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3 years ago

What is the broad name for movement of materials across a membrane

Chemistry

1 answer:

sineoko [7]3 years ago

8 0

Answer: i’m not 100% sure but might be active transport

Explanation:

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____________ limits or stops the growth of bacteria.

Kaylis [27]

Answer:

Vaccines

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3 years ago

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For the following Bronsted acid-base reaction, complete the reaction, identify the acid, base, conjugate

madam [21]

Answer:

algun moderador puede acabar a este otro este me borró respuesta que no debió borrar

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3 years ago

Rank the following aqueous solutions from highest to lowest freezing point: 0.1 m FeCl3, 0.30 m glucose (C6H12O6), 0.15 m CaCl2.

Tresset [83]

Answer:

Explanation: See images below for explanation

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3 years ago

The isomerization of methylisonitrile to acetonitrile (CH3NC(g) ???? CH3CN) is first order in CH3NC. The rate constant for the r

Hoochie [10]

Answer:

The half life is $H_{1/2}= 7333.3sec$

Explanation:

The half life of a first order reaction is mathematically represented as

$H_{1/2} = \frac{0.693}{Rate Constant }$

Substituting $9.45 * 10^{-5}s^{-1}$ for the rate constant

$H_{1/2} = \frac{0.693}{9.45*10^{-5}}$

$H_{1/2}= 7333.3sec$

5 0

3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

3 years ago