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lukranit [14]
2 years ago
11

Los neutrones tienen carga eléctrica———-y masa de ——una

Chemistry
1 answer:
3241004551 [841]2 years ago
6 0

Answer:

thanks for your points

God bless you ALWAYS

and pa follow

at pa brainlest answer please

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By hot springs, fumaroles and geysers.
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The energy produced in the stars results from
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Explanation:

Hydrogen fusion (nuclear fusion of four protons to form a helium-4 nucleus) is the dominant process that generates energy in the cores of main-sequence stars. It is also called "hydrogen burning", which should not be confused with the chemical combustion of hydrogen in an oxidizing atmosphere.

3 0
3 years ago
In the reaction
user100 [1]

Answer:

a) KOH

Explanation:

In the given balanced reaction

                         2K + 2H2O → 2KOH + H2

In the compound KOH,

The elements are K,O, and H and in the compound, there is one mole each of K , O ,and H.

So the element ratio here is 1 : 1 : 1.

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3 years ago
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A sample of an ideal gas has a volume of 2.21 L at 282 K and 1.03 atm. Calculate the pressure when the volume is 1.84 L
PIT_PIT [208]

Answer:

1.33 atm

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rearrange and make P2 the subject then solve,it should give you 1.33 atm

3 0
3 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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