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2 years ago

8

Match the landform to its description.

2 answers:

yuradex [85]2 years ago

5 0

Answer:

cinder cone: made of pieces of lava

shield volcano: wide summit, gentle slope

volcanic soil: rich in nutrients

caldera: bowl shaped depression

lava plateau: material fills in valleys

Explanation:

it’s right

nikitadnepr [17]2 years ago

5 0

Answer:

Caldera, Bowl-Shaped depression

Shield volcano, Wide summit, Gentle slope

Lava plateau, Material fills in valleys

Volcanic soil, Rich nutrients

Cinder Cone, Made of pieces of lava

Explanation:

Got it right on edge

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3 years ago

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

2 years ago

A geological sample is found to have a Pb-206/U-238 mass ratio of 0.337/1.00. Assuming there was no Pb-206 present when the samp

azamat

Answer:

$Age=2.52*10^9 years$

Explanation:

To determine the age, we need to know how much U have become Pb, so first we have to calculate the Pb moles present in a sample:

$n_{Pb}=\frac{0.337g}{206g/mol}=0.00164 mol$

$n_{U}=\frac{1 g}{238g/mol}=0.0042 mol$

The percentage of U degradation:

$P=\frac{n_{Pb}}{n_{Pb}+n_{U}}$

$P=\frac{0.00164}{0.00164+0.0042}=0.28$

Assuming that the life time is linear:

$Age=\frac{4.5*10^9 years}{0.5 life time}*0.28 life time$

$Age=2.52*10^9 years$

4 0

3 years ago

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