The Main Purpose of the US passing the Neutrality Act in 1939 on a crash and carry basis was to

1 answer:

4 0

Hope this helped!

Neutrality Act of 1939. ... Roosevelt prevailed over the isolationists, and on November 4 the Neutrality Act of 1939 was passed, allowing for arms trade with belligerent nations (Great Britain and France) on a cash-and-carry basis, thus in effect ending the arms embargo.

You might be interested in

How is synthesis and replacement reactions related?

Fittoniya [83]

Answer:

A synthesis reaction occurs when two or more reactants combine to form a single product. ... Replacement reactions occur when elements switch places in compounds. In a single replacement reaction, one element takes the place of another in a single compound. In a double replacement reaction, two compounds exchange elements.

3 0

3 years ago

A person uses a pulley to raise a flag up a 15 meter flagpole. The weight of

GREYUIT [131]

Answer:

Explanation:

I burnt my ramen noodles today :) <3 help me <3 I need somewhere to live

8 0

3 years ago

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

Answer: The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

Explanation:

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$