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notsponge [240]
1 year ago
5

What is cation and anion for CrS2

Chemistry
1 answer:
maksim [4K]1 year ago
4 0
Cation - Cr 4+
Anion - S ^ 2-

Chromium sulfide
(:
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Hey I'm a newbie any help please with this
solniwko [45]
6 A
7 A
8 B
9 B
10 A
11 C
4 0
3 years ago
I need help with this assignment
yuradex [85]

The answers for the following sums is given below.

1.Pd_{2}H_{2}

2.C_{2} H_{6}

3.C_{2} H_{2} O_{2} Cl_{2}

4.C_{3}Cl_{3} N_{3}

5.Tl_{2 } C_{4} H_{4}O_{6}

6.C_{8}H_{8}

7. N_{2}O_{5}

8.P_{4}O_{6}

9.C_{4}H_{8}  O_{2}

Explanation:

1.Given:

        Molar mass=216.8g

Molecular formula=PdH_{2}

       we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of PdH_{2}:

Pd=106.4×1=106.4u

H=1×2=2

molar mass of PdH_{2}=106.4+2=108.4

n=\frac{216.8}{106.4}

<u><em>n=2</em></u>

Molecular formula=2(PdH_{2})

Molecular formula=Pd_{2}H_{2}

Therefore the molecular formula of the compound is Pd_{2}H_{2}

2. Given:

        Molar mass=30.0g

Molecular formula=CH_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of CH_{3}:

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of CH_{3}=12.01 + 3 =15.01u

n=\frac{30.0}{15.01}

<u><em>n=2</em></u>

Molecular formula=2(CH_{3})

Molecular formula=C_{2} H_{6}

Therefore the molecular formula of the compound is C_{2} H_{6}

3. Given:

        Molar mass=129g

Molecular formula=CHOCl

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n=\frac{129}{64.5}

<u><em>n=2</em></u>

Molecular formula=2(CHOCl)

Molecular formula=C_{2} H_{2} O_{2} Cl_{2}

Therefore the molecular formula of the compound is C_{2} H_{2} O_{2} Cl_{2}

5. Given:

        Molar mass=577g

Molecular formula=TlC_{2} H_{2}O_{3}

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of TlC_{2} H_{2}O_{3} :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of TlC_{2} H_{2}O_{3}=204.3+24.02+1+48.00=278.32u

n=\frac{577}{278.32}

<u><em>n=2</em></u>

Molecular formula=2 (TlC_{2} H_{2}O_{3})

Molecular formula=Tl_{2 } C_{4} H_{4}O_{6}

Therefore the molecular formula of the compound is Tl_{2 } C_{4} H_{4}O_{6}

4. Molar mass=184.5g

Molecular formula=CClN

       we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n=\frac{184.5}{61.5}

<u><em>n=3</em></u>

Molecular formula=3 (CClN)

Molecular formula=C_{3}Cl_{3} N_{3}

Therefore the molecular formula of the compound is C_{3}Cl_{3} N_{3}

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula=C_{8}H_{8}

Molecular formula =C_{8}H_{8}

Molar mass of C_{8}H_{8}:

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of C_{8}H_{8}=96.08+8=104.08u

n=104.08÷78.0

<em><u>n=1</u></em>

Molecular formula = n(Empirical formula)

Molecular formula = 1(C_{8}H_{8})

Molecular formula =C_{8}H_{8}

Therefore the molecular formula of a compound is C_{8}H_{8}

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen=\frac{4.04}{14.01} = 0.289 moles

moles of oxygen=\frac{11.46}{15.999} =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula  of N_{2} O{5}.

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is N_{2}O_{5}.

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula=P_{2} O_{3}

molecular formula= 2(Empirical formula)

Molecular formula =2(P_{2} O_{3})

Molecular formula =P_{4}O_{6}

Therefore the molecular formula of the compound is P_{4}O_{6}

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula =C_{2} H_{4} O

Molecular formula = 2(Empirical formula)

Molecular formula =2(C_{2} H_{4} O)

Molecular formula =C_{4}H_{8}  O_{2}

Therefore the molecular formula of the compound is C_{4}H_{8}  O_{2}

4 0
3 years ago
Fill in the left side of this equilibrium constant equation for the reaction of benzoic acid (HC_6H_5CO_2) with water.
il63 [147K]

Answer:

[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka

Explanation:

The reaction of dissociation of the benzoic acid in water is given by the following equation:

C₆H₅-COOH + H₂O  ⇄  C₆H₅-COO⁻  +  H₃O⁺    (1)

The dissociation constant of an acid is the measure of the strength of an acid:

HA ⇄ A⁻ + H⁺        (2)

K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}      (3)

<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA].     </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}

I hope it helps you!          

6 0
3 years ago
Aee all sets of political views correct? i need long answer​
VMariaS [17]

Answer:

LOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOONGGGGGGGGGGGGGGGGGGGGGGGGGGG

Explanation:

8 0
3 years ago
Farmers who raise cotton once used arsenic acid, H₃AsO₄, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with
Stels [109]

Explanation:

The reaction equation will be as follows.

      H_{3}AsO_{4} \rightleftharpoons H_{2}AsO^{-}_{4} + H^{+}

Hence, the expression for K_{a} is as follows.

            K_{a} = \frac{[H_{2}SO^{-}_{4}][H^{+}]}{[H_{3}AsO_{4}]}

Let us assume that the concentration of both [H_{2}AsO^{-}_{4}] and [H^{+}] is x.

           2.5 \times 10^{-4} = \frac{x \times x}{0.5}

                          x = 0.01118034

This means that the concentration of [H^{+}] is 0.01118034.

Since, we know that the relation between pH and concentration of hydrogen ions is as follows.

              pH = -log [H^{+}]

                    = -log (0.01118034)

                    = 1.958

Thus, we can conclude that the pH of a 0.500 M solution of arsenic acid is 1.958.

6 0
3 years ago
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