All categories

3 years ago

What is the individual ions for ZnCO3

1 answer:

5 0

<span>ZnCO3 is an inorganic compound and it contains two parts; one is cation(positive part of compound) whereas other is anion(negative part of compound). Here, </span> $Zn ^{+2}$ <span> is cation and </span> $CO3 ^{-2}$ is anion. Hope this helps.

You might be interested in

Calculate the amount of mole of iron produced from the reaction of 15.9 grams of iron oxide.

Usimov [2.4K]

Answer:

0.19875

Explanation:

nFe2O3=0.099375

nFe=2nFe2O3=0.19875

3 0

3 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

3 years ago

The mechanical advantage of the wedge increases as it gets _____.

OLEGan [10]

Do you have Answer Choices

6 0

3 years ago

What makes an acid a strong acid in water makes a base or weak base

11Alexandr11 [23.1K]

It’s ph level on the ph scale. 1-3 is a strong acid, 4-6 is a weak acid, 7 is neutral, 8-10 is a weak base, and 10-14 is a strong base.

8 0

2 years ago

a compound has a molar mass of 129 g/mol if its empirical formula is C2H5N then what is the molecular formula

Elena-2011 [213]

Given :

A compound has a molar mass of 129 g/mol .

Empirical formula of compound is C₂H₅N .

To Find :

The molecular formula of the compound.

Solution :

Empirical mass of compound :

$M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + ( 1 \times 14 )\\\\M_e = 43\ gram/mol$

Now, n-factor is :

$n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3$

Multiplying each atom in the formula by 3 , we get :

Molecular Formula, C₆H₁₅N₃

3 0

2 years ago