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Aleonysh [2.5K]
3 years ago
10

What example of matter is a mixture

Chemistry
1 answer:
Natalka [10]3 years ago
5 0

Answer:

Most of the matter around us, however, consists of mixtures of pure substances. Air, wood, rocks and dirt are examples of such mixtures. Mixtures can be further classified as Homogeneous and Heterogeneous.

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Ethanol (c2h5oh) melts at -114°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9
Nimfa-mama [501]

Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

Explanation:

Temperature of Solid C_2H_5OH=-135^oC=138 K(0^oC=273K)

Melting temperature of Solid C_2H_5OH=114^oC=159 K

Temperature of liquid C_2H_5OH=-50^oC=223K

Specific heats of solid  ethanol = 0.97 J/gK

Specific heats of liquid ethanol = 2.3 J/gK

Heat required to melt the the 25 g solid C_2H_5OH at 159 K

\Delta T_1 = 159 K - 138 K = 21 K

Q_1=mc\Delta T= 25\times 0.97J/gK\times 21 K=509.25 J

Heat required to melt and raise the temperature of C_2H_5OH upto 223 K

\Delta T_2 = 223 K - 159 K  = 64 K

Q_2=mc\Delta T= 25\times 2.3J/gK\times 64 K=3680 J

Total heat to convert solid ethanol to liquid ethanol at given temperature :

Q_1+Q_2=509.25 J+3680 J=4189.25 J=4.18925 kJ (1kJ=1000J)

Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

6 0
3 years ago
P.1 is a variant of SARS-CoV2. This is the so-called "Brazil variant". It has the amino acid the substitution N501Y in the spike
Free_Kalibri [48]

Answer:

more negative

Explanation:

There are various variants of Coviid virus. The Brazilian variant P also known as Gamma variant is the third variant of the original SARS-CoV2. This variant has raised concerns since it has ability to spread more quickly then previous variants and this is more negative variant. It is assumed that Coviid variant Gamma and Delta has ability to absorb move UV light but this is not proved yet and research is underway.

7 0
3 years ago
Write the fraction of the mass of kcl produced from 1 g of k2c03​
Minchanka [31]

Mass of KCl= 1.08 g

<h3>Further explanation</h3>

Given

1 g of K₂CO₃

Required

Mass of KCl

Solution

Reaction

K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂

mol of K₂CO₃(MW=138 g/mol) :

= 1 g : 138 g/mol

= 0.00725

From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :

= 2/1 x mol K₂CO₃

= 2/1 x 0.00725

= 0.0145

Mass of KCl(MW=74.5 g/mol) :

= mol x MW

= 0.0145 x 74.5

= 1.08 g

6 0
3 years ago
How many molecules are there in 0.45 moles of so3
vivado [14]

Answer:

2.71

Explanation:

6 0
3 years ago
Copper has two naturally occurring isotopes. Cu−63 has a mass of 62.939 amu and relative abundance of 69.17%.
fiasKO [112]
The answer is 64.907 amu.

The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2

We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083

Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
   ⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
7 0
3 years ago
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