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Alexandra [31]
3 years ago
8

If a negatively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be

directed ________. Thus, the equilibrium potential for this ion would be ________ charged. outward : negatively inward : positively inward : negatively outward : positively outward : neutrally
Chemistry
1 answer:
Lera25 [3.4K]3 years ago
3 0
Outward, positively
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If a student didn’t understand the periodic table, how would you explain it to them?
morpeh [17]
Maybe use flash cards for them to understand or look up periodic table on line and show them.
6 0
3 years ago
Need help on both #19 and #20 please
ziro4ka [17]
This is the answer to question 19. If it’s not clear, send me a message.

5 0
4 years ago
A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t
kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

4 0
3 years ago
The two gases, H2 and O2, were allowed to effuse under the same conditions through a pinhole. If the rate of effusion of O2 was
TEA [102]

The rate of effusion of H₂ : 7.2 x 10⁻² m/s

<h3>Further explanation  </h3>

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or  

the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

or  

\rm M_1\times r_1^2=M_2\times r_2^2

MW₁ O₂ = 32 g/mol

MW₂ H₂ = 2 g/mol

\tt \dfrac{1.8\times 10^{-2}}{r_2}=\sqrt{\dfrac{2}{32} }\\\\\dfrac{(1.8\times 10^{-2})^2}{r_2^2}=\dfrac{2}{32}\\\\r_2=0.072=7.2\times 10^{-2}~m/s

7 0
3 years ago
A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and
Ilya [14]

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-

Molarity of the silver nitrate solution = 0.03337 M

Volume of the silver nitrate = 23.28 mL = 0.02328 L

Moles of silver nitrate = n

0.03337 M=\frac{n}{0.02328 L}

n = 0.03337 M\times 0.02328 L=0.0007768 mol

According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.

Then 0.0007768 moles of silver nitrate will react with:

\frac{1}{1}\times 0.0007768 mol=0.0007768 mol chloride ions.

In one mole of aldrin there are 6  moles of chloride ions.

Then moles of aldrin containing 0.0007768 moles chloride ions are:

\frac{0.0007768 mol}{6}=0.0001295 mol

Moles of aldrin present in the sample = 0.0001295 mol

Mass of 0.0001295 moles of aldrin present in the sample :

0.0001295 mol × 364.92 g/mol =0.04726 g

Percentage of an aldrin in the sample:

\frac{0.04726 g}{0.1064 g}\times 100=44.41\%

8 0
3 years ago
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