What mass of aluminium chloride can be obtained when 0.48mols of aluminium is completely reacted with dilute hydrochloric acid?
Equation for this reaction:
2Al + 6HCl 2AlCl3 + 3H2​

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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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.

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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:

Answer:

sorry i dont know

Explanation:

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.

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

From 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.

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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Explanation:

Carbon dioxide and water

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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.

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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Answer:

e

Explanation:

Answer:

EDTA disodium salt | C10H14N2Na2O8

Explanation:

Answer:

dog dog im not gon hold u we kinda screwed

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

n = t / t½

n = 4 / 1

Number of half-lives (n) = 4

Initial amount (N₀) = 1 g

[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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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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Answer:

it gives an atom its charge

Answer:

chlorine

Explanation:

Answer:chlorine reacts with hydrogen peroxide

Explanation:

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

The 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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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]

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Answer:

the answer is B.

Explanation:

your welcome

Answer:

the answer is B

Explanation:

i have the same exact questionn

Answer:

increase the number of gas particles

Explanation:

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

Answer:

c) single bonds, double bonds, carbon atoms and hydrogen atoms

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

Answer:BF3

Explanation:USA test prep

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

The heat of fusion of the given sample of the ethyl alcohol is 104.16 J/g.

The given parameters:

The 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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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.

Answer: Atomic number

Explanation:

I hope this helps you!

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:

#let's study