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3 years ago

14

What is the correct name for the ionic compound Ca2P5

2 answers:

maks197457 [2]3 years ago

7 0

Answer:

It is calcium phosphide.

Explanation:

Calcium phosphide is a salt like material that is composed of calcium and phosphorus in the ratio of 2:5. It has the appearance of red brown Crystalline salt.

It is use as rodenticide to kill rats.

It is also use in fire works.

Studentka2010 [4]3 years ago

3 0

Answer:

Calcium phosphorous

Explanation:

First, this compound is a mixture of the following elements:

Ca: Calcium

P: Phosphor

Now, both of these elements are united as a binary compound, the metal and a non metal. When we have a binary compound with a metal and non metal, (When the non metal is not hydrogen or oxygen) this usually becomes a salt. A bynary salt. So a binary salt, we should name like this

In this, we should name first the metal and then the non metal with the sufije ous. In this case:

Calcium phosphorous.

Then the name of this salt is that.

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Covalent compounds occur------

irga5000 [103]

Answer:

Only between nonmetals

Explanation:

Covalent compounds are compounds that are formed as a result of covalent bonding between nonmetal atoms.
Covalent compounds occur only between non metals, where atoms of the nonmetals involved join to form a covalent bond as a result of sharing of electrons.
<u><em>Electrons shared during the formation of covalent bond, may be from one atom involved or each atom contributing equally to the bond formation.</em></u>
An example of covalent compounds include, phosphorus (v) oxide, methane, water, CO2,etc.

4 0

3 years ago

Q8. The titration of 15.00 mL of HBr solution of unknown concentration requires 18.44 mL of a 0.100 M KOH solution to reach the

lara [203]

Answer: 0.123 M

Explanation:

According to the neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HBr$ solution = ?

$V_1$ = volume of $HBr$ solution = 15.00 ml

$M_2$ = molarity of $KOH$ solution = 0.100 M

$V_2$ = volume of $KOH$ solution = 18.44 ml

$n_1$ = valency of $HBr$ = 1

$n_2$ = valency of $KOH$ = 1

$1\times M_1\times 15.00=1\times 0.100\times 18.44$

$M_1=0.123$

Therefore, the concentration of the unknown HBr solution is 0.123 M

8 0

3 years ago

Calculate the mass of 12.044*10^23atom of carbon

Keith_Richards [23]

Answer: 24 gram is a final Answer.

Explanation:1 mol of carbon weight is 12 gram then

12.044×1023 contain 2 mole carbon so

4 0

3 years ago

How many grams of ethyl alcohol (C2H5OH) are in a 1500 ml solution which is 1.54M?

Sedbober [7]

Answer:

The answer is 1.54molar mass.

7 0

3 years ago

Read 2 more answers

A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2

Ymorist [56]

Number of moles is defined as the ratio of given mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = $\frac{0.01761 g}{44.01 g/mol}$

= $0.0004001 mol$

Mass of carbon = number of moles of carbon dioxide \times molar mass of carbon

= $0.0004001 mol\times 12.011 g/mol$

= $0.004806 g$

Number of moles of water= $\frac{0.00481 g}{18 g/mol}$

= $2.672\times 10^{-4}$

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = $2\times 2.672\times 10^{-4}$

= $5.34\times 10^{-4}$

Mass of hydrogen = $5.34\times 10^{-4}\times \times 1.008 g/mol$

= $5.34\times 10^{-4} g$

Mass of oxygen = $0.001175-(5.38\times 10^{-4}g+0.004806 g)$

= $0.006405 g$

Number of moles of oxygen = $\frac{0.006405 g}{15.999 g/mol}$

= $0.000400$

Now,

$C_{0.0004001} H_{0.000534} O_{0.000400}$

Divide the smallest number to get the whole number,

$C_{\frac{0.0004001}{0.000400}} H_{\frac{0.000534}{0.000400}} O_{\frac{0.000400}{0.000400}}$

we get,

$C_{1} H_{1.33} O_{1}$

Now, multiply all the subscript by 3 to get the whole number,

$C_{3} H_{4} O_{3}$ (empirical fomula)

Molar mass of the compound = $3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol$

= $88.062 g/mol$

Divide given molar mass of the compound with the molar mass of the compound.

= $\frac{176.1 g/mol}{88.062 g/mol}$

= $1.999\simeq 2$

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

$C_{6}H_{8}O_{6}$

Hence, empirical formula is $C_{3}H_{4}O_{3}$ and molecular formula is $C_{6}H_{8}O_{6}$

8 0

3 years ago