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alex41 [277]
3 years ago
5

How many sig figs are in the measurment 10010m

Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
3 0

Answer:

5

Explanation:

They all matter unless it is a decimal.

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How many Dalmatian puppies are in Pongo and Perdita’s litter?
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 Litter is the<span> birth of multiple offspring at one time. </span>The number of puppies in a litter often varies with the size of the dog, bigger dogs give birth to more puppies than smaller dogs. Pongo and Perdita are the parents in the story of 101<span>Dalmatians. There are 15 puppies in Pongo and Perdita's litter. These dogs are later kidnapped and Pongo and Perdita start searching them.</span>
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3 years ago
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IS THISS RIGHTTTTT???
shtirl [24]

Answer: No

Explanation: For it to be a divergent boundary, the arrows would have to be pointing in opposite directions. (one points left, one points right).

3 0
3 years ago
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What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso
Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247

The pH of the buffer can be known as

pH = pK_{a} + log[\frac{[A-]}{[HA]}}]

The concentration of [A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\

Similarly, the concentration of [HA] = \frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236

So, the pH of the buffer is 6.1236.

5 0
4 years ago
What is the limiting factor in determining the accumulation of siliceous ooze/calcareous ooze, respectively?
zhannawk [14.2K]

Answer:

productivity and water depth

Explanation:

The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.  

The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.

Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.

8 0
3 years ago
Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the sys
8090 [49]

Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}.

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both [\mathrm{SO_2\, (g)}] and [\mathrm{O_2\, (g)}] will increase if the pressure is increased through compression. However, because \rm SO_2\, (g) and \rm O_2\, (g) have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient Q.

As a result, the increase in pressure will have no impact on the value of Q\!. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.

8 0
3 years ago
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