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3 years ago

Which describes the oxidizing agent in a chemical reaction?

1 answer:

7 0

The reagent in the chemical reaction which oxidizes other reactants but itself reduces are called oxidizing agent.
Hope this helps, ask doubt in comment section below if you have one.

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Fluids that have a lower viscosity than water

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Syrup, molasses, and honey have a lower viscosity than water

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3 years ago

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For question #12, use the following picture:<br><br> 12. What element is this? How do you know?

labwork [276]

Answer: The element shown in the image is Helium (He).

Explanation: We are given a image of an atom having protons, neutrons and electrons.

Number of protons as shown in image = 2

Number of neutrons as shown in image = 2

Number of electron as shown in image = 2

Atomic number = Number of protons = Number of electrons

Atomic number of the element = 2

Atomic Mass = Number of protons + Number of neutrons

Atomic mass = 2 + 2 = 4

The element having Atomic number = 2 and mass number = 4 is Helium.

Element = $^4_2\textrm{He}$

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3 years ago

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50 examples word equation with balanced chemical equarion

gulaghasi [49]

This question may only be ansewered by frequent mattrrs

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2 years ago

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0 125 kPa

At t = teq 125 - 2x 4x x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$

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3 years ago

The burning of a sample of propane generated 1 04.6 kJ of heat. All of this heat was used to heat 500.0 g of water that had an i

Paul [167]

Answer: 70.0°C

Explanation:

Quantity of heat = Mass * Specific heat * Change in temperature

Quantity of heat = 104.6 KJ

Mass = 500.0 g

Specific heat of water is 4.18 J/g°C

Change in temperature assuming final temperature is x = x - 20

Units should be in grams and joules:

104,600 = 500 * 4.18 * (x - 20)

104,600 = 2,090 * (x - 20)

x - 20 = 104,600/2,090

x = 104,600/2,090 + 20

x = 69.8

= 70.0°C

3 0

3 years ago