Step
: To find the mass of a single washer, divide the total measured mass by the number of washers
. Step
: Convert kilograms to milligrams knowing
; thus
resulting in the final answer
.
The stronger the attraction between elements, the shorter the bond length becomes; conversely, a weaker attraction results in a longer bond length. This attraction arises from differences in their electronegativities, which is the capacity of an element to draw electrons toward itself. According to periodic trends, electronegativity rises as you move left to right and bottom to top on the periodic table. Therefore, the order from the most electronegative to the least is: Cl > Br > I. As a result, the sequence by bond length from shortest to longest is: C-Cl > C-Br > C-I.
Step 1: Convert density from g/mL to g/L; 0.807 g/mL is equivalent to 807 g/L. Step 2: Calculate Moles of N₂; Density = Mass / Volume, or Mass = Density × Volume. Plugging in values, Mass = 807 g/L × 1 L gives us Mass = 807 g. Similarly, Moles = Mass / M.mass, which leads to Moles = 807 g / 28 g.mol⁻¹, giving us Moles = 28.82 moles. Step 3: Apply the Ideal Gas Law to determine Volume of gas occupied; P V = n R T, thus V = n R T / P. Remember to convert temperature to Kelvin (25 °C + 273 = 298 K). Hence, V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm, resulting in V = 704.76 L.
Density is calculated as mass divided by volume.
Step one:
Convert m³ to ml.
1 m³ = 1,000,000 ml
0.250 m³ x 1,000,000 = 250,000 ml
Step two: Convert mg to g.
1 mg = 0.001 g, hence 4.25 x 10^8 mg equals 0.459 g.
Consequently, the density comes out to be 0.459 g/250,000 = 1.836 x 10^-6 g/ml.
Answer:
Joe correctly mixed the solution.
Explanation:
When evaluating both procedures, it's evident that both Jennifer and Joe weighed the same amount of potassium phosphate, which isn’t the variable here.
The difference is that Jennifer added the solid to 1.0 liters of water, resulting in a final volume greater than 1.0 L, thus her concentration will be lower than 1.0 M.
Joe's solution has a final volume of 1.0 L, which is why his preparation is accurate.
