Rank the types of visible light in order from longest to shortest wavelength?

Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

GarryVolchara [31]

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

6 0

3 years ago

A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B = 21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A: M = m1 *0.8662 O = m1 *0.1338

compound B: M = m2 *0.90666 O = m2 *0.09334

compound C: M = m3 *0.92832 O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338 m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48 ⇒ m2 = 227.74 g

0.92832m3 = 206.48 ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0

3 years ago

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

3 years ago

Explain why there was a reaction between zinc metal and a salt solution of copper sulphate please help me

natali 33 [55]

Answer:

When zinc is added to copper sulphate (CUSO4) solution due to more reactivity of zinc, cooper is replaced by the zinc and forms zinc sulphate. During the process, the colour of the solution changes from blue to colourless.

6 0

2 years ago

Suppose a solution is described as concentrated. Which of the following statements can be concluded? Select the correct answer b

Novosadov [1.4K]

Answer:

last option: none of the above.

Explanation:

Describing a solution as concentrated tells that the solution has a relative large concentration, but it is a qualitative description, not a quantitative one, so this does not tell really how concentrated the solution is. This is, the term concentrated is a kind of vague; it just lets you know that the solution is not very diluted, but, as said initially, that there is a relative large amount (concentration) of solute.

One conclusion, of course, is that the solute is soluble: else the solution were not concentrated.

On the other hand, the terms saturated and supersaturated to define a solution are specific.

A saturated solution has all the solute that certain amount of solvent can contain, at a given temperature. A supersaturated solution has more solute dissolved than the saturated solution at the same temperature; superstaturation is a very unstable condition.

From above, there is no way that you can conclude whether a solution is supersaturated or not from the statement that a solution is concentrated, so the answer is none of the above.

5 0

3 years ago