1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Margarita [4]
3 years ago
11

The valences of metal x,y and z are 1,2 and 3 respectively. What are the formulae of their;a) hydroxides, b) sulphates, c) hydro

gen, d) carbonates, e) nitrates, f) phosphates
Chemistry
1 answer:
Rina8888 [55]3 years ago
6 0

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

1. Hydroxides.

An hydroxide is formed when an oxyde of a metal reacts with water. When this happens, the general molecular formula is:

Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

Hope this helps

You might be interested in
Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol
Sonbull [250]

<u>Answer:</u> The value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

                N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>            1.30       1.65

<u>At eqllm:</u>       1.30-x    1.65-3x             2x

Evaluating the value of 'x'

\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol

The expression of K_c for above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

Equilibrium moles of nitrogen gas = (1.30-x)=(1.30-0.05)=1.25mol

Equilibrium moles of hydrogen gas = (1.65-x)=(1.65-0.05)=1.60mol

Putting values in above expression, we get:

K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}

Calculating the K_c' for the given chemical equation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2

Hence, the value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

8 0
3 years ago
Read 2 more answers
Determine the Density of a mass = 2000g and volume = 80ml
sergey [27]

Answer:

Explanation:

D = m/v

D = 2000/80 = 25

5 0
2 years ago
You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt
Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:

pKa_1=-Log[6.9x10^-^3]=2.16

pKa_2=-Log[6.2x10^-^8]=7.20

pKa_3=-Log[4.8x10^-^13]=12.31

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which  H_2PO_4^-^1 is the <u>acid</u> and HPO_4^-^2 is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.

8 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500=193000C is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

14\times 1000g=14000g of copper is deposited by =\frac{193000}{63.5}\times 14000=42551181 C

To calculate the time required, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

40.0=\frac{42551181 C}{t}\\\\t=1063779sec

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, 1063779s\times \frac{1hr}{3600s}=295hr

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0
3 years ago
Pls help Would be much appreciated:)
Pie

Answer:

Ok so,  b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...

i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced

ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.

iii. Calculate the standard potential (voltage) of the cell

Look up the reduction potential,

E

⁰

red

, for the reduction half-reaction in a table of reduction potentials

Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,

E

⁰

ox

=

-

E

⁰

red

.

iv. What kind of electrochemical cell is this? Explain your answer.

All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells

1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery

Explanation:

3 0
2 years ago
Other questions:
  • Under which circumstance might a gas decrease in volume when heated? A)The gas is held constant at STP. B)The gas remains under
    7·1 answer
  • In an experiment, an unknown gas effuses at one-half the speed of oxygen gas, which has a molar mass of 32 g/mol. which might be
    5·1 answer
  • Zinc is more active than cobalt and iron but less active than aluminum. Cobalt is more active than nickel but less active than i
    9·1 answer
  • Starting with benzene as the only aromatic compound
    8·1 answer
  • Help asap!!!!!!!!<br><br> How much net force is required to accelerate a 50.00 kg mass at 4.00 m/s??
    13·1 answer
  • In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How
    14·1 answer
  • What is back titration?​
    15·1 answer
  • How many moles of lithium are in 18.2 grams of lithium
    9·2 answers
  • El siguiente bloque de masa 2 kg acelera a razón<br>de 4 m/s, calcula el módulo de la fuerza F​
    10·1 answer
  • How to change τhe color of silver mercury to red???
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!