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3 years ago

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Solid substance A has a melting point of 100°C. Liquid substance B has a freezing point of 110°C. For each substance, identify i

ts state of matter and describe the motion of its particles when the substance is at 115°C.

1 answer:

9966 [12]3 years ago

4 0

Answer:

both will be at liquid state. the particles will move rapidly in all directions and will collide with other particles in random motion

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PLS HELP What is one of the benefits that scientists have discovered from organisms' daily exposure to radioisotopes? 1, being a

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Being able to carbon date fossils

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5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

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Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

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Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

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The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

<u>Answer:</u> The empirical formula for the given compound is $CaCN_2$

<u>Explanation:</u>

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

<u>For calculating the mass of nitrogen:</u>

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

<u>For calculating the mass of calcium:</u>

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

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Which of the following has the greatest electronegativity difference between the bonded atoms? View Available Hint(s) Which of t

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Answer: A group 1 alkali metal bonded to fluoride, such as LiF.

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself. The size of an atom increases as we move down the group because a new shell is added and electron gets added up.

1. A strong acid made of hydrogen and a halogen, such as HCl : A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms. Electronegativity difference = electronegativity of chlorine - electronegativity of hydrogen = 3-2.1= 0.9

2. A group 1 alkali metal bonded to fluoride, such as LiF: Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal.

Electronegativity difference = electronegativity of fluorine - electronegativity of lithium= 4-1= 3

3. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO: A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

Electronegativity difference = electronegativity of oxygen - electronegativity of carbon= 3.5-2.5= 1.0

4. A diatomic gas, such as nitrogen $(N_2)$ : Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms.

Electronegativity difference = 0

Thus the greatest electronegativity difference between the bonded atoms is in LiF.

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