seawater contains 0.0065% (by mass) of bromine. how many grams of bromine are ther in 1.oo l of seawater? the density of seawater is 1.025 g/ml.
The seawater's mass is 1.025 g/ml * 1000ml = 1025g
the bromine's mass is .000065 * 1025 = 0.066625 g
To determine the grams of bromine in 1.00 L of seawater, we need to consider the density of seawater and the percentage by mass of bromine.
Step 1: Calculate the mass of 1.00 L of seawater.
Given the density of seawater is 1.025 g/mL, we can calculate the mass using the following formula:
mass = volume × density
mass = 1.00 L × 1.025 g/mL
Step 2: Convert the mass of seawater to grams.
We need to convert the mass of seawater from milliliters (mL) to grams (g) since the given percentage is by mass. Since 1 mL is equal to 1 gram, the conversion is straightforward.
Step 3: Calculate the grams of bromine.
The percentage by mass of bromine in seawater is 0.0065%. To find the grams of bromine, we multiply the mass of seawater (in grams) by the percentage of bromine:
grams of bromine = mass of seawater × 0.0065
Let's plug in the values and calculate the answer.
mass = 1.00 L × 1.025 g/mL = 1.025 g of seawater
grams of bromine = 1.025 g × 0.0065 = 0.0066625 g
Therefore, there are approximately 0.0066625 grams of bromine in 1.00 L of seawater.
To find the grams of bromine in 1.00 L of seawater, you will need to calculate it using the given information.
First, we need to determine the mass of 1.00 L of seawater. The density of seawater is given as 1.025 g/mL, which means that for every 1 mL of seawater, the mass is 1.025 grams.
Since we have 1.00 L of seawater, we can convert it to mL by multiplying it by 1000 (1 L = 1000 mL):
1.00 L * 1000 mL/L = 1000 mL
Now, we can calculate the mass of 1.00 L of seawater:
Mass of seawater = Density * Volume
Mass of seawater = 1.025 g/mL * 1000 mL
Mass of seawater = 1025 grams
Next, we can calculate the grams of bromine in the seawater. We are given that seawater contains 0.0065% (by mass) of bromine.
To find the grams of bromine, we can use the following formula:
Grams of bromine = Mass of seawater * Percentage of bromine (calculated as a decimal)
Percentage of bromine = 0.0065% = 0.0065/100 = 0.000065 (as a decimal)
Grams of bromine = 1025 g * 0.000065
Grams of bromine ≈ 0.066625 grams
Therefore, there are approximately 0.066625 grams of bromine in 1.00 L of seawater.