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miskamm [114]
3 years ago
14

LOOK AT PICTURE PLEASEE

Chemistry
2 answers:
daser333 [38]3 years ago
7 0

Answer:

Its B

Explanation:

ozzi3 years ago
3 0

Answer: The answer is sodium chloride.

Explanation: Since salt is the most dissolvable, sodium/substance, then that makes salt/sodium chloride, the easiest, to break down.

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What type of reaction is Mg + 2Ag -> mg2 + 2Ag
bulgar [2K]

Answer:

double replacement ...i think ^^

i hopefully its right

4 0
3 years ago
How many moles are in 12 gr of magnesium?
myrzilka [38]

Answer: 0.5 mole Mg

Explanation: solution:

12 g Mg x 1 mole Mg / 24 g Mg

= 0.5 mole Mg

5 0
4 years ago
Calculate the weight of 3.491 into 10 to the power 19 molecules of cl2​
Rom4ik [11]

Answer:

The mass of 3.491 × 10¹⁹ molecules of Cl₂  of Cl₂ is 4.11 × 10⁻³ grams

Explanation:

The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles

Therefore, we have;

One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂

3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂

The mass of one mole of Cl₂ = 70.906 g/mol

The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams

Therefore;

The mass of 3.491 × 10¹⁹ molecules of Cl₂  of Cl₂ = 4.11 × 10⁻³ grams.

4 0
3 years ago
You pimp air into your volleyball the ball mass is?
Mumz [18]

Answer: When it comes to maintaining your volleyball ball, there are some things you should get accustomed to doing. One of the most important is knowing how to inflate it. The professionals keep the psi of a ball at 0.3 to 0.325 for indoor balls and at 0.175 to 0.225 for beach, or outdoor, balls.

Explanation:

did this help?

4 0
3 years ago
Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a
NISA [10]

Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is  ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0
3 years ago
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