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VLD [36.1K]
3 years ago
12

Which determines the state of matter for any material? A. the size of the atoms B. the hardness of the material C. the behavior

of the molecules D. the weight of the material
Chemistry
2 answers:
snow_tiger [21]3 years ago
7 0
I believe C. The behaviour of molecules determines the state of matter for any material.
aalyn [17]3 years ago
6 0
Out of the choices given, the behavior of the molecules determines the state of matter for any material. The correct answer will be C. 
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Name four products of incomplete combustion<br><br> One :<br> Two :<br> Three :<br> Four :
melamori03 [73]

Answer:

Carbon Monoxide / Carbon Dioxide / Sulfur and Nitrogen Dioxide

Explanation:

8 0
4 years ago
You need to produce a buffer solution that has a pH of 5.26. You already have a solution that contains 10. mmol (millimoles) of
Andreas93 [3]

The question is incomplete, complete question is :

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 acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pka of acetic acid is 4.74.

Answer:

33.11 millimoles of acetate we will need to add to this solution.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

Where :

tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acid

[salt] = Concentration of salt

[Acid] = Concentration of salt

We have:

pH = 5.26

pK_a=4.74

[salt] =[CH_3COO^-] = ?

[acid] = [CH_3COOH]=10.0 mmol

5.26=4.74+\log(\frac{[CH_3COO^-]}{[10.0 mmol]})

[CH_3COO^-]=33.11 mmol

33.11 millimoles of acetate we will need to add to this solution.

3 0
3 years ago
What is the atomic mass for copper?
Lena [83]
The answer is: 63.546
8 0
4 years ago
Certain metal oxide has the formula MO where M denotes the metal. A 73.35−g sample of the compound is strongly heated in an atmo
pshichka [43]

Answer:

atomic mass 112.4 and Cadmium (Cd)

Explanation:

You have 73.35 g of MO.

After the reaction the O is removed and you only have M which the mass is 64.21 g.

With that you can calculate the mass of O removed:

Mass of O = 9.14 g

Mass = AM * moles ; (AM : Atomic Mass)

9.14 = 16 * moles

moles = 9.14 / 16

moles = 0.57125

The formula of the metal oxide is MO, meaning it has 1 mole of M per mole of O. 73.35 had 0.57125 moles of O, then it also had 0.57125 moles of M, and the remaining mass of 64.21 g represents those moles

Mass = AM * moles ; (AM : Atomic Mass)

64.21 = AM * 0.57125

AM = 64.21/0.57125 = 112.4 amu

The metal with an atomic mass of 112.4 is Cadmium (Cd), therefore the metal oxide is CdO

7 0
3 years ago
The density of a gas is 1.35 L what is the volume of the gas when the mass is 12.5 g
mart [117]

Answer:

<h2>9.26 L</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{12.5}{1.35}  \\  = 9.259259...

We have the final answer as

<h3>9.26 L</h3>

Hope this helps you

8 0
3 years ago
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