The ___ blends into outer space.

Why is it that the decomposition of nitroglycerin can result in so

patriot [66]

Answer:

with the molecular formula C3H5(ONO2)3, has a high nitrogen content (18.5 percent) and contains sufficient oxygen atoms to oxidize the carbon and hydrogen atoms while nitrogen is being liberated, so that it is one of the most powerful explosives known.

Explanation:

NTG reduces preload via venous dilation, and achieves modest afterload reduction via arterial dilation. These effects result in decreased myocardial oxygen demand. In addition, NTG induces coronary vasodilation, thereby increasing oxygen delivery.

3 0

3 years ago

Name two compounds in unpolluted air?

goblinko [34]

Nitrogen and oxygen are in unpolluted air

6 0

3 years ago

g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0

3 years ago

A sample of gas X occupies 10 mº at a pressure of 120 kPa.

Mashutka [201]

Answer:

The new pressure of the gas comes out to be 400 KPa.

Explanation:

Initial volume of gas = V = $10\textrm{ m}^{3}$

Initial pressure of gas = P = 120 KPa

Final volume of gas = V' = $3\textrm{ m}^{3}$

Assuming temperature to be kept constant.

Assuming final pressure of the gas to be P' KPa

$PV = P'V' \\120\textrm{ KPa}\times 10\textrm{ m}^{3} = \textrm{P'}\times 3\textrm{ m}^{3} \\\textrm{P'} = 400\textrm{ KPa}$

New pressure of gas = 400 KPa

5 0

3 years ago

The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 10−8 L/mol/s. What is the instantaneou

pochemuha

Answer: Rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For a reaction : $A\rightarrow products$

$Rate=k[A]^x$

k= rate constant

x = order of the reaction = 2

$Rate=4.71\times 10^{-8}L/mol/s\times {5.55\times 10^{-4}M}^2$

$Rate=1.45\times 10^{-14}$

Thus rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

6 0

3 years ago