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tiny-mole [99]
3 years ago
12

The following reaction is a reaction. fission fusion

Chemistry
2 answers:
Serggg [28]3 years ago
5 0

Answer:

FUSION

Explanation:

ahrayia [7]3 years ago
3 0
Fusion is the correct answer
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What would the formula be for this model above?
Daniel [21]

Answer:

CH4

Explanation:

A= CH4

8 0
3 years ago
What is the mass, in grams, of a sample of 6.98 × 1024 atoms of magnesium (Mg)? Show your work or explain the steps that you use
CaHeK987 [17]

Hello!

* First Step: to know Avogrado's Law

We know that by the Law of Avogrado, for each mole of substance we have 6.02*10²³ atoms, if:

** Second Step: to know the molar mass of the solute

The molar mass of of magnesium = 24.30 g/mol

*** Third step: make the ratio mass / mol with atoms

1 mol we have 6.02*10²³ atoms

1 mole of Mg we have 24.30 g

Then we have:

24.30 g ------------- 6.02*10²³ atoms

x ----------------------- 6.98*10^24 atoms

\dfrac{24.30}{x} = \dfrac{6.02*10^{23}}{6.98*10^{24}}

multiply cross

6.02*10^{23}*x = 24.30*6.98*10^{24}

6.02*10^{23}\:x = 1.69614*10^{26}

x = \dfrac{1.69614*10^{26}}{6.02*10^{23}}

\boxed{\boxed{x \approx 281.75\:grams}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

7 0
3 years ago
CH4(g)+O2(g)→CO2(g)+H2O(g). What mass of water is produced from the complete combustion of 5.00×10−3 g of methane?
bixtya [17]

Answer:

1gram of water

Explanation:

First balance the equation

Ch4+2O2>CO2+2H2O

Ratio is 1:2

500×10-3what about 2?

500×10-3×2=1g

1g of water

4 0
3 years ago
Please help me in my TEST
Natali [406]

Answer:

D) Please look below at the cart to cheak your answer!

Hope this helps! mark me brainliset!

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4 0
4 years ago
In another experiment, the student titrated 50.0 mL of 0.100 M HC,H,O, with
____ [38]

Answer:

Eqv Pt pH = 8.73

Explanation:

    HOAc                   +            NaOH            =>            NaOAc              + H₂O

50ml(0.10M HOAc)  +  50ml(0.10M NaOH) => 100ml(0.05M NaOAc) + H₂O

For neutralized system, 100ml of 0.05M NaOAc remains

NaOAc => Na⁺ + OAc⁻

Na⁺ + H₂O => No Rxn

          OAc⁻  +  H₂O  => HOAc + OH⁻

C(i)   0.05M       -----        0M      0M

ΔC        -x           -----         +x        +x

C(f)    0.05-x      

       ≅ 0.05M    -----          x          x

Kb = Kw/Ka = [HOAc][OH⁻]/[OAc⁻] = 1 X 10⁻¹⁴/1.7 X 10⁻⁵ = (x)(x)/(0.05M)

=> x = [OH⁻] = SqrRt(0.05 x 10⁻¹⁴/1.7 x 10⁻⁵) = 5.42 x 10⁻⁶M

=> pOH = -log[OH⁻] = -log(5.42 x 10⁻⁶) = 5.27

pH + pOH = 14 => pH = 14 - pOH = 14 - 5.27 = 8.73 Eqv Pt pH

8 0
3 years ago
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