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3 years ago

When copper (ii) carbonate is heated if forms copper (ii) oxide and carbon dioxide?

Chemistry

1 answer:

Marina CMI [18]3 years ago

4 0

Is this the answer you are looking for CuCO3=> CuO + CO2.

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A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

2 years ago

3. Which of the following most likely describes an isotope? *

Nimfa-mama [501]

I think the answer might be B but i’m not positive

5 0

2 years ago

The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p

earnstyle [38]

Answer:

The frequency is $f = 2,01 * 10^{15} \ Hz$

Explanation:

From the question we are told that

The energy required to ionize boron is $E_b = 801 KJ/mol$

Generally the ionization energy of boron pre atom is mathematically represented as

$E_a = \frac{E_b}{N_A}$

Here $N_A$ is the Avogadro's constant with value $N_A = 6.022*10^{23}$

$E_a = \frac{801}{6.022*10^{23}}$

=> $E_a = 1.330 *10^{-18} \ J/atom$

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

$E = hf = E_a$

=> $hf = E_a$

Here h is the Planks constant with value $h = 6.626 *10^{-34}$

$f = \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}$

=> $f = 2,01 * 10^{15} \ Hz$

8 0

2 years ago

When 100 mL of 1.0 M Na3PO4 is mixed with 100 mL of 1.0 M AgNO3, a yellow precipitate forms and [Ag ] becomes negligibly small.

lana [24]

Answer:

$Na3PO4 + 3AgNO3 -------> Ag3PO4 + 3NaNO3$

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M

The answer is A) PO43- < NO3- < Na+

Explanation:

Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-

We have 100mL of each reactant with the same concentration for both (1.0 M) so:

(0.1)(1)(3)= 0.3 mol Na+

(0.1)(1)= 0.1 mol NO3-

so PO43- < NO3- < Na+

5 0

3 years ago

Convert 112°C to Kelvin.

podryga [215]

112°C is 385.15 Kelvin

5 0

2 years ago

Read 2 more answers