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If a particular ore contains 58.6 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosp

rjkz [21]

Answer is: mass of the ore is 8.54kg.<span>

</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.

5 0

3 years ago

If the solubility of a N30 gas is 2.26g/l at 1.26atm of pressure, what is the solubility of

natali 33 [55]

The answer for your question is 0.27g/L

5 0

3 years ago

A gas has a pressure of 8.5atm and occupies 24L at 25∘C. What volume (in liters) will the gas occupy if the pressure is increase

Vesna [10]

The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L

From the question given above, the following data were obtained:

Initial pressure (P₁) = 8.5 atm

Initial volume (V₁) = 24 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

Final pressure (P₂) = 13.5 atm

Final temperature (T₂) = 15 °C = 15 + 273 = 288 K

<h3>Final volume (V₂) =? </h3>

The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298} = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\$

Cross multiply

298 × 13.5 × V₂ = 204 × 288

4023 × V₂ = 58752

Divide both side by 4023

$V_{2} = \frac{58752}{4023}\\\\$

Therefore, the final volume of the gas is 15 L

Learn more: brainly.com/question/25547148

3 0

2 years ago

Suppose there was a release of 1 mole of Alpha emission particle and 1 mole of Beta emission particles and both particles are ac

Vitek1552 [10]

Answer:

DT the best and they I'm on to work use as some help new thread so this communication contains you would appreciate everything about getting started using some new version from here or email the day car rental house has come home today was automatically and your company based website has anyone that a big hugs bhi to

Explanation:

DT the best and they I'm on to work use as some help new thread so this communication contains you would appreciate everything about getting started using some new version from here or email the day car rental house has come home today was automatically and your company based website has anyone that a big hugs bhi to

6 0

3 years ago

what is the pH of a solution prepared from solid, neutral 2-nitrophenol providing a fromal concentration of 0.0353M, given that

REY [17]

Answer:

pH = 4.34

Explanation:

pH= -1/2(logKa) -1/2(log C)

= -1/2( log 5.98*10^-8) -1/2(log 0.0353)

=-1/2(-7.22)-1/2(-1.45)

=3.61+0.725= 4.34

7 0

3 years ago