Anomalous data on a graph would show up as say a very high or very low value which does not fit in with the normal values which may be background values.If it was a straight line graph then the anomalous point would plot well above or below the line or if it was a bar graph ie a histogram it would be much higher or lower than the surrounding data. In mineral exploration, anomalies are looked for in say geophysics or geochemistry data values for high or low magnetism or conductivity or high chemical values indicating the presence of valuable minerals at that point.

6 0

3 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

3 years ago

Given the balanced equation representing a reaction:

sleet_krkn [62]

4Al(s) + 3O2(g) --> 2Al2O3(s) This is the balanced.
From the equation:

4 moles of Al required 3 moles of O2 to produce 2 moles of Al2O3

3 moles of O2 reacted with 4 moles of Al to produce 2 moles of Al2O3
1 mole of O2 reacted with 4/3 moles of Al to produce 2/3 moles of Al2O3 (Divide by 3)
4.5 moles of O2 reacted with (4/3 *4.5) moles of Al to produce (2/3*4.5) moles of Al2O3

4.5 moles of O2 reacted with 6moles of Al to produce 3moles of Al2O3

(3) is the answer. 6 mol of Al.

7 0

3 years ago

If Marie and Calvin dissolve 50 grams of KBr in 100 grams of water at 90oC, the solution is

hram777 [196]

Answer:

The solution is 50 %wt

Explanation:

50% wt is a sort of concentration and means, that 50 g of solute (in this case, the potassium bromide) dissolved in 100 g of water.

It is the same to say, that there are 50g of KBr for every 100g of H₂O

8 0

3 years ago