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Lostsunrise [7]
3 years ago
7

Which is the chemical formula for zinc phosphate

Chemistry
1 answer:
Orlov [11]3 years ago
3 0
The Chemical fomulla for Zinc Phosphate

Zn3(PO4)2
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What is the predominant intermolecular force in the liquid state of each of these compounds: hydrogen fluoride (HF), carbon tetr
telo118 [61]

Answer:

HF - hydrogen bonding

CBr4 - Dispersion

NF3 - Dipole-dipole

Explanation:

Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom such as fluorine, chlorine nitrogen, oxygen etc. Hence the dominant intermolecular force in HF is hydrogen bonding.

CBr4 is nonpolar because the molecule is tetrahedral and the individual C-Br dipole moments cancel out leaving the molecule with a zero dipole moment hence the dominant intermolecular force are the dispersion forces.

NF3 has a resultant dipole moment hence the molecules are held together by dipole-dipole interaction.

4 0
3 years ago
Oxygen gas was produced in a reaction and collected over water. A 136.1 mL mL sample of gas was collected over water at 25C and
Xelga [282]

Answer:

Explanation:

We shall find volume of gas at NTP or at 273 K , 760 mm of Hg .

Pressure of given gas = 1.06 x 760 mm of Hg less vapor pressure of water .

= 805.6 - 23.76 = 781.84 mm of Hg

For it we use gas law formula ,

P₁V₁ / T₁ = P₂V₂ / T₂

781.84 x 136.1 / ( 273 + 25 ) = 760 x V₂ / 273

= 128.26 mL .

= 128.26  x 10⁻³ L .

22.4 L of oxygen will have mass of 32 g

128.26 x 10⁻³ L of oxygen will have mass of 32 x 128.26 x 10⁻³ / 22.4 g

= 183.22 mg .

4 0
3 years ago
1 point
lawyer [7]

Answer:

B

Explanation:

you're moving the decimal 8 spots to the left so it can only be B

6 0
3 years ago
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
Read 2 more answers
Atoms that have a positive or negative electrical charge are called ________.​
hjlf

Explanation:

They are called or known as cations

4 0
3 years ago
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