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2 years ago

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Which of these statements explains the relationship among elements, compounds, and mixtures? A. Elements are made of compounds a

nd mixtures. B. Both compounds and mixtures are made up of elements. C. Mixtures are made of elements, and elements are made of compounds. D. Compounds are made of mixtures, and mixtures are made of elements.

2 answers:

Nady [450]2 years ago

7 0

Answer:

Its B, both compounds and mixtures are made up of elements.

sweet-ann [11.9K]2 years ago

5 0

The answer is answer D

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The diagram below shows part of the rock cycle. (6 points)

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Answer:

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Explanation:

Assuming this is a cycle, the volcanic eruption would lead back to rock B; rocks formed by volcanic eruptions are considered Igneous.

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Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002

marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

c

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

$pH=-\log[H^+]$

$=-\log[0.0001901 M]=3.72$

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3 years ago

What has to happen in order for dew to form on grass?

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2 years ago

"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the

soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 46.7 g

$m_2$ = mass of cold water = 45.33 g

$T_{final}$ = final temperature = 59.4°C

$T_1$ = initial temperature of hot water = 80.6°C

$T_2$ = initial temperature of cold water = 40.6°C

$c_1$ = specific heat of hot water = 4.184 J/g°C

$c_2$ = specific heat of cold water = 4.184 J/g°C

$c_3$ = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

$46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)$

$c_3=30.68J/g^oC$

Hence, the specific heat of calorimeter is 30.68 J/g°C

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3 years ago