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Ghella [55]
3 years ago
14

Fog is an example of solution collide suspension

Chemistry
2 answers:
solniwko [45]3 years ago
6 0
Collide I am pretty sure :)
IrinaVladis [17]3 years ago
4 0
I'd say it's suspension, but I'm not 100% sure...

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A newly discovered element, Z, has two naturally occurring isotopes. 90.3 percent of the sample is an isotope with a mass of 267
blsea [12.9K]
atomic mass=percentage of isotope a * mass of  isotope a + percentage of isotope b * mass of  isotope b+...+percentage of isotope n * mass of isotope n.

Data:
mass of isotope₁=267.8 u
percentage of isotope₁=90.3%

mass of isotope₂=270.9 u
percentage of isotope₂=9.7%

Therefore:

atomic mass=(0.903)(267.8 u)+(0.097)(270.9 u)=
=241.8234 u + 26.2773 u≈268.1 u

Answer: the mass atomic of this element would be 268.1 u
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3 years ago
An acid is:
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An acid is a proton donor
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3 years ago
Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
avanturin [10]
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate
</span>
5 0
3 years ago
If 85.0 L of helium at 29.0°C is compressed to 32.0 L at constant pressure, what is the new temperature?
Feliz [49]

Answer:

113.69°k

Explanation:

V1=85L of helium                         V2=32L

T1= 29°C +273= 302°K                 T2=?

     T2=<u>TIV2</u>

              V1

    T2=<u>(302)(32)</u>= <u>9664</u>

                85           85

    T2=  113.69°K

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2 years ago
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The correct option is A.
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2 years ago
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