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Luden [163]
3 years ago
13

PLS HELP ME I HAVE TILL 12

Chemistry
1 answer:
MariettaO [177]3 years ago
3 0

Answer:

alright bud lets see hmm.... the answer is a. 90.5kpa

Explanation:

92.3 kPa - 1.82 kPa = 90.5 kPa

keep up your hope also corrected me if a am wrong in anyway! :)

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Do your cells have access to everything they need over the first 4 or 5 hours that you are lost? Yes or No?
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Explanation: cell in the body is enclosed by a cell (Plasma) membrane. The cell membrane separates the material outside the cell, extracellular, from the material inside the cell, intracellular. ... All materials within a cell must have access to the cell membrane (the cell's boundary) for the needed exchange

**Answer**: The answer would be Yes I believe

4 0
3 years ago
A sample of gas (1.9 mol) is in a flask at 21 °c and 697 mm hg. the flask is opened and more gas is added to the flask. the new
garik1379 [7]

To solve this problem, we assume ideal gas so that we can use the formula:

PV = nRT

since the volume of the flask is constant and R is universal gas constant, so we can say:

n1 T1 / P1 = n2 T2 / P2

 

1.9 mol * (21 + 273 K) / 697 mm Hg = n2 * (26 + 273 K) / 841 mm Hg

<span>n2 = 2.25 moles</span>

8 0
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Balance this equation K+b2o3=k2o+b
AVprozaik [17]

Answer:

Explanation:

2K and 2b

5 0
3 years ago
What does the subscript 2 indicate in the formula o2?
Tpy6a [65]

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6 0
3 years ago
c. The reaction Br2 (l) --&gt; Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a
egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

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

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

6 0
3 years ago
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