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Marrrta [24]
3 years ago
7

Captain Hook had a nasty encounter with a crocodile in Never-Never Land. As a result of the battle, he lost his hand to the croc

, which also swallowed an alarm clock. Fortunately for Hook, the loud ticking warned him of the hungry croc’s approach. Unfortunately for Hook, any clock’s ticking now ushers in a full-blown anxiety attack.
Chemistry
1 answer:
kati45 [8]3 years ago
4 0

Answer:

What's your Question men

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Which of these is an example of a physical change?
trasher [3.6K]
Ain't is the correct answer
8 0
4 years ago
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If the concentration of phosphate in the cytosol is 2.0 mM and the concentration of phosphate in the surrounding fluid is 0.1 mM
Allushta [10]
<span>The correct option is C. The concentration of phosphate inside the cytosol is already greater than the concentration of phosphate in the surrounding fluid, yet, the cell still want to move more phosphate into the cell. To do this, energy is needed to move the phosphate ions against the concentration gradient, so the type of transportation requires is ACTIVE TRANSPORT.</span><span />
8 0
3 years ago
30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at
Zanzabum

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

6 0
3 years ago
I/3 of 12 = 2/3 of<br>please help at 11 o'clock am it needs to be done ​
Nostrana [21]

Answer:

6

Explanation:

This question is seeking for an equivalent of one expression. The first expression says 1/3 of 12, which means 1/3 × 12 = 4. To find 2/3 of a number that will give us 4, we say

Since; 1/3 of 12 is 4

Then, 4 = 2/3 of x, where x is the number, which when multiplied by two-third (2/3) will result to 4.

4 = 2/3 × x

4 = 2x/3

Cross multiply

2x = 3 × 4

2x = 12

x = 12/2

x = 6.

This means that; 1/3 of 12 = 2/3 of 6

4 0
3 years ago
If I have an unknown quantity of gas at a pressure of 380 mmHg, a volume of 25 L, and a temperature of
Bess [88]

Answer:

                     0.5077 moles

Explanation:

Data Given:

                Moles  =  n  =  <u>???</u>

                Temperature  =  T  =  300 K

                 Pressure  =  P  =  380 mmHg = 0.50 atm

                 Volume  =  V  =  25 L

Formula Used:

Let's assume that the hydrogen gas in balloon is acting as an Ideal gas, the according to Ideal Gas Equation,

                 P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹

Solving Equation for n,

                 n  =  P V / RT

Putting Values,

                 n  =  (0.50 atm × 25 L) / (0.082057 atm.L.mol⁻¹.K⁻¹ × 300 K)

                 n  = 0.5077 moles

6 0
3 years ago
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