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alisha [4.7K]
3 years ago
6

I’m about to die of cuteness overload Naruto, sasuke, sakura, and Kakashi-sensei

Chemistry
2 answers:
Gnoma [55]3 years ago
7 0

Answer:

dfadskjfkhasdljkfasdhfaljksfh so cuttttteeee :3

;w;""'

vlabodo [156]3 years ago
6 0
I am so glad i have now seen this picture. my life is now complete
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A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal
devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0
3 years ago
Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.
aksik [14]
Molten barium chloride is separeted into two species :
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

<span>The anode is positive and the cathode is negative.</span>

7 0
3 years ago
Can someone help me with these two questions?
guapka [62]

Answer:

no

Explanation:

I can't help

5 0
3 years ago
what is the maximum amount of moles of P2O5 that can theoretically be made from 136 g of P4 and excess oxygen
zhuklara [117]
We are given with
136 g P4
excess oxygen

The complete combustion reaction is
P4 + 5O2 => 2P2O5
Converting the amount of P4 to moles
136/123.9 = 1.098 moles

Using stoichiometry
moles P2O5 = 1.098 x 2 = 2.195 moles P2O5
6 0
3 years ago
Read 2 more answers
Azithromycin suspension dosing is 10 mg/kg/d and is available in 200mg/5ml doses. how many ml would a 25kg child receive daily?
bekas [8.4K]
The answer to this question is 6.25ml

To answer this question, you need to calculate the azithromycin drug doses for this patient. The calculation would be: 25kg * 10mg/kg/d= 250mg/d

Then multiply the doses with the available drug. It would be:
250 mg/d / (200mg/5ml)= 6.25ml/d
8 0
3 years ago
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