All categories

4 years ago

What examples does - Making Materials - Moving Materials - Removing Waste

Chemistry

1 answer:

Sidana [21]4 years ago

7 0

Answer:

You remove waste as a gas (carbon dioxide), as a liquid (urine and sweat), and as a solid. Excretion is the process of removing wastes and excess water from the body. Recall that carbon dioxide travels through the blood and is transferred to the lungs where it is exhaled

Moving Materials include aluminum foil tape, shipping tape, box tape and duct tape. Moving labels take the guesswork out of the moving process.

You might be interested in

Phosphorus-32 has a half-life of 14.3 days. how many grams of phosphorus-32 remain after 71.5 days if you start with 4.00 grams?

almond37 [142]

Answer : 0.125 g

Explanation : If you are calculating the half life of P-32 which has half life of 14.3 days and you want to calculate how many grams are remaining after 71.3 days if yo are starting with 4 g?

starting you can calculate the number of half life P-32 undergoes first,

which is n = 71.5 / 14.3 = 5 days.

now, to calculate the remaining mass,

$m_{remaining} = (\frac{initial mass}{2 X 2X 2X2 X2 } ) = (\frac{initial mass}{ 2^{5} } )$

= $= \frac{4 g }{ 2^{5} } = 0.125 g$

7 0

3 years ago

ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 grams of salt is dissolved.

SSSSS [86.1K]

Answer:

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

Explanation:

Given:

Tank A contain $V_{1} = 150$ lit

Rate $\alpha = 5 \frac{L}{min}$

Dissolved salt $A = 10$ gm

Salt pumped in one minute is $4 \frac{L}{min}$

Salt pumped out is $\frac{5L}{150L} = \frac{1}{30}$ of initial amount added salt.

To find $A(t)$

$\frac{dA}{dt} = Rate _{in} - Rate _{out}$

$A' = 5 - \frac{A}{30}$

$A' + \frac{A}{30} = 5$

Solving above equation,

$I .F = e^{\int\limits {p} \, dt }$

$y = e^{\int\limits {\frac{1}{30} } \, dt }$

$y = e^{\frac{t}{30} }$

$(Ae^{\frac{t}{30} } )' = 5 e^{\frac{t}{30} } + c$

Integrating on both side,

$Ae^{\frac{t}{30} } = 5 \times 30 e^{\frac{t}{30} } +c$

Add $e^{-\frac{t}{30} }$ on above equation,

$A = 150 + ce^{-\frac{t}{30} }$

Here given in question, $A(t=0) = 10$

$10 =150 +c$

$c = -140$

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

$A(t) = 150-140 e^{-\frac{t}{30} }$

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

7 0

3 years ago

I NEED ASAP PLEASE!!

miss Akunina [59]

Answer: freezing and evaporation for sure

Explanation:

7 0

4 years ago

A saline solution contains 72.3 grams of NaCl in 500.0 mL if water. What is the concentration (Molarity) of this solution?

valina [46]

Answer:

2.47 M

Explanation:

Molarity is (mol of solute)/(L of solvent). To find molarity, convert grams of NaCl to moles of NaCl and milliliters of water to liters of water.

72.3 g ÷ 58.44 g/mol = 1.2372... mol

500 mL ÷ 1000 mL = 0.5 L

Molarity = 1.2372/0.5 = 2.47 M

7 0

3 years ago

How does convection occur in the troposhere

adell [148]

Answer:The surface heated air expands as it warms, becomes less dense than surrounding cooler air and rises as buoyant and turbulent bubbles. This is convection and is the main process by which the troposphere mixes and heats. Although convection stirs and mixes the troposphere, the higher it is the colder it becomes.

Explanation: Read this and you'll get your answer~! I hope i helped you~! Have an GREAT day too~! <\3

7 0

4 years ago