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Dmitry [639]
3 years ago
11

What is the chemical equation for this?

Chemistry
1 answer:
Vlad1618 [11]3 years ago
3 0

Answer:

J

Explanation:

Chlorine Gas is Cl2 so if it isnt on the left as that its wrong.

Then on the products see that Fe III means Fe is +3 charge so we have to have 3 Cl to balance it to 0 since each cl as -1 charge.

FInally check if the equation is balanced J is maybe H but its to blurry to tell

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Which substance is an ionic solid? A. LiCl B. HCl C. Ne D. Fe
andrew11 [14]

Answer:

A. LiCl

Explanation:

Lithium chloride is an ionic compound but it also has some covalent character due to the very small size of lithium metal.

7 0
3 years ago
This is the chemical formula for talc (the main ingredient in talcum powder):
MAXImum [283]

Answer : 6.41 moles of silicon are in the sample.

Solution : Given,

Moles of magnesium = 9.62 moles

The given chemical formula for talc is, Mg_3Si_2O_{52}OH_2

From the given chemical formula we conclude that

The stoichiometry of magnesium and silicon is 3 : 2

3 moles of magnesium is equivalent to 2 moles of silicon

9.62 moles of magnesium will be equivalent to \frac{2}{3}\times 9.62=6.41 moles of silicon

Therefore, the number of silicon present in sample are 6.41 moles.

7 0
4 years ago
3.87 moles of iron bromide yielded10.4 moles of sodium bromide
Roman55 [17]

Answer: 89.57 %

<em>Hope it helps</em>

7 0
4 years ago
When of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing poi
Nesterboy [21]

The given question is incomplete. The complete question is:

When 282. g of glycine (C2H5NO2) are dissolved in 950. g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 282. g of ammonium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X.

Answer:  the van't Hoff factor for ammonium chloride is 1.74

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point  

K_f = freezing point constant = ?

i = 1 ( for non electrolyte)

m= molality

8.2^0C=1\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (glycine) = 75.07 g/mol

Mass of solute (glycine) = 282 g

8.2^0C=1\times K_f\times \frac{282g}{75.07g/mol\times 0.95kg}

K_f=2.07

ii) 20.0^0C=i\times \times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of solute (ammonium chloride) = 53.49 g/mol

Mass of solute (ammonium chloride) = 282 g

20.0^0C=i\times 2.07\times \frac{282g}{53.49g/mol\times 0.95kg}

i=1.74

Thus the van't Hoff factor for ammonium chloride is 1.74

4 0
3 years ago
AACGTACGATCGATGCACATGCATGGCTACGC
Pachacha [2.7K]

Explanation:

if u have biology A=T and G=C

id.k what ur question is supposed to mean..

7 0
3 years ago
Read 2 more answers
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