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Montano1993 [528]

2 years ago

12

Discuss the phenomena of sunrise and sunset seen on Earth and describe how they will be different from the perspective of outer

space.

1 answer:

Monica [59]2 years ago

8 0

Answer:

its very dark and scary

Explanation:

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A reaction occurs in which hydrogen-1 and hydrogen-2 form helium-3. Is this a chemical reaction or a nuclear reaction, and how d

olga nikolaevna [1]

Cause -3 is -3sjsnsnsnsnsnnsnsnsnnsnsnsnsn

7 0

3 years ago

7. How many chlorine atoms are in a 34.2 g sample of dichlorine pentoxide?

Reptile [31]

Answer:

The answer to your question is 1.36 x 10²³ atoms

Explanation:

Data

number of atoms = ?

mass of the sample = 34.2 g

Molecule = Cl₂O₅

Process

1.- Calculate the molar mass of Cl₂O₅

Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g

2.- Calculate the atoms of Cl₂O₅

151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms

34.2 g of Cl₂O₅ ------------ x

x = (34.2 x 6.023 x 10²³) / 151

x = 1.36 x 10²³ atoms

4 0

3 years ago

Suppose an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron. Comple

AfilCa [17]

Explanation:

An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.

Symbol of an alpha particle is $^{4}_{2}\alpha$ . Whereas a neutron is represented by a symbol $^{1}_{0}n$ , that is, it has zero protons and only 1 neutron.

Therefore, reaction equation when an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron is as follows.

$^{27}_{13}Al + ^{1}_{0}n \rightarrow ^{30}_{15}P + ^{1}_{0}n$

5 0

3 years ago

What is symbol, molecule, electrovalent compound?

jok3333 [9.3K]

Answer:

the abbreviation form of full name is called symbol.

the smallest unit of cimpound is called molecule.

8 0

3 years ago

In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2

allsm [11]

Answer:

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

Explanation:

$SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)$

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Balance the charge by adding 2 electrons on product side:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-$ ....[1]

Reduction :

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)$

Balance O by adding water on required side:

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

Now, balance H by adding $H^+$ on the required side:

$SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

At last balance the charge by adding electrons on the side where positive charge is more:

$SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)$ ..[2]

Adding [1] and [2]:

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

4 0

3 years ago