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3 years ago

Logs in a fire, and charged batteries are examples of?

Chemistry

1 answer:

vlabodo [156]3 years ago

7 0

Answer:

Logs in a fire is an example of something burning, also changing its state of form. I think thats chemical change; going from wood to ashes. And charged batteries means they haven't been used yet or they were just charged, resulting in physical change.

Explanation:

I just learned this in science.

I hope this helps!

You might be interested in

How many protons and electrons are present in the mg2+ ion? 1. 14 protons, 14 electrons 2. 12 protons, 10 electrons 3. 12 proton

Nataly_w [17]

There are 12 protons and 10 electrons in a Mg2+ ion, the normal amount of neutrons is 12.

3 0

3 years ago

Rank these acids according to their expected pKa values.

givi [52]

Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO- + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

Finally, we can conclude that according to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH is the weakest acid, so the highest pKa value.

3 0

2 years ago

A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is

Vlada [557]

Answer:

$c=0.127\ J/g^{\circ} C$

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

$Q=mc\Delta T$

c is specific heat of the metal

$c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C$

So, the specific heat of metal is $0.127\ J/g^{\circ} C$ .

4 0

2 years ago

.....................?

liraira [26]

Answer:

Look at the periodic table of elements

Explanation:

Group 1 is the most reactive and to the right

Group 18 is least reactive and all the way to the left. The rest are in between. The groups go vertical.

You'd be able to solve with this information youre welcome

4 0

2 years ago

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

prisoha [69]

Answer:

2.76 × 10⁻¹¹

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹: 0 0 -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

$\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K$

T = (25.0 + 273.15) K = 298.15 K

$\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}$

3 0

3 years ago