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2 years ago

Given the data table below, which metal will take the least amount of

Chemistry

1 answer:

Lerok [7]2 years ago

8 0

Answer:

Barium

Explanation:

The specific heat of a material is a property defined as the amount of energy required to increase the temperature of 1g of this material in 1°C = 1K

In this way, the metal that takes the least amount of energy to increase its temperature is the metal that has the lowest specific heat.

In the problem, this metal is Barium (Ba = 0.201J/gK)

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2 years ago

0.0027432 using scientific notation

alexira [117]

Answer: 2.7423 × 10^-3

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3 years ago

Determine the mass of the following.

Luba_88 [7]

Answer:

a ) 1876.14 grams CaBr2

b ) 19.78 grams N2

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Explanation:

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2 years ago

PLS HELP!! <br> Balance this equation: __H2+ 2 O2---> 2 H2O

alekssr [168]

Answer:

2H₂ + O₂ → 2H₂O

Explanation:

The expression of the equation is given as:

_H₂ + 2O₂ → 2H₂O

Now for expression above,

Reactants Products

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O 4 2

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2H₂ + O₂ → 2H₂O

6 0

2 years ago

The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)

Karo-lina-s [1.5K]

Answer:

b. Second order in NO and first order in O₂.

Explanation:

A. The mechanism

$\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)$

B. The rate expressions

$-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]$

The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.

C. Assume the first step is an equilibrium

If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.

$\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}$

D. Substitute this concentration into the rate law

$\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]$

The reaction is second order in NO and first order in O₂.

8 0

3 years ago