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3 years ago

Which three groups of the periodic table contain the most elements classified as metalloids (semimetals)?

2 answers:

8 0

The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).

If you like my answer, please vote me a 'brainliest' - trying to improve my rank :-)

andrey2020 [161]3 years ago

5 0

Answer:

The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).

If you like my answer, please vote me a 'brainliest' - trying to improve my rank :-)

Explanation:

what they said

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Nadusha1986 [10]

Answer: 20 mg Te-99 remains after 12 hours.

Explanation: N(t) = N(0)*(1/2)^(t/t1/2)

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N(t) = 20 mg remains after 12 hours

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2 years ago

One isotope of carbon has 6 protons and 6 neutrons. The number of protons and neutrons of a second isotope of carbon would be __

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D. 6 protons and 7 neutrons

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3 years ago

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Which of the following statements about the carbon cycle is NOT true?

nalin [4]

Answer:

Plants consume carbon through transpiration

Explanation:

In transpiration, plants lose water vapor through the stomata in their leaves. No carbon is involved in transpiration, which has an outbound direction. Nothing can be consumed through the stomata when vapor is going out of the plant. It´s like trying to get in through the exit.

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3 years ago

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A tau lepton decays into an electron, an electron antineutrino and a tau neutrino. Write out this reaction in symbolic (equation

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2 years ago

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

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3 years ago