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Anestetic [448]
3 years ago
10

HELP PLEASEEEEEEEE HELPPPPPPP (IMAGE)

Chemistry
1 answer:
lapo4ka [179]3 years ago
5 0
1. Liquid
2. Gas
3. Solid
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Give the % composition for each element found in tin (IV) nitride.
drek231 [11]

32.3699% Tin (Sn)

15.2774% Nitrogen (N)

52.3527% Oxygen (O)

4 0
3 years ago
What color are the stars on the diagram with the lowest surface temperatures?
Luba_88 [7]

Answer:

red supergiants is the answer

3 0
2 years ago
Luminosity is the amount of energy emitted by a star each second. Stars radiate light over a broad range of frequencies in the e
lapo4ka [179]
Luminosity is the amount of energy emitted by a star each second. Stars radiate light over a broad range of frequencies in the electromagnetic spectrum, <span>from the low energy radio waves to high energy gamma rays. Energy emitted from stars is the result of fusion of gases within its core</span>
7 0
4 years ago
A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N
vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

                 2NH₃(g)   ⇄     N₂(g)     +     3H₂(g)

Initially       0.0733

React         0.0733α          α/2                3/2α

Eq     0.0733 - 0.0733α    α/2                0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

3 0
3 years ago
what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio
koban [17]

Answer:

Cu~+~PtCl_2->Pt~+~CuCl_2

Explanation:

In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case Pt^+^2) and the anion is the atom at the right of the name (in this case Cl^-). With this in mind, the <u>formula would be</u> PtCl_2.

Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u> Cu. So, we have as reagents:

Cu~+~PtCl_2->

The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:

A~+~BC->AB~+~C

With this in mind, the reaction would be:

Cu~+~PtCl_2->Pt~+~CuCl_2

I hope it helps!

5 0
3 years ago
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