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3 years ago

Is facilitated diffusion passive or active transport?

Chemistry

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

Its passive

Explanation:

Facilitated diffusion is a type of passive transport that allows substances to cross membranes with the assistance of special transport proteins.

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What is the independent and dependent variable?

tigry1 [53]

Independent is the variable that you change
Dependent is the thing you measure

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3 years ago

What mass of aluminum chloride could be made from 8.1 g of aluminum and 4.2 L of chlorine at STP?

posledela

In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,

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4 years ago

The amount of a chemical solution is measured to be 2 liters.

irinina [24]

2 liters may be 1.5 to 1.9 rounded up to 2 or 2.1 or 2.4 rounded down to 2.

2 - 1.5 = 0.5

percent error = (absolute error / quantity) * 100

percent error = 0.5/2 * 100% = 0.25 * 100% = 25%

Choice C. 25%.

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3 years ago

What determines an element's chemical properties

Alik [6]

The three factors determine the chemical properties of an element:
The number and arrangement of electrons in an atom
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3 years ago

Read 2 more answers

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

4 0

2 years ago