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4 years ago

Is a diamond a homogeneous or heterogeneous mixture?

Chemistry

2 answers:

Ivanshal [37]4 years ago

7 0

Answer:

It's a heterogeneous mixture.

Explanation:

Diamond is made of just one element: carbon. Each carbon atom in diamond is connected to four other carbon atoms, in a crystal that extends on and on. There are other forms of pure carbon where the atoms are bonded differently, notably charcoal and graphite.

Klio2033 [76]4 years ago

4 0

Answer:

It's a heterogeneous mixture.

Explanation:

Diamond is made of just one element: carbon.

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A sample of hydrated tin (II) chloride (SnCl2) has a mass of 4.90 g. When it is dehydrated, it has a mass of 4.10 g. Which is th

Usimov [2.4K]

Answer:

SnCl₂·2 H₂O.

Explanation:

Relative atomic mass data from a modern periodic table:

Sn: 118.710;
Cl: 35.45;
H: 1.008;
O: 15.999.

How many moles of SnCl₂ formula units in this sample?

The first mass 4.90 grams contain both the SnCl₂ formula units and a number of water molecules. Luckily, the mass of the dehydrated salt 4.10 grams contains only SnCl₂.

Formula mass of tin (II) chloride SnCl₂:

$M(\rm SnCl_2) = 118.710 + 2\times 35.45 = 189.610\; g\cdot mol^{-1}$ .

Number of moles of tin (II) chloride SnCl₂ formula units in this sample:

$\displaystyle n(\mathrm{SnCl_2}) = \frac{m}{M} = \rm \frac{4.10\; g}{189.610\; g\cdot mol^{-1}} = 0.0216233\; mol$ .

How many moles of water molecules H₂O in this sample?

Water of crystallization exist as H₂O molecules in typical hydrated salts. The molar mass of these molecules will be:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g\cdot mol^{-1}$ .

The mass of water in the hydrated salt is the same as the mass that is lost when the water molecules are removed and the salt is dehydrated.

In other words,

$\begin{aligned}m(\text{Water of Hydration})&=m(\text{Hydrated Sample}) - m(\text{Anhydrous Sample}) \\ & = \rm 4.90\; g - 4.10\; g \\ &= \rm 0.80\; g\end{aligned}$ .

$\displaystyle n(\mathrm{H_2O}) = \frac{m}{M} = \rm \frac{0.80\; g}{18.015\; g\cdot mol^{-1}} = 0.0444074\; mol$ .

What's the coefficient in front of water in the formula of this hydrated salt? In other words, how many water molecules are there in the compound for each SnCl₂ formula unit?

$\displaystyle \frac{n(\mathrm{H_2O})}{n(\mathrm{SnCl_2})} = 2.05 \approx 2$ .

There are approximately two water molecules for each SnCl₂ formula unit. The formula of this compound shall thus be $\rm SnCl_2 \cdot 2H_2 O$ .

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3 years ago

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Ostrovityanka [42]

Static electricity jrbfbfjfbfbbfbfjfngbffjfbfbffjnfnfnf

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3 years ago

When 45 g of an alloy, at 25°C, are dropped into 100.0g of water, the alloy absorbs 956J of heat. If the temperature of the allo

Novosadov [1.4K]

You can use this formula to help:

$c = \frac{q}{m \: \times \: change \: in \: t}$

Where:

C = specific heat

q = heat

m = mass

t = temperature

What we know:

C = unknown

q = 956 J

m = 45 g

change in t = 12°C because 37°C - 25°C = 12°C

Plug known values into the formula:

C = 956 J / (45 g) (12°C) and we are left with a specific heat of 1.77J/g°C

Now, convert Joules to calories and then you get:

Answer: A. 0.423 cal/g°C

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3 years ago

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ryzh [129]

Weather satellites and Doppler radar, helps the process of looking over a large area, an does the network of weather observations.

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3 years ago

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Question 6 The mineral barite, (BaSO.) has a ke of 1.1 x 10" at 25°C. Calculate the solubility of barium sulfate in water, in: 6

Sav [38]

Explanation:

(6.1). The reaction equation will be as follows.

$BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO^{2-}_{4}(aq)$

Assuming the value of $K_{sp}$ as $1.1 \times 10^{-10}$ and let the solubility of each specie involved in this reaction is "s". The expression for $K_{sp}$ will be as follows.