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Alex17521 [72]
3 years ago
15

What do you do if your Bf call you selfish -_-

Chemistry
2 answers:
VLD [36.1K]3 years ago
7 0

Answer:

depends on what he called u selfish for

....

ryzh [129]3 years ago
4 0

Answer:

cry?!

Explanation:

You might be interested in
Chemistry question.
grandymaker [24]
Hello!

Percentage error = (true value - measured value)/true value x 100%

Percentage error = (27.7-27.0)/27.7 x100% = 25.3%
8 0
3 years ago
Consider the reaction H2(g) + I2(g) Double headed arrow. HI(g) with an equilibrium constant of 46.3 and a reaction quotient of 5
Alika [10]

Answer:

The reaction shifts to the left.

Explanation:

Equilibrium constant (K) = 46.3

Reaction Quotient (Q) = 525

The relationship between Q and K with their implications are given as;

K = Q (No net reaction)

K > Q (Reaction shifts to the right)

K < Q (Reaction shifts to the left)

Since in this question, Q (525) > K (46.3)

The reaction shifts to the left.

5 0
2 years ago
What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intrac
Ganezh [65]

The equilibrium membrane potential is 41.9 mV.

To calculate the membrane potential, we use the <em>Nernst Equation</em>:

<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}

where

• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions

• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

• <em>T</em> = the Kelvin temperature

• <em>z</em> = the charge on the ion (+1)

• <em>F </em>= the Faraday constant [96 485  C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

• [Na]_o = the concentration of Na^(+) outside the cell

• [Na]_i = the concentration of Na^(+) inside the cell

∴ <em>V</em>_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

4 0
3 years ago
In the equation below
yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)}  + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

  • The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
  • The ratio between NH₃ and Cu is 2:3.
  • The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

  • The amount of N₂ produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}

  • The amount of Cu produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}

  • And the amount of H₂O produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

8 0
2 years ago
What can happen if you have heated glass next to cool or cold water?
Effectus [21]
The class can break or if you put it in cold water it can cool down down fast.
6 0
2 years ago
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