Answer:J , the answer is super giants
Answer:
Decreasing the volume of solvent in the solution of molecule A
Explanation:
We know that one of the factors that affect the rate of reaction is the concentration of the reactants. The greater the concentration of reactants, the faster the rate of reaction (the greater the frequency of collision between reactants).
Hence, when we decrease the volume of solvent in the solution of molecule A, the concentration of the solution increases and consequently more particles of molecule A are available to collide with particles of molecule B resulting in a higher rate of reaction.
Answer:
0.07172 L = 7.172 mL.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
Answer:
442.3 mL
Explanation:
Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:
Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:
Then you can solve for the volume of the solution:
Hope it helps!
Answer:
gde
Explanation:
We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.
Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.
