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Anna71 [15]
3 years ago
11

Please help me with this it’s for my bell ringer

Chemistry
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

first one is 5

Explanation:

Reactants are on the left side. Products on the right. Each capital letter represents a new element that is being put there. Hopefully that makes sense.

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Pls help 2-3 tyvm ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎
Triss [41]

Answer:

I could I guess?

But like only 1 lol

Explanation:

6 0
3 years ago
What is the molarity of a solution containing 55.8 g of mgcl2 dissolved in 1.00 l of solution?
MakcuM [25]
The answer is 0.59 M.

Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l

So, 1 mol has 95.2 g/l.

Our solution contains 55.8g in 1 l  of solution, which is 55.8 g/l

Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
7 0
3 years ago
Read 2 more answers
Explain how a light can be seen when sent through a colloid mixture, but not a solution.
Free_Kalibri [48]
I think the answer is that is lights up?
5 0
3 years ago
Which bond will each pair of elements form?
jok3333 [9.3K]

Explanation: Sodium is a metal with a low electronegativity it will form an ionic bond with a non metal with a high electronegativity.

Both Florine and Sulfur are non metals with high electronegativity.

Neon is a noble gas family VIII A and normally doesn't form any bonds at all.

<u><em>MARK ME BRAINLIST</em></u>

7 0
3 years ago
n a gas-phase equilibrium mixture of SbCl5, SbCl3, and Cl2 at 500 K, pSbCl5 = 0.17 bar and pSbCl3 = 0.22 bar. Calculate the equi
Lorico [155]

Answer:

2.7 × 10⁻⁴ bar

Explanation:

Let's consider the following reaction at equilibrium.

SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)

The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.

Kp = pSbCl₃ × pCl₂ / pSbCl₅

pCl₂ = Kp × pSbCl₅ / pSbCl₃

pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22

pCl₂ = 2.7 × 10⁻⁴ bar

7 0
3 years ago
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