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Vesnalui [34]
3 years ago
10

Describe a homogeneous mixture

Chemistry
2 answers:
hoa [83]3 years ago
8 0
Its when to chemicals are the same and can be taken out of but the product is still the same
ANTONII [103]3 years ago
6 0
It's a mixture that is the same through out. NO clumps or layers. It can have multiple elements/substances combined within. 

Hope this helps!
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Suppose that 4.8 L of methane at a pressure
Ghella [55]

Answer:

972.3 Torr

Explanation:

P2=P1V1/V2

You can check this by knowing that P and V at constant T have an an inverse relationship. Hence, this is correct.

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2 years ago
Drag each tile to the correct location.
cestrela7 [59]
Carbon, helium, and sodium are monoatomic elements.

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3 0
2 years ago
Read 2 more answers
Isotonic compound have high melting point. Why ? Name the ions present in CaS.
yuradex [85]

Explanation:

Because a large amount of energy is required to break the strong inter-ionic attraction.

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3 0
3 years ago
Which one has more mass a bag of cotton balls or a bag of nails and explain
Mama L [17]
You would think that the bag of nails would have more mass but their masses are identical. <span>If you were to put them both in a vacuum chamber and let them fall from a great height, they would fall the same speed. The vacuum chamber would suck all of the air out of the cotton balls, thus making it heavier and weigh the same as the bag of nails.

Hopefully this is helpful and makes sense.</span>
6 0
2 years ago
What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

Best regards!

7 0
2 years ago
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