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3 years ago

What are the elements in group 3-12 called

Chemistry

1 answer:

Wittaler [7]3 years ago

5 0

Answer:

Transition Metals

Explanation:

The elements in groups 3-12 are called Transition Metals. These groups contain metals that usually form multiple cations. All other groups on the table (1, 2, 13-18) are called Main Group Elements.

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What is the temperature shown on the thermometer below?

olchik [2.2K]

The temperature is -8 F

Answer D

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2 years ago

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The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st

Andrei [34K]

First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 = $\frac{[H^+][HC2O4^-]}{[H2C2O4]}$ = x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6

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3 years ago

How many molecules are in 4.62 moles of nitric acid (HNO3)?

Lena [83]

Answer:

2.8x10^24

Explanation:

To convert moles to molecules, multiply the number of moles by Avagadro's number (6.02x10^23. Round if required.

4.62mol × 6.02x10^23 = 2.8x10^24

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3 years ago

What is the phase change in which a solid turns into a liquid

STatiana [176]

Answer:

melting

Explanation:

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3 years ago

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

$\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\$

$\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\$

$\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\$

$\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}$

$E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\$

$E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}$

3 0

3 years ago