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Snowcat [4.5K]
3 years ago
13

Describe the ph changes expected if an acid is used to neutralize a base with ph12

Chemistry
1 answer:
ad-work [718]3 years ago
4 0
It becomes neutral because acid neutralises the alkaline base
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How much of the sulfur burnt is normally oxidized to hexavalent state?
jeyben [28]

I feel that it is D but I am not 100% sure sadly.

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what is always true of a combustion reaction? oxygen gas must be one of the reactants carbon monoxide will be produced the react
cestrela7 [59]
The correct answer should be
<span>carbon monoxide will be produced

Combustions release heat, they don't absorb it. Oxygen is not necessary as other gasses can also do it, while chlorine is not necessary.</span>
6 0
3 years ago
When compared to the standard hydrogen electrode, zinc has a reduction potential of -0.762 volts and copper a reduction potentia
Vladimir [108]

Answer:

When compared to the standard hydrogen electrode, zinc has a reduction potential of -0.762 volts and copper a reduction potential of + 0.342 volts. What is the reduction potential when zinc and copper are connected to one another? Standard Santand Inc ар -40 INC IM IM Z IM ಅವಳ ದdಂಗpa volts 0.420- volts 0.420- volts 0.420- volts 1.104+

Explanation:

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4 0
3 years ago
In the reaction 2co(g) + o2(g) â 2co2(g), what is the ratio of moles of oxygen used to moles of co2 produced? select one:
Bess [88]
<span>2CO(g)           +          1O2(g) --->          2CO2(g)
2 mol CO                1 mol O2             2 mol CO2

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7 0
3 years ago
An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic fi
arsen [322]

Explanation:

In a magnetic field, the radius of the charged particle is as follows.

             r = \frac{mv}{qB}

where,   m = mass,      v = velocity

              q = charge,    B = magnetic field

Therefore, q will be calculated as follows.

         q = \frac{mv}{rB}

            = \frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}

            = \frac{34.98 \times 10^{-22}}{0.01092}

            = 3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}

            = +2e

Thus, we can conclude that the charge of the ionized atom is +2e.

5 0
3 years ago
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