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4 years ago

A mixture which contains more than one phase is called .

Chemistry

2 answers:

kirza4 [7]4 years ago

6 0

This is a heterogeneous mixture.

Ghella [55]4 years ago

4 0

It is called a heterogeneous mixture..

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Which of the following molecules has/have the same empirical formula as acetic acid (HC2O2H3)? (a) Formaldehyde (b) Benzaldehyde

chubhunter [2.5K]

Answer: Option (a) is the correct answer.

Explanation:

A chemical formula which represents proportions between the elements present in a compound is known as empirical formula.

When chemical or molecular formula is able to be reduced to smaller whole-number ratio of the elements then both molecular and empirical formulas are considered to be the same.

Relation between molecular formula and empirical formula is as follows.

Molecular formula = n × Empirical formula

where, n is an integer.

Chemical formula of acetic acid is $CH_{3}COOH$ .

Chemical formula of formaldehyde is $CH_{2}O$ . It is the smaller whole-number ratio of molecular formula of acetic acid.

Chemical formula of Benzaldehyde is $C_{6}H_{5}COOH$ .

Chemical formula of glucose is $C_{6}H_{12}O_{6}$ .

Thus, we can conclude that formaldehyde molecules has/have the same empirical formula as acetic acid ( $HC_{2}O_{2}H_{3}$ ).

5 0

4 years ago

Complete combustion of 1.00 g of the hydrocarbon pagodane gives 3.38 g carbon dioxide. what is the empirical formula of pagodane

Blababa [14]

Combustion of any hrdrocarbon yields carbon dioxide and water such that the hydrogen and carbon are derived from the hydrocarbon.
1 mole of carbon dioxide has a mass out of which 12 g is the mass of carbon, therefore, 3.38 g of carbon will contain (3.38 × 12)/44 = 0.922 g of carbon.
Thus, since the hydrocarbon has a mass of 1 g, then the mass of hydrogen will be (1-0.922 g) = 0.078 g.
To get the empirical formula we divide the mass of each element by the atomic mass to get the number of moles.
Carbon= 0.922/12 = 0.0768 moles
Hydrogen = 0.078/1 = 0.078 moles
Then we get the ratio of the moles of carbon : hydrogen
= 0.0768 : 0.078
= 1 : 1.016
≈ 1: 1
Therefore the empirical formula of the hydrocarbon will be CH

6 0

4 years ago

Can someone please help me with this ?

ICE Princess25 [194]

Answer:

Sorry i really dont know goodluck

Explanation:

4 0

3 years ago

Hydrogen-3 has a half-life of 12.32 years. A sample of H-3 weighing 3.02 grams is left for 15.0 years. What will the final weigh

kaheart [24]

Answer:

$\large \boxed{\text{1.38 g}}$

Explanation:

Two important formulas in radioactive decay are

$(1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt$

1. Calculate the decay constant k

$\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{12.32 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{12.32 yr}}\\\\& = & \text{0.056 26 yr}^{-1}\\\end{array}$

2. Calculate the mass remaining

$\begin{array}{rcl}\ln \left(\dfrac{A_{0}}{A}\right) &= &kt \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &\text{0.056 26 yr}^{-1}\times \text{15 yr} \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &0.8439 \\\\\dfrac{\text{3.20 g}}{A} &= &e^{0.8439} \\\\\dfrac{\text{3.20 g}}{A}&= &2.325 \\\\A &= &\dfrac{\text{3.20 g}}{2.325}\\\\&= & \textbf{1.38 g}\\\end{array}\\\text{The final mass of the sample will be $\large \boxed{\textbf{1.38 g}}$}$

8 0

3 years ago

If the coefficient 2 is placed in front of the product tetraiodine nonaoxide (I4O9), then, how many atoms of each element must b

gogolik [260]

Answer:

Number of iodine atoms = 2 × 4 = 8

Number of oxygen atoms = 2 × 9 = 18

Explanation:

Coefficient 2 is placed in front of the product tetraiodine nonaoxide:

= $2I_4O_9$

Number of iodine atom in a single molecule tetraiodine nonaoxide = 4

Number of oxygen atoms in a single molecule tetraiodine nonaoxide = 9

Coefficient 2 is placed in front of the product tetraiodine nonaoxide:

= $2I_4O_9$

Number of iodine atoms = 2 × 4 = 8

Number of oxygen atoms = 2 × 9 = 18

4 0

3 years ago