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JulsSmile [24]
3 years ago
12

Solve for v in the formula d = v + at

Chemistry
2 answers:
lyudmila [28]3 years ago
7 0

Answer:

d = v + at \\ d - at = v \:  \: or \:  \: v = d - at

Anna [14]3 years ago
6 0

Answer: v=d−at

Explanation: I don't know if this will help

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what is the anode material used in the extraction of aluminium by electrolysis and why it needs to be changed at interval​
pishuonlain [190]

Answer:

the anode material is graphite(replaceable graphite rods).

The oxygen gas burns away the anode as Carbon dioxide therefore the anode must be replaced continuously.

3 0
3 years ago
Which of the following is the end product of a natural<br> radioactive series?
Rainbow [258]

Answer:

Lead

Explanation:

Lead (Pb) is the end product of every natural radioactive series.

8 0
3 years ago
A neutral atom with the electron configuration 2-8-6 would most likely form a bond with an atom having the configuration
Fittoniya [83]

Answer:

The configuration of the atom would be 2-8-2.

Explanation:

Any atom of an element combines with other element to complete its octet and become stable.

The electron configuration of the given atom is 2-8-6. That means the atom has 6 electrons in its outermost shell. To become stable the atom should have 8 electrons in its outermost shell. The given atom has 6 electrons so it either lose 6 electrons or gain 2 electrons to complete its octet.

But we know the atom having 5,6,7 electrons in its outermost shell they do not lose, they gain either 3 or 2 or 1 electrons to complete its octet.

So we say that atom with the electron configuration 2-8-6 bond with the atom having electron configuration 2-8-2.

8 0
3 years ago
What mass of carbon dioxide is produced from the complete combustion of 6.00×10−3 g of methane?
Crazy boy [7]

The answer for the following problem is explained below.

  • <u><em>Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.</em></u>

Explanation:

Given:

mass of methane = 6.00 × 10^-3 grams

CH_{4} + O_{2} → CO_{2} + H_{2}O

Firstly balance the following equation:

Before balancing the equation:

CH_{4} + O_{2} → CO_{2}  + H_{2} O

After balancing the equation:

CH_{4}  + 2O_{2} → CO_{2} + 2 H_{2} O

where;

CH_{4}  represents methane molecule

O_{2} represents oxygen molecule

CO_{2}  represents carbon dioxide molecule

H_{2}O  represents water molecule

 

       CH_{4} +2 O_{2} → CO_{2} + 2H_{2}O

        16  grams of methane       →        44 grams of carbon dioxide

        6 × 10^-3 grams of methane →          ?

                 = \frac{44*6.00*10^{-3} }{16}

          = 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.

<u><em>Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.</em></u>

5 0
3 years ago
Caffeine belongs to a large class of nitrogen-containing natural products called ____________.
Sati [7]

Answer:

I Dont know sorry.........

8 0
3 years ago
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