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Marrrta [24]
3 years ago
13

Look up the molecular weight of Lithium diisoproylamine and determine the amount of g of base needed for this experiment. Round

to the hundredths place
Chemistry
1 answer:
Bezzdna [24]3 years ago
7 0

Answer:

0.72 g

Explanation:

Note that, the mass in gram of any substance is obtained from the relationship;

Mass= number of moles × molar mass

From the sequence of the reaction, the number of moles of Lithium diisopropylamide is 6.75mmol which is the same as 6.75 × 10^-3 moles

Molar mass of base= 107.1233 g/mol

Hence mass of the base is given by;

6.75 × 10^-3 moles × 107.1233 g/mol = 0.72 g

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Is caffeine additive?
Romashka-Z-Leto [24]

Answer:

yes

Explanation:

yes

7 0
3 years ago
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A sample of gaseous methyl ether has a mass of 8.12g and occupies a volume of 3.96L at STP. What is the molecular weight of meth
Naya [18.7K]
Molar volume is when 1 mol of any gas occupies 22.4 L at STP.
Methyl ether has a mass of 8.12 g,
Volume occupied - 3.96 L
If 22.4 L occupied by 1 mol of gas
Then 3.96 L occupied by 1/22.4 x 3.96 = 0.176 mol of gas
The mass of 0.176 mol = 8.12 g
Molar mass is mass of 1 mol
Therefore mass of 1 mol = 8.12/0.176 = 46.1
Molecular weight is 46.1 g/mol
5 0
3 years ago
Identify the hybridization of the c atom in ch2br2.
Oliga [24]
Hello!

The hybridization of the C atom in CH₂Br₂ is sp3

When bonding, the orbitals "s" and "p" from C atoms interact to form hybridized orbitals. If the C atom has 4 sigma bonds, as is the case in CH₂Br₂, there are 4 hybridized orbitals required, so 1 "s" orbital and 3 "p" orbitals hybridize to form an sp3 hybrid orbital. This orbital has tetrahedral geometry and the bond angle is 109,5 °.

Have a nice day!
7 0
3 years ago
Read 2 more answers
Why are large crystals of sodium chloride used instead of small<br> crystals when making ice cream
matrenka [14]

Answer:

Sodium chloride is added to the ice to lower the freezing point of the ice. Large crystals dissolve more slowly than small crystals. This allows time for the ice cream to freeze more evenly.

Explanation:

6 0
2 years ago
If you have a 1.0 L buffer containing 0.208 M NaHSO3 and 0.134 M Na2SO3, what is the pH of the solution after addition of 50.0 m
Vlada [557]

Answer:

pH = 7.233

Explanation:

Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.

NaHSO₃ reacts with NaOH thus:

NaHSO₃ + NaOH → Na₂SO₃ + H₂O

50.0 mL of 1.00 M NaOH are:

0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced  0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:

NaHSO₃: 0.208 mol - 0.050 mol = <em>0.158 mol</em>

Na₂SO₃: 0.134 mol + 0.050 mol = <em>0.184 mol</em>

<em> </em>

As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:

pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]

pH = 7.167 + log₁₀ [0.184] / [0.158]

<em>pH = 7.233</em>

6 0
2 years ago
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