Answer:- 0.134 seconds
Solution:- The speed is given as 
and the circumference is 24900 miles which is same as the distance light have to covered. It asks to calculate the time required to cover this distance by the light.
Unit conversion is needed from miles to meters since the speed is given in meters per second.
1 mile = 1609.34 meters
Thus, 
= 40072566 meters
Now, 
Rearranged for time, that gives: 
Inserting the values:
= 0.134 seconds
Hence, light would take 0.134 seconds to traverse the indicated distance. The answer without the unit is 0.134.
Answer:
Explanation:
Given data:
Initial temperature T₁ = 25.2°C = 298.2K
Initial pressure P₁ = 0.6atm
Final temperature = 72.4°C = 345.4K
What we need to find:
Final pressure = ?
To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:
Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.
Now we can substitute the known variables:
P₂ = 0.7atm
1) To express 0.89% m/v, it equals 0.89 grams of NaCl per 100 ml of solution.
This corresponds to 8.9 grams of NaCl in 1000 ml of solution, or 8.9 grams in 1 liter.
2) Molarity is represented as M = moles of solute / liters of solution.
Thus, we need to determine the moles in 8.9 grams of NaCl.
3) The molar mass of NaCl is calculated as 23.0 g/mol + 35.5 g/mol = 58.5 g/mol.
4) Therefore, the number of moles of NaCl calculates as mass / molar mass = 8.9 g / 58.5 g/mol = 0.152 moles.
5) Consequently, M = 0.152 moles of NaCl / 1 liter of solution = 0.152 M.
Answer: 0.152 M
Answer:
Please review the following responses
Explanation:
1) A solution of 100. mL contains 19.5 g of NaCl (3.3M)
2) 100. mL of NaCl solution at 3.00 M (3 M)
3) A solution of 150. mL holds 19.5 g of NaCl (2.2 M)
4) The concentrations of beakers 1 and 5 are identical (1.5M)
Molar mass of NaCl = 23 + 36 = 59 g
For beaker number 3:
59 g -------------- 1 mol
19.5 g ------------- x
x = 19.5 x 1/59 = 0.33 mol
Molarity (M) = 0.33 mol/0.150 l = 2.2 M
For beaker number 4:
Molarity (M) = 0.33mol/0.10 l = 3.3 M
For beaker number 5:
Molarity (M) = 0.450/0.3 = 1.5 M
