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xz_007 [3.2K]
4 years ago
10

Two or more substances in variable proportions, where the composition is variable throughout are a a homogeneous mixture. b a so

lution. c an amorphous solid. d a heterogeneous mixture. e a compound.
Chemistry
1 answer:
REY [17]4 years ago
4 0

<u>Answer:</u> In heterogeneous mixture, the composition remains variable throughout.

<u>Explanation:</u>

For the given options:

Homogeneous mixtures are defined as the mixtures that appears uniformly throughout the solution and the particle size or shapes are not different. The composition remains constant throughout.

A solution is defined as the solution in which the components gets completely dissolved in it. Particles are evenly spread in these solutions. These solution does not scatter light falling on it. The composition remains constant throughout.

An amorphous solid is defined as the solid in which the constituent particles of the matter are arranged in the random manner. The composition remains constant throughout.

A heterogeneous mixture is defined as the mixture in which component are unevenly spread throughout the solution. The size and shape of the particles differ in these mixtures. These mixture scatter the light falling on it.  The composition do not remains constant throughout.

Compound is defined as the chemical specie which is formed by the combination of two or more different type of atoms. <u>For Example:</u> H_2O,O_2 etc...The composition remains constant throughout.

Hence, in heterogeneous mixture, the composition remains variable throughout.

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PRACTICE
Stels [109]

Answer:

35.8 u

Explanation:

The atomic mass of Cl is the weighted average of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its percent abundance).

Atomic mass of Cl-35 = 17p + 18n = 17 × 1.007 u + 18 × 1.009 u

= 17.119 u + 18.162 u = 35.28 u

Atomic mass of Cl-37 = 17p + 20n = 17 × 1.007 u + 20 × 1.009 u

= 17.119 u + 20.180 u = 37.30 u

Set up a table for easy calculation.

0.755 × 35.28 u =  26.64  u

0.245 × 37.30 u =    9.138 u

             TOTAL = 35.8     u

Note: The actual atomic mass of Cl is 35.45 u.

The calculated value above is incorrect because

(a) the given isotopic percentages are incorrect and

(b) the protons and neutrons have less mass when they are in the nucleus than when they are free. Thus, the calculated masses of Cl-35 and Cl-37 are too high.

7 0
4 years ago
Read 2 more answers
Air is made up of different gases, such as oxygen, nitrogen, and car
tia_tia [17]

The correct option would be that oxygen, nitrogen, and carbon dioxide cannot react with one another.

<h3>Why air components cannot react</h3>

The components of atmospheric air, nitrogen, oxygen, carbon dioxide, etc., cannot react with one another because there is not enough energy in the atmosphere to set the reaction rolling.

For a reaction to take place, there must be enough energy to break the bonds in each air component. This is why the air components will not spontaneously react with one another, except during special events such as lightning and thunder.

More on air components can be found here: brainly.com/question/17288850

#SPJ1

7 0
2 years ago
You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a
balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH  +  H₂O  ⇄   CH₃COO⁻  +  H₃O⁺     Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0
3 years ago
Motorcyclists must make a bigger adjustment in speed than other drivers when __________.
goblinko [34]
When you encounter a storm drain, gravel surface or pothole. 
6 0
3 years ago
For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t
Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) 2.136197(10^{25}) Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = 6.02.(10^{23}) elemental units

For example :

1 mol of oxygen atoms = 6.02.(10^{23}) oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

molarmass_{(Al)}=26.982\frac{g}{mol}

molarmass_{(S)}=32.065 \frac{g}{mol}

molarmass_{(O)}=15.999\frac{g}{mol}

If we sum all the molar masses =(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

2.663 kg=2663g

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

x=\frac{2663}{75.046}mol=35.485 mol

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = 6.02(10^{23})

35.485 moles of Al have (35.485)(6.02)(10^{23})=2.136197(10^{25}) Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

x=\frac{(2663).(15.999)}{75.046} g

x = 567.723 g of O

6 0
3 years ago
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