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2 years ago

How do you classify a metal

Chemistry

1 answer:

Alenkinab [10]2 years ago

8 0

Answer:

They are solid (with the exception of mercury, Hg, a liquid).

They are shiny, good conductors of electricity and heat.

They are ductile (they can be drawn into thin wires).

They are malleable (they can be easily hammered into very thin sheets).

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A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

kaheart [24]

The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L = 1.600M

In this case, we must set up the ICE table as shown;

Br2(g) + I2(g) ↔ 2IBr(g)

I 0.6 1.6 0

C -x -x +2x

E 0.6 - x 1.6 - x 1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 - 0.595 = 0.005

[I2] = 1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

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4 0

1 year ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

Option d: 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0

3 years ago

You have 16.7 grams of hydrogen and 15.4 grams of oxygen in a synthesis rxn. Which is the limiting reagent?

sleet_krkn [62]

Answer:

oxygen is limiting reactant

Explanation:

Given data:

Mass of hydrogen = 16.7 g

Mass of oxygen = 15.4 g

Limiting reactant = ?

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 16.7 g/ 2 g/mol

Number of moles = 8.35 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 15.4 g/ 32 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of both reactant with product,

H₂ : H₂O

2 : 2

8.35 : 8.35

O₂ : H₂O

1 : 2

0.48 : 2×0.48 = 0.96 mol

The number of moles of water produced by oxygen are less so it will limiting reactant.

3 0

3 years ago

The The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k=0.029 W/ m measured temperature differ

vfiekz [6]

Answer:

Heat flux = 13.92 W/m2

Rate of heat transfer throug the 3m x 3m sheet = 125.28 W

The thermal resistance of the 3x3m sheet is 0.0958 K/W

Explanation:

The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

$q=-k*A*\frac{\Delta T}{\Delta X}\\\\q=-0.029\frac{W}{m*K}*(3m*3m)*\frac{12K}{0.025m} =125.28W$

The heat flux can be defined as the amount of heat flow by unit of area.

Using the previous calculation, we can estimate the heat flux:

$heat \, flux=\frac{q}{A}=\frac{125.28 W}{9 m^{2} } =13.92 W/m^{2}$

It can also be calculated as:

$q/A=-k*\frac{\Delta T}{\Delta X}$

The thermal resistance can be expressed as

$\Delta T=R_t*Q\\R_t=\Delta T/Q=\frac{\Delta X}{k*A}$

For the 3m x 3m sheet, the thermal resistance is

$R_t = \frac{\Delta X}{k*A}=\frac{0.025m}{0.029W/mK*9m^{2}}=0.0958 \, K/W$

4 0

3 years ago

Which two events are apart of the rock cycle

JulsSmile [24]

Answer:

The key processes of the rock cycle are crystallization, erosion and sedimentation, and metamorphism.

Explanation:

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3 0

2 years ago

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