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3 years ago

Identify each of the following compounds as binary or ternary. Then write the chemical formula for each ionic compound.

1 answer:

7 0

Binary compounds have 2 different elements, and ternary compounds have 3
Nickel(III)oxide: binary, Ni2O3
Copper (II)iodide: binary, CuI2
Tin(IV) nitride: binary, Sn3N4
Chromium (II)bromide: binary, CrBr2
<span>Iron(III)phosphide: binary, FeP</span>

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How many moles of water are in 24.3 grams of water?

shtirl [24]

Answer:

Explanation: n=m/M(molar mass)

n=24.3 grams/(16+2x1.008)grams/moles(molar mass of H2O)

n=24.3grams/18.016grams/moles

n=1.35moles

8 0

3 years ago

Read 2 more answers

One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will

saw5 [17]

Explanation:

The given data is as follows.

Boiling point of water ( $T^{o}_{b}) = 100^{o}C$ = (100 + 273) K = 323 K,

Boiling point of solution ( $T_{b}) = 101.24^{o}C$ = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

$\Delta T_{b} = (T_{b} - T^{o}_{b})$

= 374.24 K - 323 K

= 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

Let molar mass of the solute is x grams.

Therefore, Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

m = $\frac{288 g \times 1000}{x g \times 90}$

= $\frac{3200}{x}$

As, $\Delta T_{b} = k_{b} \times molality$

$1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}$

x = $\frac{0.512 ^{o}C/m \times 3200}{1.24}$

= 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is $C_{n}H_{2n}O_{n}$ .

As, its empirical formula is $CH_{2}O$ and mass is 30 g/mol. Hence, calculate the value of n as follows.

n = $\frac{\text{Molecular mass}}{\text{Empirical mass}}$

= $\frac{1321.29 g}{30 g/mol}$

= 44 mol

Thus, we can conclude that the formula of given material is $C_{44}H_{88}O_{44}$ .

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3 years ago

A trophic level is a number that tells you the position an organism occupies in a food chain or food pyramid.

svet-max [94.6K]

Answer:

true

Explanation:

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3 years ago

This element is a metalloid of Group IV. It has ____ valence electrons on the third energy level. Its name is ____, symbol____.

babymother [125]

Answer: 4 valence electrons, Silicon, Si, 14, 28.0855, 14

Explanation:

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3 years ago

Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

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1 year ago