220 grams of sugar would be in 2 liters of orange juice
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
learn more about Ions:
brainly.com/question/13692734
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Answer:
11812.58 meters = 11.81258 Km
Explanation:
hope that helps
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
27,586
<h3>
Further explanation</h3>
<u>Given:</u>
A single gold atom has a diameter of 
From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately 
<u>Question:</u>
How many atoms thick were Rutherford's foil?
<u>The Process:</u>
Convert thickness from mm to cm.

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.


Therefore, we get an atomic thickness of 27,586 atoms.
<u>Notes:</u>
- In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
- One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
- In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.
<h3>Learn more</h3>
- The energy density of the stored energy brainly.com/question/9617400
- The theoretical density of platinum which has the FCC crystal structure. brainly.com/question/5048216
- Compound microscope brainly.com/question/4000241
Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment