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3 years ago

Can someone help me with this crossword?

Chemistry

1 answer:

erastovalidia [21]3 years ago

8 0

All of this involdes cell of different types

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How does volvox differ from other protists

Dovator [93]

It differs because it is a unicellular plant. Many scientists claim that Volvoxes are protists while many claim that they are plants, and protists are not plants.

7 0

3 years ago

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

7 0

3 years ago

how do I do this I’m confused

Naily [24]

by using pro.gram..

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${\huge{\orange{\underline{\underline{﹌﹌﹌﹌﹌﹌﹌﹌﹌♪}}}}}$

${\huge{\purple{\underline{\underline{✎﹏﹏﹏﹏﹏﹏﹏﹏﹏﹏}}}}}$

4 0

3 years ago

216 J of energy is required to raise the temperature of a piece of aluminum from 15.0º C to to 35º C.

ddd [48]

Answer: 12g

Explanation:

The amount of energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = 216 joules

Mass of aluminium = ? (let unknown value be Z)

C = 0.90 JºC-1g-1

Φ = (Final temperature - Initial temperature)

= 35°C - 15°C = 20°C

Then, Q = MCΦ

216 J = Z x 0.90 JºC-1g-1 x 20°C

216 J = Z x 18 J°g-1

Z = (216J/18 J°g-1)

Z = 12g

Thus, the mass of the aluminium is 12grams

8 0

3 years ago

What is the value for (delta)G at 1000 K if (delta)H = -220 kJ/mol and (delta)S = -0.05 kJ/(molK)?

tankabanditka [31]

The system is isothermal, so we use the formula:
(delta)G = (delta)H - T (delta) S

Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ

6 0

3 years ago

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