Answers
The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.
The mass in grams of 1.50 x 10²³ atoms of Mg is 5.98 g.
The mass in grams of 4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.
The mass in grams of 8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.
The mass in grams of 25 atoms of W is 3.1 x 10⁻²¹ g.
The mass in grams of 1 atom of Au is 3.27 x 10⁻²² g.
What is the mass in grams of 3.011 x 10²³ atoms F?
The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;
6.023 x 10²³ atoms = 19 g of F
3.011 x 10²³ atoms F = ?
= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)
= 9.5 g
The mass in grams of 1.50 x 10²³ atoms of Mg is calculated as follows;
6.023 x 10²³ atoms = 24g of Mg
1.5 x 10²³ atoms F = ?
= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)
= 5.98 g
The mass in grams of 4.50 x 10¹² atoms of Cl is calculated as follows;
6.023 x 10²³ atoms = 35.5 g of Cl
4.5 x 10²³ atoms Cl = ?
= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)
= 2.65 x 10⁻¹⁰ g
The mass in grams of 8.42 x 10¹⁸ atoms of Br is calculated as follows;
6.023 x 10²³ atoms = 80 g of Br
8.42 x 10¹⁸ atoms Br = ?
= (8.42 x 10¹⁸ x 80 g)/(6.023 x 10²³)
= 1.12 x 10⁻³ g
The mass in grams of 25 atoms of W is calculated as follows;
6.023 x 10²³ atoms = 74 g of W
25 atoms W = ?
= (25 x 74 g)/(6.023 x 10²³)
= 3.1 x 10⁻²¹ g
The mass in grams of 1 atom of Au is calculated as follows;
6.023 x 10²³ atoms = 197 g of Au
1 atom of Au = ?
= (1 x 197 g)/(6.023 x 10²³)
= 3.27 x 10⁻²² g
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Final answer:
This solution provides the calculations necessary to convert the number of atoms of various elements (smallest particle of an element) to grams. It does so by using the molar mass of each element and Avogadro's number.
Explanation:
The mass of atoms can be determined by using Avogadro's number (6.022 x 1023 atoms/mol) and the molar mass of the specific element (g/mol). We use these to create a conversion factor and multiply by the number of atoms given.
- For F (fluorine), which has a molar mass of about 18.9984 g/mol, 3.011 x 1023 atoms F is 9.00 g F.
- For Mg (magnesium), with molar mass of about 24.3050 g/mol, 1.5 x 1023 atoms Mg is 6.07 g Mg.
- For Cl (chlorine), with molar mass of about 35.453 g/mol, 4.50 x 1012 atoms Cl is 2.67 x 10-10 g Cl.
- For Br (bromine), with molar mass about 79.904 g/mol, 8.42 x 1018 atoms Br is 0.12 g Br.
- For W (tungsten), with molar mass about 183.84 g/mol, 25 atoms W is 7.65 x 10-22 g W.
- For Au (gold), with molar mass about 197.0 g/mol, 1 atom Au is 3.28 x 10-22 g Au.
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Related Questions
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A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2 (aq) and Mg(s), producing Cd(s) and Mg2 (aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.
What are the names of the following compounds: FeCl HNO NaSO SO
If the enthalpy change of a reaction in a flask is δh = +67 kj, what can be said about the reaction?
Answers
Answer:
38.96383282 amu
Explanation:
39.0983 = (40.9618 0.067302) + ( ?
(1-0.067302)
39.0983 = 2.756811064 + ( ? 0.932698)
subtract 2.756811064 from both sides
36.34148894 = ( ? 0.932698)
divide both sides by 0.932698
? = 38.96383282 amu
Answer:
38.96383282 amu
Explanation:
39.0983 = (40.9618 0.067302) + ( ? (1-0.067302)
39.0983 = 2.756811064 + ( ? 0.932698)
subtract 2.756811064 from both sides
36.34148894 = ( ? 0.932698)
divide both sides by 0.932698
? = 38.96383282 amu
Answers
Answer:
4,25% v/v H3PO4
Explanation:
The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:
%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%
In other words, we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition. Which in this case is 1 L (1000 mL).
We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:
Ci*Vi=Cf*Vf
Where:
Ci, is the initial concentration of the substance.
Vi, the initial volume of the substance
Cf, the final concentration reached after dilution
Vf, the final volume of the solution at which the substance was diluted
In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:
Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v
Note that being up and down in the division, the mL unit is canceled to result in% v / v.
(b) The boiling point of AsH₃ from the boiling points of PH₃ (- 87.4°C) and SbH₃ (-17.1°C) (actual value = -55°C)
Answers
Answer:
a) Approximate boiling point of HBr = -60.15 °C
b) Approximate boiling point of AsH₃ = -52.25 °C
Explanation:
Döbereiner stated that some elements could be arranged in groups of 3 similar elements ( known as "triads) , and the element of the middle ( elements are ordered with respect to their atomic mass) would have properties between the other 2 ( the average value)
a) In the first case the triad would be the halogen triad ( Cl , Br and I ) . And according to Döbereiner , the boiling point of HBr should be the average of HCl and HI . Therefore
Approximate boiling point of HBr = [(- 84.9°C) + (-35.4°C)]/2 = -60.15 °C
b) Simmilarly for AsH₃ , PH₃ and SbH₃ , the boiling point of AsH₃ would be
Approximate boiling point of AsH₃ = [(- 87.4°C) + (-17.1°C)]/2 = -52.25 °C
b. 72.8 g c2h6o in 2.34 l of solution
c. 12.87 mg ki in 112.4 ml of solution
Answers
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M
Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M
Q3)
Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M
The molarities of the given solutions: (a). 0.38 mol of LiNO₃ in 6.14 L of solution has a molarity of 0.062 M. (b). 72.8 g of C₂H₆O in 2.34 L of solution has a molarity of 0.675 M. (c). 12.87 mg of KI in 112.4 mL of solution has a molarity of 0.000688 M.
To calculate the molarity (M) of a solution, you can use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
a. 0.38 moles of LiNO₃ in 6.14 L of solution:
Molarity (M) = 0.38 moles / 6.14 L = 0.062 M
b. 72.8 grams of C₂H₆O (ethyl alcohol) in 2.34 L of solution:
First, you need to convert grams to moles using the molar mass of C₂H₆O.
Molar mass of C₂H₆O = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.08 g/mol
Now, calculate moles of C₂H₆O:
moles = 72.8 g / 46.08 g/mol = 1.58 moles
Molarity (M) = 1.58 moles / 2.34 L = 0.675 M
c. 12.87 mg of KI in 112.4 mL of solution:
First, convert milligrams to grams (1 g = 1000 mg):
12.87 mg = 12.87 g (since 12.87 mg / 1000 = 0.01287 g)
Now, convert mL to liters (1 L = 1000 mL):
112.4 mL = 0.1124 L
Calculate moles of KI:
Molar mass of KI = 39.10 g/mol (for K) + 126.90 g/mol (for I) = 166.00 g/mol
moles = 0.01287 g / 166.00 g/mol = 7.75 × 10⁻⁵ moles
Molarity (M) = (7.75 × 10⁻⁵ moles) / 0.1124 L = 0.000688 M
So, the molarities of the solutions are as follows:
a. 0.062 M
b. 0.675 M
c. 0.000688 M
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Answers
Nuclear reaction: ¹⁶O + p⁺→ ¹³N + α (alpha particle).
Alpha decay is radioactive decay in which an atomic nucleus emitsan alpha particle (helium nucleus) and transforms into an atomwith an atomic number that is reduced by two and massnumber that is reduced by four.
When oxygen-16 gain one proton, atomic mass is 17, but when lose alpha particleatomic mass reduces by four to 13.
Answers
Answer:
I belive the answer is A but the image quality is hard to tell.
Explanation:
As pressure increases the higher the melting point of rock becomes making it harder to melt, thats why molten rocks brought to the surface melt because of the change in pressure.