Answer: The students should take 49.67 g acid acetic in benzene solution. With significant digits 4.967X10 g.
Explanation: If 220 g solution 30.2% w/w acetic acid has 66.44 g acid acetic (220 g solution * 30.2%), 49.67 g solution 30.2% acetic acid has 15 g acetic acid (with a simple three rule, 15 g acetic acid*220 g solution/66.44 g acetic acid)
Plants and animals use glucose as a soluble, easily distributed form of chemical energy which can be 'burnt' in the cytoplasm and mitochondria to release carbon dioxide, water and energy. Pure monosaccharides, such as glucose, attract water.
Answer:
n= 2.55 moles
Explanation:
Using the formula of ideal gas law:
PV = nRT
nRT=PV
n= PV/RT
n= number of moles
R= Avogadro constant = 0.0821
T= Temperature in K => ºC + 273.15 K
V= volume in L
P= pressure in atm
n= (1 atm)(57.20 L) / (0.0821)(237.15 K)
n= 2.50 moles
Answer: The correct option is D.
Explanation:
In option A:
The names of both the ions are correct but the representation of lithium ion is wrong. Lithium forms a cation and here lithium anion is shown. So, this option is not correct.
In option B:
Here, the representation and name of
is written wrong. This ion has a representation of
and is named as oxalate. So, this is not correct.
In option C:
The representation of both the ions is correct but the name of
is chloride, not chlorate. So. this is not a correct option.
In option D:
The representation and names of both the ions are correct.
Hence, the correct option is D.
Answer:
45.2%
Explanation:
To calculate the percent of an element in a compound we divide the molar mass of the element by the compound and multiply that by 100
First lets find the molar mass of Fluoride
Looking at the periodic table Fluoride has a molecular mass of 18.998 g
Now we need to find the molecular mass of NaF
Looking at a periodic table, Sodium (Na) has a molecular mass of 22.990g and Fluoride has a molecular mass of 18.998 so NaF has a molecular mass of 22.990(1) + 18.998(1) = 41.988g
Now we divide the mass of fluoride by the mass of sodium fluoride and multiply that by 100 to find the percentage of fluoride that is present in NaF
Mass of Fluoride = 18.998g
Mass of Sodium Fluoride = 41.988g
Percentage of fluoride present in NaF = (18.998g / 41.988g) * 100 = 45.2%