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3 years ago

In a comparison of metals to nonmetals,metals tend to have

Chemistry

1 answer:

Georgia [21]3 years ago

3 0

They conduct electricity. They tend to have conductors of electricity.

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What are the two types of alleles? A: Dominant and recessive B: genotype and phenotype C: Gene and chromosome or D: Gene and DNA

ololo11 [35]

Answer:

dominant and recessive

Explanation:

4 0

3 years ago

What is the surface area of the given figure?

eduard

Answer:

the answer will be 2,280 cm>2

3 0

2 years ago

Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp

Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

Equilibrium when T = 100 °C

Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must add heat to boil water

ΔS > 0 (positive), because changing from a liquid to a gas increases the disorder

ΔG = 0, because the liquid-vapour equilibrium process is at equilibrium at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.

If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.

If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0

3 years ago

n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form

Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

$pK_a=8.0$

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}$

Now put all the given values in this expression, we get:

$6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}$

$\frac{[Deprotonated]}{[Protonated]}=0.01$

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

$\frac{[Protonated]}{[Deprotonated]}=100$

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0

3 years ago

If you want to lift a 30-kg box to a height of 1 m, how much work will it take?

natta225 [31]

Explanation:

workdone = force x distance

force = mass x acceleration

30 x 10 = 300N

300N x 1m

workdone= 300J

7 0

3 years ago