<span>1.02x10^2 ml
Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this
m0*v0 = m1*v1
where
m0, v0 = molarity and volume of original solution
m1, m1 = molarity and volume of final solution.
Solve for v0, then substitute the known values and calculate:
m0*v0 = m1*v1
v0 = (1.75 M * 500 ml)/8.61 M
v0 = (1.75 M * 500 ml)/8.61 M
V0 = 101.6260163
Rounding to 3 significant figures gives 102 ml.
So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.</span>
Steps of a scientific investigation include identifying a research question or problem, forming a hypothesis, gathering evidence, analyzing evidence, deciding whether the evidence supports the hypothesis, drawing conclusions, and communicating the results.
Answer:
We can solve this by the method of which i solved your one question earlier
so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372
which is almost = 41.26 moles
so to get one mole of C12H25NaSO4 we need one mole of C12H26O
so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O
so the mass of C12H26O = 41.26× its molar mass
C12H26O = 41.26×186.34
= 7688.38 gm!!
so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!
Again i d k wether it's right or wrong but i tried my best hope it helped you!!
Freezing point, boiling point, melting point, smell, attraction or repulsion to magnets, colour change, and many more examples.
Answer:
The atomic mass of element is 65.5 amu.
Explanation:
Given data:
Abundance of X-63 = 50.000%
Atomic mass of X-63 = 63.00 amu
Atomic mass of X-68 = 68.00 amu
Atomic mass of element = ?
Solution:
Abundance of X-68 = 100-50 = 50%
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50×63)+(50×68) /100
Average atomic mass = 3150 + 3400 / 100
Average atomic mass = 6550 / 100
Average atomic mass = 65.5 amu.
The atomic mass of element is 65.5 amu.
