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MariettaO
9 days ago
9

What element from space is pulled by gravity and turn into a protostar?

Chemistry
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How many molecules are in 0.25 grams of dinitrogen pentoxide?
eduard [2782]
To determine the answer, you need to understand the formula for converting grams to moles, which will then lead you to the number of molecules.
The result is 2 moles of N2O5. The process is as follows:
(0.25 g N2O5) (1 mol/ 108 g)=2.31 molecules
Thus, the final answer is 2 molecules.
4 0
1 month ago
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What is the mass of 22.4 L of H2 at STP?
eduard [2782]

A. 1.01 is the accurate result

Because

The formula used is Pv= nRT

P=1 atm

V= 22.4 L

N= x

R= 0.0821

T= 273 K (since it’s standard temperature)

Thus, (1)(22.4)=(x)(0.0821)(273)

X= 1.001

7 0
2 months ago
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Select the option that correctly completes this statement: If the reaction takes place at a constant volume in a thermally insul
KiRa [2933]

Answer:

The temperature of the gas rises.

Explanation:

This is classified as an ISOCHORIC process where the volume remains unchanged. There is no work done by the system.

The gas only receives internal energy from the heat transferred to it from the surroundings.

In this situation, the pressure also increases.

8 0
2 months ago
What is the symbol for the isotope of 58 co that possesses 33 neutrons?
KiRa [2933]
The element with atomic number 58 is Cerium, meaning its symbol should be Ce rather than Co, which belongs to Cobalt with atomic number 27. Therefore, the notation for isotopes consists of the element's symbol accompanied by a superscript and a subscript, properly aligned. The superscript indicates the mass number.

Mass number = protons + neutrons = 58 + 33 = 91

The subscript denotes the atomic number, which is 58. This notation is illustrated in the attached image.

6 0
3 months ago
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The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan
eduard [2782]

Response: The rate constant at 525 K is, 0.0606M^{-1}s^{-1}

Rationale:

Based on the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant when 701K = 2.57M^{-1}s^{-1}

K_2 = rate constant when 525K =?

Ea = activation energy for the process = 1.5\times 10^2kJ/mol=1.5\times 10^5J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 701 K

T_2 = final temperature = 525 K

Substituting the provided values into this formula yields:

\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]

K_2=0.0606M^{-1}s^{-1}

Thus, the rate constant at 525 K is, 0.0606M^{-1}s^{-1}

8 0
2 months ago
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