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3 years ago

The chemical agent or hazardous material that interferes with the body's ability to transfer oxygen to the cells is:

Chemistry

2 answers:

faust18 [17]3 years ago

8 0

Answer:

B. Hydrogen cyanide

Explanation:

Hello,

In this case, hydrogen cyanide whose formula is HCN and also called prussic acid, is a colorless, venomous and flammable liquid that boils near to room temperature. Now, since it is lighter than air, it rapidly disperses up into the atmosphere, capacity that allows it to poison people as done during the world war two by interfering the body's capacity to transfer oxygen to the cells due to its lightness.

Best regards.

zheka24 [161]3 years ago

7 0

Phosgene its a my friend used in the war

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3 years ago

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3 years ago

How many moles are in 1.5x1023 molecules of Na2SO4?

german

One mole of Na2SO4 is 6.022 * 10^(23) molecules. We can divide this into the quantity in the question to find a value of 1.5/6.022 = 0.2491 moles. Rounded to two significant figures and put in scientific notation, we can rewrite this quantity as 2.5 * 10^(-1) moles

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2 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

The formula can be written for the calculation of moles as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

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