Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
The volume of a box with length 25 cm, height 25 cm and width 1.0 m is 0.0625m³.
Volume of the box which is a cuboid can be calculated by multiplying the length, breadth and height of the given box.
Volume of the box is given by the product of the length of the box, Height of the box and Breadth or width of the box.
Since, the box is a cuboid, hence the formula is given by the products of length, breadth and height.
Given,
length of the box= 25cm = 0.25m
Height of the box =25cm = 0.25m
width of the box= 1m
Volume = length × width × height of box
Volume = 0.25 × 0.25 × 1
Volume = 0.0625m³
The volume of the box is 0.0625m³.
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Answer:
The compound which are boh soluble in water and hexane is
B. Ethanol and 1-propanol
Explanation:
The compounds ethanol and 1-propanol are soluble in both hexane and water.
It is soluble in water as both consists of polar end due to hydrogen bonding present in the -OH functional group.
and both are soluble in hexane as they contain a non polar end and the alliphatic hydrocarbon chain in them.
The solubility of alcohols varies in increasing order as the hydrocarbon chain increases. And becaue of this it becomes more non polar.
Non polar properties decreases for branched molecules.
so, the correct option is ethanol and 1-propanol.
Answer:
When cells become damaged or die the body makes new cells to replace them. This process is called cell division. One cell doubles by dividing into two. Two cells become four and so on.
Explanation:
corrected question: A chemist adds 135mL of a 0.21M zinc nitrate solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Round your answer to significant digits.
Answer:
5.37g
Explanation:
0.21M means ; 0.21mol/dm³
1dm³=1L , so we can say 0.21mol/L
if 0.21mol of Zinc nitrate is contained in 1L of water
x will be contained in 135mL of water
x= 0.21*135*10³/1
=0.02835moles
number of moles= mass/ molar mass
mass= number of moles *molar mas
molar mass of Zn(NO₃)₂=189.36 g/mol
mass= 0.02835 *189.36
mass=5.37g
