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kumpel [21]
2 years ago
13

Determine if the phrases describe a solid, liquid, or gas.

Chemistry
2 answers:
Fynjy0 [20]2 years ago
4 0
Solid is the state in which matter maintains a fixed volume and shape; liquid is the state in which matter adapts to the shape of its container but varies only slightly in volume; and gas is the state in which matter expands to occupy the volume and shape of its container.
galina1969 [7]2 years ago
3 0

Answer:

there are no phrases

Explanation:

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What mass of sucrose c12h22o11 is needed to make 300 ml of a 0.50m solution?
zepelin [54]
Use the concentration to obtain the moles. I am assuming you mean to write capital M. because little m means molality. 

So, first convert the ml into Liters and then into moles, then moles to grams using the molar mass (just adding the values of each atom from the periodic table. )

Molar mass= 12 (12.0) + 22 (1.01)+ 11 (16.0)= 342 grams/mole

300 ml (1 liter/ 1000 mL) x (0.50 moles/ 1 Liter) x (342 grams/ 1 mole)= 51.3 grams


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Câu 3: Nêu phương pháp hóa học nhận biết các dung dịch không màu sau: (a) HCI, NaOH, H2SO,, NaCl. (b) HNO3. KOH, BaCl. Ca(OH)2.
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Bases turn litmus paper what color?<br><br> a) red<br> b) blue<br> c) yellow<br> d) pink
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Compare the volume of 14.1 g of helium to 14.1 g of argon gas (under identical conditions).
s2008m [1.1K]
The Volumes can be calculated from Masses by using following Formula,

                                        Density  =  Mass / Volume
Solving for Volume,
                                        Volume  =  Mass / Density

Mass of Both Gases  =  14.1 g

Density of Argon at S.T.P  =  1.784 g/L

Density of Helium at S.T.P  =  0.179 g/L

For Argon:
                                        Volume  =  14.1 g / 1.784 g/L

                                        Volume  =  7.90 L

For Helium:
                                        Volume  =  14.1 g / 0.179 g/L

                                        Volume  =  78.77 L
4 0
3 years ago
What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you
Brut [27]
Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) =  4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9 

CaCO3(s)→ Ca +2(aq)    +  CO3-2(aq)  

2) equation 2 for Ka1 = 4.7 x 10^-7

 H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

 HCO3-(aq) + H2O(l) → CO3-2 (aq)  + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s)  +  H+(aq)  → Ca2+(aq)   + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l)  Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq)  +  CO3-2(aq)   Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq)  ↔ Ca2+ (aq) + HCO3-(aq)   K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142 
6 0
3 years ago
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