].0.0.0.0.0.0.4.6=0.00000046[/tex]
Hope this helps
crude oil is a liquid kind of___________
what is the answer?
#let's study
HELP!! what are the usual products of combustion reactions?
Explanation:
Carbon dioxide and water
I hope it helps
Answer:
The usual products of combustion reactions are carbon dioxide and water.
Explanation:
Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.
What is the temperature, in Celsius, of 20.8 g of carbon dioxide (CO2) in a 575 mL container under 1.5 atm of pressure?
Answer:
Ideal Gas Law Calculator. Easily calculate the pressure, volume, temperature or quantity in moles of a gas using this combined gas law calculator (Boyle's law calculator, Charles's law calculator, Avogadro's law calculator and Gay Lussac's law calculator in one).Supports a variety of input metrics such as Celsius, Fahrenheit, Kelvin, Pascals, bars, atmospheres, and volume in both metric and
Can someone help me out with this question and explain me the answer too!!! Please
Answer:
sorry i dont know
Explanation:
The structure of Disodium edta
Answer:
EDTA disodium salt | C10H14N2Na2O8
Explanation:
Which statement best describes mechanical energy?
A.the energy of heat and magnetism
B.the energy of electricity and motion
C.the energy of moving parts and motion
D.the energy of electricity and heat
Answer:
the answer is B.
Explanation:
your welcome
Answer:
the answer is B
Explanation:
i have the same exact questionn
The Difference between the number of protons and electrons give an atom it’s _________.
Answer:
it gives an atom its charge
How many moles of potassium iodide, KI, are required to precipitate all of the lead (II) ion from 25.0 mL of a 1.6 M Pb(NO3)2 solution? (First, write a balanced equation for the reaction.)
0.04 moles of iodide is required to precipitate all the lead II ions from 25.0 mL of a 1.6 M Pb(NO3)2 solution.
The reaction equation is;
2KI(aq) + Pb(NO3)2(aq) -------> PbI2(s) + 2KNO3(aq)
The net ionic equation is;
Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)
Number of moles of Pb(NO3)2 = 25/1000 L × 1.6 M = 0.04 moles
Number of moles of Pb^2+ = 0.04 moles /2 = 0.02 moles
Since 2 moles of iodide reacts with 1 mole of Pb^2+
x moles of iodide reacts with 0.02 moles of Pb^2+
x = 2 moles × 0.02 moles/ 1 mole
x = 0.04 moles of iodide
Hence, 0.04 moles of iodide is required to precipitate all the lead II ions from 25.0 mL of a 1.6 M Pb(NO3)2 solution.
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How do we know flowing water is the geologic process that formed the channel on Mars?
Evidence that water was once present on a planet is evidence that the planet may once have had living organisms. When landforms on different rocky planets look similar, it is evidence that they may have been formed by the same geologic process. The channel on Mars may have been caused by flowing water or flowing lava.
Hopes this helps :)
10) Explain which substance(s) would you expect to be a gas at standard
temperature and pressure (STP).
NI3
BF3
PC13
A) BF3
Its molecules are nonpolar and, therefore, are subject only to
dispersion forces.
B) NI3
Its molecules are polar and possess dipole-dipole interactions
therefore allowing them to exist as a gas.
C) PCL3
Its molecules are slightly polar and, therefore, possess dipole-
dipole interactions. However, they are weak enough to allow it
to exist as a gas.
D) Nlz and BF3
Both molecules are very light in mass. Therefore, they have
weak dispersion forces which causes them to be attracted very
weakly and exist as a gas.
Answer:BF3
Explanation:USA test prep
What identifies the number of protons in the nucleus of an atom?
Answer: Atomic number
Explanation:
I hope this helps you!
How many grams are 0.300 moles of glucose, C6H12O6?
Answer:
Explanation:
Firstly, let us calculate the molar mass of the glucose. To find the molar mass we need to add the masses of individual elements which constitute one glucose molecule.
Now, we know that
Molar mass of Carbon C=12gmol−1
Molar mass of Hydrogen H=1gmol−1
Molar mass of oxygen O=16gmol−1
Therefore, Molar Mass of glucose (C6H12O6) can be calculated as shown below
⇒ 6×12+12×1+6×16
⇒72+12+96⇒180gmol−1
Given mass of glucose = 300g
Now, we can calculate the number of moles in given mass of glucose, using the below formula,
Using the Formula numberofmoles=givenmassmolarmass we get
numberofmoles=300180 = 1.7 moles or 2 moles (approx.)
Hence the number of moles present in 300 g of glucose is 1.7 moles or 2 moles approximately.
You are given 1.091 grams of a white powder and told that it is a mixture of potassium carbonate and sodium carbonate. You are asked to determine the percent composition by mass of the sample. You add some of the sample to 10.00 mL of 0.8903 M nitric acid until you reach the equivalence point. When you have added enough carbonate to completely react with the acid, you reweigh your sample and find that the mass is 0.573 g. Calculate the mass of the sample that reacted with the nitric acid. Calculate the moles of nitric acid that reacted with the sample.
The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.
Molar mass of sodium carbonate = 106 g/mol
Molar mass of potassium carbonate = 138 g/mol
Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles
Mass of HNO3 = 0.008903 moles × 63 g/mol = 0.56 g
Mass of sample added = 1.091 g
Mass of sample left over = 0.573 g
Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g
The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.
Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O
106g of Na2CO3 reacts with 126g of HNO3
x g of Na2CO3 reacts with (126 × x/106)g of HNO3
K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O
138 g of K2CO3 reacts with 126 g of HNO3
(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g
Total mass of HNO3 used;
1.19x + 0.47 + 0.91x = 0.56
2.1x + 0.47 = 0.56
2.1x = 0.56 - 0.47
2.1x = 0.09
x = 0.09/2.1
x = 0.0043 g
Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g
Mass percent of K2CO3 = 0.5137 g/ 1.091 g × 100/1 = 47.1%
Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%
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helpp
What mass of carbon dioxide will be produced if
144 g of carbon react with 384 g of oxygen gas
according to the equation C+02 → CO2?
The mass of the carbon dioxide that will be produced when 144 g of carbon react with 384 g of oxygen gas is 528 g
We'll begin by calculating the masses of C and O₂ that reacted and the mass of CO₂ produced from the balanced equation.
C + O₂ —> CO₂
Molar mass of C = 12 g/mol
Mass of C from the balanced equation = 1 × 12 = 12 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 1 × 32 = 32 g
Molar mass of CO₂ = 12 + (2×16) = 44 g/mol
Mass of CO₂ from the balanced equation = 1 ×44 = 44 g
SUMMARY:
From the balanced equation above,
12 g of C reacted with 32 g of O₂ to produce 44 g of CO₂
Next, we shall determine the limiting reactantFrom the balanced equation above,
, 12 g of C reacted with 32 g of O₂.
Therefore,
144 g of C will react with = (144 × 32)/12 = 384 g of O₂.
From the calculations made above, we can see that both C and O₂ are sufficient for the reaction. Thus, C and O₂ are both limiting reactants.
Finally, we shall determine the mass of CO₂ obtained from the reaction.From the balanced equation above,
balanced equation above, 12 g of C reacted to produce 44 g of CO₂.
Therefore,
144 g of C will react to produce = (144 × 44)/12 = 528 g of CO₂.
Thus, the mass of CO₂ obtained from the reaction is 528 g
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What is it called when you have hydrogen peroxide that just eventually turns into water
Answer:
chlorine
Explanation:
Answer:chlorine reacts with hydrogen peroxide
Explanation:
A 4.0 L flask containing N2 at 15 atm is connected to a 4.0 L flask containing H2 at 7.0 atm and the gases are allowed to mix. What is the mole fraction of N2
The mole fraction of N₂ after the mixture of 4.0 L of N₂ at 15 atm with 4.0 L of H₂ at 7.0 atm is 0.68.
We can calculate the mole fraction of N₂ with the following equation:
[tex] X_{N_{2}} = \frac{n_{N_{2}}}{n_{t}} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{H_{2}}} [/tex] (1)
The number of moles of N₂ and H₂ can be found with the ideal gas law:
[tex] PV = nRT [/tex]
Where:
P: is the pressure
R: is the gas constant
T: is the temperature
V: is the volume
For nitrogen gas we have:
[tex] n_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{RT} [/tex] (2)
And for hydrogen:
[tex] n_{H_{2}} = \frac{P_{H_{2}}V_{H_{2}}}{RT} [/tex] (3)
After entering equations (2) and (3) into (1), we get:
[tex] X_{N_{2}} = \frac{\frac{P_{N_{2}}V_{N_{2}}}{RT}}{\frac{P_{N_{2}}V_{N_{2}}}{RT} + \frac{P_{H_{2}}V_{H_{2}}}{RT}} [/tex]
Since RT are constants, we have:
[tex] X_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{P_{N_{2}}V_{N_{2}} + P_{H_{2}}V_{H_{2}}} [/tex]
We know that:
[tex] P_{N_{2}} = 15 atm[/tex]
[tex] V_{N_{2}} = 4.0 L[/tex]
[tex] P_{H_{2}} = 7.0 atm[/tex]
[tex] V_{H_{2}} = 4.0 L[/tex]
so:
[tex] X_{N_{2}} = \frac{15 atm*4.0 L}{15 atm*4.0 L + 7.0 amt*4.0 L} = 0.68 [/tex]
Therefore, the mole fraction of N₂ is 0.68.
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how does agriculture contribute to the global population in relation to health and food supplies?
Answer:
We present a quantitative analysis of global and regional food supply to reveal the flows of calories, protein and the micro-nutrients vitamin A, iron and zinc, from production through to human consumption and other end points. We quantify the extent to which reductions in the amount of human-edible crops fed to animals and, less importantly, reductions in waste, could increase food supply. The current production of crops is sufficient to provide enough food for the projected global population of 9.7 billion in 2050, although very significant changes to the socio-economic conditions of many (ensuring access to the global food supply) and radical changes to the dietary choices of most (replacing most meat and dairy with plant-based alternatives, and greater acceptance of human-edible crops currently fed to animals, especially maize, as directly-consumed human food) would be required. Under all scenarios, the scope for biofuel production is limited. Our analysis finds no nutritional case for feeding human-edible crops to animals, which reduces calorie and protein supplies. If society continues on a ‘business-as-usual’ dietary trajectory, a 119% increase in edible crops grown will be required by 2050.
Explanation:
Which of the following is a way to increase pressure on a gas?
Answer:
increase the number of gas particles
Explanation:
why do you continuously gain exactly the amount of mass you consume with each meal
Answer:
It depends on how much the calories and fat there is in the meals and if you don't get enough physical activity of the same amount of food that you eat it can lead to weight gain
Explanation:
A particular reaction has a DH o value of -164 kJ and DS o of -185 J/K at 298 K. Calculate DG o at 617 K in kJ (with 3 significant digits), assuming that DH o and DS o hardly change with temperature
Answer:
e
Explanation:
7. What's the structure of methane, CH,? Is it polar or non-polar?
O A. Trigonal planar, polar
O B. Tetrahedral, polar
O C. Trigonal planar, non-polar
O D. Tetrahedral, non-polar
Answer:
The answer is D.
Explanation:
Methane (CH4) is a non-polar hydrocarbon molecule made of a single carbon atom and four hydrogen atoms. It is found in nature in the form of natural gas. It is non-polar because the difference in electronegativities between carbon and hydrogen is not sufficient to produce a polarized chemical connection between these two elements.
The methane molecule has a tetrahedral structure in terms of molecular geometry. Because the molecule is not planar, the angles between the H-C-H bonds are 109.5°, which is more than the 90° that they would be if the molecule were planar.
Explanation:
tetrahedral, non polar
Please help asap chemistry worksheet doesn’t make sense
Answer:
dog dog im not gon hold u we kinda screwed
Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if an additional 0.182mol of methane is added to the container under constant temperature and pressure? Give your answer in three significant figures.
The final volume of the methane gas in the container is 6.67 L.
The given parameters;
initial volume of gas in the container, V₁ = 2.65 Linitial number of moles of gas, n₁ = 0.12 moladditional concentration, n = 0.182 molThe total number of moles of gas in the container is calculated as follows;
[tex]n_t = 0.12 + 0.182 = 0.302 \ mol[/tex]
The final volume of gas in the container is calculated as follows;
[tex]PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L[/tex]
Thus, the final volume of the methane gas in the container is 6.67 L.
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