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storchak [24]
3 years ago
6

During a lab, you heat 1.62 g of a CoCl2 hydrate over a Bunsen burner. After heating, the final mass of the dehydrated compound

is 0.88 g. Determine the formula of the hydrate and also give the full name of the hydrate.
PLZZZ!! Need this ASAP
Chemistry
1 answer:
sashaice [31]3 years ago
8 0
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O) 
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
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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
If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2
Vlad [161]

Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

1 mole of O2 occupy 22.4L at stp.

Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.

Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:

Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

3 0
3 years ago
I don't know how to solve this​
Misha Larkins [42]

For the first part, use the question M=mol/vol (liters)

To do this, you have the given 1.6 M solution

divide the 360g by the molar mass of ethanol (44.07) to get moles

360/44.07=8.16 mol

so

1.6M = 8.16 mol/x vol

volume: 5.1 Liters

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abruzzese [7]
I think it’s mass number 12
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What is soil? where do the water and nutrients found in the soil come from?
Liono4ka [1.6K]

Answer:

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We get these from the plants that grow, and the animals that we eat that eat the plants. Most soils have a large supply of nutrients in them, and they get taken up by plants when plants absorb water. Soils need to be healthy to grow large quantities of plants, and animals need plants to grow strong.

Explanation:

Plz mark brainliest thanks

6 0
3 years ago
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