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Katyanochek1 [597]
3 years ago
7

- prediction

Chemistry
1 answer:
kolezko [41]3 years ago
8 0

Answer:

theory

Explanation:

Fits all criteria

- Hope that helps! Please let me know if you need further explanation.

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A ninja motorbike travelling at 55 m/s takes 10 sec to come to rest . What is its deceleration rate ?
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Speed=Distance

Time

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Abdid is an astronomer who has been observing objects that orbit the Sun in the asteroid belt. He finds a previously undiscovere
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Since it is round and it is the asteroid belt it is most likely that he found a dwarf planet 
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how many H+ ions are in a solution with a pH of 2 and how many H+ ions are in a solution with a pH of 6?
LiRa [457]

Answer:

i) pH = 2

pH = -log(H+)

:- (H+) = 10^(-2)

:- (H+) = 0.01 M

ii) pH = 6

pH = -log(H+)

:- (H+) = 10^(-6)

:- (H+) = 0.000001 M

Explanation:

By definition: pH = -log(H+).

Given your pH, solve for the H+ using the the following log rule:

if a = (+/-) log (b) then

b = 10^((+/-) a).

Also remember unit of concentration is molar (M)

3 0
3 years ago
Describe the heat transfer that occurs as ocean currents move within their gyres.
Aleks04 [339]

Answer:

Ocean currents!!

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3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0
3 years ago
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