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Vanyuwa [196]
3 years ago
13

Which compound is impossible?A. ZnsB. Na2OC. Ca3P2D.LiCl2​

Chemistry
1 answer:
Irina18 [472]3 years ago
8 0
B. Na2O



Explanation:
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Why can't polyatomic ions ever stand alone?
Basile [38]
The prefix "poly-" means "many," and "atomic" refers to atoms, so a polyatomic ion is made of 2+ atoms covalently bonded together. These ions are charged, which is why they can never stand alone. 
7 0
3 years ago
According to the reaction represented by the unbalanced equation above, how many moles of so2(g) are required to react completel
melomori [17]
2SO₂ + O₂ = 2SO₃

n(O₂)=1 mol

n(SO₂)=2n(O₂)

n(SO₂)=2 mol
6 0
3 years ago
Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of
ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

  • P= 1.10 atm
  • V= 26.5 L
  • n=?
  • R=0.082057 \frac{atm*L}{mol*K}
  • T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* 0.082057 \frac{atm*L}{mol*K}*295 K

Solving:

n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}

mass of NaN₃=52.008 grams

<u><em>52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.</em></u>

6 0
4 years ago
Read 2 more answers
A standard solution contained 0.8 mg/mL. A student took 2 mL of the standard solution and added 10 mL of water. What is the new/
galben [10]
To get the concentration of the second solution let us use the following formulae

C1V1=C2V2 where C1 is concentration of first solution and V1 is the volume of solution first solution. on the other hand C2 is the concentration of second solution and V2 is the volume of second solution.

therefore

0.8×2=(2+10)×C2
   1.6 =12×C2
1.6/12=C2
C2     = 0.1333mg/mL
7 0
3 years ago
Read 2 more answers
A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
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