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maxonik [38]
3 years ago
10

Convert the density of water, 1.00 g/mL to lb/ft3. Provide only the numerical value in your answer with the correct number of si

gnificant figures.
Chemistry
1 answer:
PtichkaEL [24]3 years ago
3 0

Answer:

The density of water 1.00 g/ml=62.43 lbs/ft^3.

Explanation:

Density of water = 1.00 g/mL

1 lb = 453.592 g

1 g=\frac{1}{453.592} lbs

ft^3=28316.8 mL

1 mL=\frac{1}{28316.8} ft^3

Density of the water in lb/ft^3:

\frac{1.00 g}{1 mL}=\frac{1.00}{453.592 } lb\times \frac{28316.8 }{1 ft^3}

=62.43 lbs/ft^3

The density of water 1.00 g/ml=62.43 lbs/ft^3.

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Elemental calcium reacts with water at room temperature to form calcium hydroxide and hydrogen gas as shown in this equation. So
dem82 [27]
Reaction of Elemental Calcium with Water is as follow,

                        Ca  +  2 H₂O    →    Ca(OH)₂  +  H₂↑

When Sodium Metal is reacted with Water it produces NaOH and eliminates Hydrogen gas as shown in following equation,

                         2 Na  +  2 H₂O    →    2 NaOH  +  H₂↑

In both cases corresponding Hydroxides along with Hydrogen gas is produced.
8 0
3 years ago
Read 2 more answers
Any object that orbits around a larger object is called what
valkas [14]

I am not 100% sure but I think it is circulation/orbitiation.


4 0
3 years ago
Which is not an indication of a chemical change
sasho [114]
Did you have any choices??

8 0
3 years ago
For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Rudik [331]

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

8 0
3 years ago
What is the hybridization and shape for an XeF6 2+ molecule?
Anna11 [10]

Answer:

Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²

Explanation:

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²

Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48

Ca. #Substrate e⁻ = 6F = 6(8) = 48

#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs

#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs

BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²    

XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)      

8 0
3 years ago
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