Equations on chemistry??

Vanadium (IV) carbonate

Anna71 [15]

Answer:

Explanation:

1) Vanadium (IV) → V⁺⁴

Carbonate → CO₃⁻²

So , Vanadium (IV) Carbonate = V₂(CO₃)₄ or V(CO₃)₂

2) Tin (II) = Sn⁺²

Nitrite = NO₂⁻

So, Tin (II) Nitrate = Sn(NO₂)₂

3) Cobalt (III) = Co⁺³

Oxide = O⁻²

So , Cobalt (III) Oxide = Co₂O₃

4) Titanium (II) = Tn⁺²

Acetate = CH₃COO⁻

So , Titanium (II) Acetate = Tn(CH₃COO)₂ or Tn(C₂H₃O₂)₂

5) Vanadium (V) = V⁺⁵

Sulfide = S⁻²

So , Vanadium (V) Sulfide = V₂S₅

6) Chromium (III) = Cr⁺³

Hydroxide = OH⁻

So , Chromium (III) Hydroxide = Cr(OH)₃

7) Lithium = Li⁺

Iodide = I⁻

So , Lithium Iodide = LiI

8) Lead (II) = Pb⁺²

Nitride = N⁻³

So , Lead (II) Nitride = Pb₃N₂

9) Silver = Ag⁺

Bromide = Br⁻

So , Silver Bromide = AgBr

5 0

3 years ago

How many atoms are present in 4.0 Mol of sodium

LekaFEV [45]

1 mole = 6.02 x 10^23 atoms (of any element)

So we are going to take our known value of 4 moles, and multiply by 6.02 x 10^23 (Avogadro's number) and we will get the number of atoms that are in 4 moles.

(4.0 moles of Na) x (6.02 x 10^23) / (1 mole) = 2.4 x 10^24 atoms of Na
There are 2.4 x 10^24 atoms of Na in 4.0 moles.

8 0

3 years ago

What do you measure in an experiment to determine reaction rate?

Dovator [93]

D. Time
Just took the test

6 0

3 years ago

Read 2 more answers

UGRENT! Please help showing all work

Bess [88]

Answer:

wait I AM TRYING..................

this is limiting reactant

6 0

3 years ago

An iron ball is dropped onto a pavement from a height H. If half of the heat generated goes into warming the ball, find the temp

mr Goodwill [35]

The given question is incomplete. The complete question is as follows.

A 10 kg iron ball is dropped onto a pavement from a height of 100 m. If half of the heat generated goes into warming the ball,find the temperature increase of the ball. (In Sl units, the specific heat capacity of iron is 450 J/kg.C degree).

Explanation:

In the given case, when the ball comes in contact with the pavement then its kinetic energy will be equal to potential energy of the ball before it was dropped. Half the K.E heats the ball.

As K.E = P.E = mgh

So, mgh = $10 \times 10 \times 100$ = 10,000 joules