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Lerok [7]
3 years ago
10

Predict the solubility of the following substances in water and place them in the appropriate bins

Chemistry
1 answer:
Shkiper50 [21]3 years ago
5 0
<span> NaNO3: Soluble 2. AgBr: Insoluble 3. NH4OH: Soluble 4. Ag2CO3: Insoluble 5. NH4Br: Soluble 6. BaSO4: Insoluble 7. Pb(OH)2: Insoluble 8. PbCO3: Insoluble</span>
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What word or two-word phrase best describes the shape of the hydrogen cyanide ( hcn ) molecule?
elena-14-01-66 [18.8K]

Answer : Linear

Explanation : Hydrogen Cyanide (HCN) when drwan in the Lewis diagram shows carbon atom at the center with no lone electron pairs.

The carbon and nitrogen atoms are bonded through a triple bond which counts as "one electron pair".

The molecule has two electron pairs in all and appears to be linear.

Also, according to the VSEPR theory; the electron clouds on atoms around the carbon will try to repel each other.

They will get pushed apart, which gives HCN molecule a linear molecular geometry or shape.

The bond angle that is developed will be 180 degrees since it has a linear molecular geometry of HCN. The hybridisation observed in this molecule is SP.

5 0
3 years ago
Read 2 more answers
CaCO3(s) ⇄ CaO(s) + CO2(g) 0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K.
Montano1993 [528]

Answer:

The final mass of CaCO3 is 10.68 grams

Explanation:

Step 1: Data given

Number of moles CaCO3 = 0.100 moles

Number of moles CaO = 0.100 moles

Volume = 10.0 L

When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.250 atm of CO2 is added, while keeping the temperature constant

Step 2: The balanced equation

CaCO3(s) <==> CaO(s) + CO2(g)

Step 3: Calculate moles of CO2

n = PV/RT

⇒n = the initial number of moles CO2 = TO BE DETERMINED

⇒P = the pressure of CO2 at theequilibrium = 0.220 atm

⇒V = the volume of the container = 7.0 L

⇒R = the gas constant = 0.08206 L*atm / mol * K

⇒T = the temperature = 385 K

n = 0.220*7.0/(0.08206*385) = 0.0487 (mol)

this is the amount of CaCO3 which has been converted to CaO before pumping-in additional 0.225 atm CO2(g).

Step 4: Calculate moles CaCO3

After adding additional 0.250 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm.  All this additional CO2 would completely convert to CaCO3:

n = PV/RT = 0.250*7.0/(0.08206*385) = 0.0554 moles

The total CaCO3 after equilibrium is reestablished is:

0.100 - 0.0487+ 0.0554 = 0.1067 mol

Step 5: Calculate mass CaCO3

Mass CaCO3 = 0.1067 moles * 100.09 g/mol

Mass CaCO3 = 10.68 grams

The final mass of CaCO3 is 10.68 grams

8 0
3 years ago
Suppose 11.4g of ammonium chloride is dissolved in 250 ml of a 0.3M aqueous solution of potassium carbonate. Calculate the final
Sindrei [870]

Answer:

The molarity of the final ammonium cation is 0.252M

Explanation:

<u>Step 1:</u> Data given

Mass of ammonium chloride (NH4Cl) = 11.4 grams

Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L

<u>Step 2:</u> The balanced equation

2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3

<u>Step 3:</u> Calculate moles of (NH4)Cl

moles (NH4)Cl = 11.4 grams /53.49 g/mol

Moles (NH4)Cl = 0.213 moles

<u>Step 4: </u>Calculate moles of K2CO3

Moles K2CO3 = Molarity * Volume

Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles

<u>Step 5:</u> Calculate moles (NH4)Cl at the equilibrium

For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3

(NH4)2CO3l will dissolve in 2NH4+ + CO32-

Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles

Moles NH4+ = moles (NH4)Cl = 0.063 moles

<u>Step 6:</u> Calculate Molarity of NH4+

Molarity = Moles / volume

Molarity of NH4+ = 0.063 moles / 0.250 L

Molarity of NH4+ = 0.252 M

The molarity of the final ammonium cation is 0.252M

5 0
3 years ago
Which correctly lists the types of matter, in terms of their atoms, in order from least tightly packed to most tightly packed? s
alukav5142 [94]

Answer:

gas, liquid, solid

Explanation:

As you go from gas to solid, the atoms get closer together and the kinetic energy decreases.

3 0
3 years ago
Read 2 more answers
A long (i.e., L &gt;&gt; D) stainless steel rod 6.4 mm in diameter is initially at a uniform temperature of 25°C and is suddenly
sp2606 [1]

Answer:

68.4 seconds

Explanation:

Assume that the temperature of the stainless steel rod will reach 120 C in t seconds. According to lumped – capacity analysis use the following expression to determine the value of t

T- T∞/T₀ – T∞ = e^-(hA/ρcV)t, where T is the required temperature, T∞ is the surface temperature, T₀ is initial temperature of the rod, h is the convection heat transfer coefficient, A is surface are, ρ is density of stainless steel, c is specific head and V is the volume of the rod.

Area of rod = π x d x L and Volume of rod is (π/4) x d² L

T - T∞/T₀ – T∞ = e^-(h x π x d x L/ρc(π/4 x d² x L))t

T - T∞/T₀ – T∞ = e^-(4h/(ρ x c x d))t, d is the radius of the log and L is the length of rod

<em>The value of c and ρ for stainless steel can be obtained from the table of properties of metals</em>

<u>Substitute 120 C for T, 150 C for T∞, 25 C for T₀, 7817 kg/m³ for ρ, 460 J/kg.C for c, 6.4 x 10⁻³ m for d, 120 W/m².C for h in the above equation</u>

T - T∞/T₀ – T∞ = e^-(4h/ρ x c x d))t

120 – 150/25 – 150 = e ^ -(4 x 120/7817 x 460 x 6.4 x 10⁻³)t

0.24 = e^-0.02086t

<u>Take natural log on both sides</u>

㏑0.24 = -0.02086t

T = 68.4 seconds

Thus, the temperature of the stainless steel will reach 120 C in 68.4s

5 0
3 years ago
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