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vagabundo [1.1K]

3 years ago

5

26 Classify each chemical equation as a combination/synthesis, decomposition, single replacement, double replacement, or combust

ion.

• Pb(s) + Hg2SO4(s) PbSO4(s) +2 Hg(l)
• NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)
• 2 H20(l)2 H2(g) + 02(g)

2 answers:

eimsori [14]3 years ago

7 0

1. Single replacement
2. Double replacement
3. Decomposition

romanna [79]3 years ago

3 0

double replacement: is the answer on E.D.G

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What characteristics or characteristics contribute to the attraction between earth and the moon

OLEGan [10]

Answer:

The moon's gravity pulls at the Earth, causing predictable rises and falls in sea levels known as tides. To a much smaller extent, tides also occur in lakes, the atmosphere, and within Earth's crust. High tides are when water bulges upward, and low tides are when water drops down.

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3 years ago

Read 2 more answers

Energy in a vacuum travels by ___.<br> A. radiation<br> B. conduction<br> C. convection

mariarad [96]

I think it’s radiation, that’s what i got from looking it up

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

Isotopes of elements have different:

Morgarella [4.7K]

Isotopes of elements have same atomic numbers and different mass number(atomic masses).

7 0

2 years ago

2) 2KClO3 --> 2KCl + 3O2

aleksandrvk [35]

$2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2$

a)

$2 \text{ mols of KClO}_3 \equiv 3 \text{ mols of O}_2$

$19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5 \text{ mols of O}_2$

$\boxed{19 \text{ mols of KClO}_3 \equiv 28,5 \text{ mols of O}_2}$

b)

$2 \text{ mols of KClO}_3 \equiv 2 \text{ mols of KCl}$

$62 \text{ mol of KClO}_3 \equiv 62 \text{ mol of KCl}$

Using the atomic mass given in the periodic table:

$62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$62\cdot122,5 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$7595 \text{ g of KClO}_3 \equiv 62 \text{ mol of KCl}$

$\boxed{7,595 \text{ kg of KClO}_3 \equiv 62 \text{ mol of KCl}}$

c)

$2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3$

$3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}$

Using the atomic mass given in the periodic table:

$3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5) \text{ g of KCl}$

$96\text{ g of O}_2 \equiv 149\text{ g of KCl}$

$\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}$

$\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}$

This result is an aproximation.

8 0

2 years ago