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2 years ago

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Lars observes a substance to be a solid and to float in water at room temperature (23°C). Based on the given properties, which s

ubstance is the most likely identity of this sample?
Substance Melting Point (°C) Boiling Point (°C) Density at Room Temperature (g/cm3)
carbon tetrachloride -22.9 76.7 1.59
cetyl alcohol 49.3 344 0.811
dichlorobenzene 53.5 174 1.25
sulfur hexafluoride -64 -50.8 0.00617
A.
carbon tetrachloride
B.
cetyl alcohol
C.
dichlorobenzene
D.
sulfur hexafluoride

1 answer:

Triss [41]2 years ago

4 0

Answer:

D. Sulfur Hexafluride

Explanation:

above it says the substance floats above water at room temperature and lists some substances and their density at room temp!
we know that the density of water is 1.0 so the substance in order for it to float has to be less than 1.0 and the densities for Sulfer Hexa, are all less than 1!!

I hope this helped !!

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The gas in a sealed container has an absolute pressure of 125.4 kilopascals. If the air around the container is at a pressure of

AlexFokin [52]

Answer: C. 25.6 kPa

Explanation:

The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.

As such, the formula will be,

PG = PT – PA

Where,

PG is Gauge Pressure

PT is Absolute Pressure

PA is Atmospheric Pressure

Inputted in the formula,

PG = 125.4 - 99.8

PG = 25.6 kPa

The gauge pressure inside the container is 25.6kPa which is option C.

4 0

2 years ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

Calculate the percent composition for each of the elements in Na3PO4. A 5.00 gram sample of an oxide of lead PbxOy contains 4.33

Shkiper50 [21]

Answer:

1. Percentage composition of: Na = 42%; P = 19.0%; O = 39%

2. Simplest formula of compound is PbO₂

3. (i) 2Cu(NO₃) ---> 2CuO + 2NO₂ + 3O₂

(ii) 2C₂H₆ + 7O₂ ---> 4CO₂ + 6H₂O

(iii) Mg₃N₂ + 6H₂O ---> 3Mg(OH)₂ + 2NH₃

4. 48 g of MG will react with 2 moles of Cl₂

5. 0.288 g of SO2 will be produced from the combustion of 0.331 g P₄S₃ in excess O₂

6. 12.8 g of nitric oxide can be produced from the reaction of 8.00 g NH₃ with 17.0 g O₂

7. The stock acid solution should be diluted to 6000 mL or 6.0 L

Explanation:

The full explanation is found in the attachments below

6 0

2 years ago

If a particular ore contains 55.4 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosp

gizmo_the_mogwai [7]

The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.

8 0

2 years ago

0.50 mol A, 0.60 mol B, and 0.90 mol C are reacted according to the following reaction

algol [13]

Reactant C is the limiting reactant in this scenario.

Explanation:

The reactant in the balanced chemical reaction which gives the smaller amount or moles of product is the limiting reagent.

Balanced chemical reaction is:

A + 2B + 3C → 2D + E

number of moles

A = 0.50 mole

B = 0.60 moles

C = 0.90 moles

Taking A as the reactant

1 mole of A reacted to form 2 moles of D

0.50 moles of A will produce $\frac{2}{1}$ = $\frac{x}{0.50}$

thus 0.50 moles of A will produce 1 mole of D

Taking B as the reactant

2 moles of B reacted to form 2 moles of D

0.60 moles of B reacted to form x moles of D

$\frac{2}{2}$ = $\frac{x}{0.6}$

x = 2 moles of D is produced.

Taking C as the reactant:

3 moles of C reacted to form 2 moles of D

O.9 moles of C reacted to form x moles of D

$\frac{2}{3}$ = $\frac{x}{0.9}$

= 0.60 moles of D is formed.

Thus C is the limiting reagent in the given reaction as it produces smallest mass of product.

5 0

1 year ago