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yuradex [85]
2 years ago
11

Please help me figure this out! God bless you <3

Chemistry
1 answer:
statuscvo [17]2 years ago
3 0
South ????????????????????????
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At 333 k, which of the pairs of gases below would have the most nearly identical rates of effusion?
rjkz [21]
E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
a) M(N</span>₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
8 0
3 years ago
When electrodes are used to record the electrocardiogram, an electrolyte gel is usually put between them and the surface of the
klemol [59]

Answer:

The equivalent circuit for the electrode while the electrolyte gel is fresh

From the uploaded diagram the part A is the electrolyte, the part part B is the electrolyte gel when is fresh and the part C is the surface of the skin

Now as the electrolyte gel start to dry out the resistance R_s of the gel begins to increase and this starts to limit the flow of current . Now when the gel is then completely dried out  the resistance of the gel R_s then increases to infinity  and this in turn cut off flow of current.

The diagram illustrating this is shown on the second uploaded image

Explanation:

5 0
3 years ago
How many valence electrons would an element with atomic number 113 have?
marshall27 [118]

How does the law of conservation of mass apply  to this reaction: C2H4 + O2 → H2O + CO2?

4 0
3 years ago
About 6 × 109 g of gold is thought to be dissolved in the oceans of the world. If the total volume of the oceans is 1.5 × 1021 L
Lelechka [254]

First, determine the number of moles of gold.

Number of moles  = \frac{given mass in g}{molar mass}

Given mass of gold  =6 \times 10^{9} g

Molar mass of gold  = 196.97 g/mol

Put the values,

Number of moles of gold  = \frac{6 \times 10^{9} g}{196.97 g/mol}

= 0.03046\times  10^{9} mole or 3.046\times  10^{7} moles

Now, molarity  = \frac{moles of solute}{volume of the solution in liters}

Put the values, volume of ocean  =1.5 \times 10^{21} L

Molarity = \frac{3.046\times 10^{7} moles}{1.5 \times 10^{21} L}

= 2.03\times 10^{-14} M

Thus, average molar concentration = 2.03\times 10^{-14} M




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3 years ago
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