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2 years ago

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Una masa de hidrogeno H2 ocupa un volumen de 12 litros a 730 mmHg. ¿Cuál es el volumen del gas a 1200 mmHg, si la temperatura pe

rmanece constante?

1 answer:

Alex2 years ago

4 0

Answer:

ENGLISH PLEASE

Explanation:

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If 3.8 moles of zinc metal react with 6.5 moles of silver nitrate, how many moles of silver metal can be formed, and how many mo

bulgar [2K]

<u>Answer:</u> 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

<u>Explanation:</u>

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = $\frac{1}{2}\times 6.5=3.25moles$ of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = $\frac{2}{2}\times 6.5=6.5moles$ of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

7 0

2 years ago

Please help. 1.600x10^8 / 8.000x10^11

sweet-ann [11.9K]

Answer:

2x10^-4 = 2E-4

Explanation:

3 0

2 years ago

Read 2 more answers

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

Degger [83]

Answer : The % of (+) limonene isomer = 79%

The % of (-) limonene isomer = 0%

The % of enantiomeric excess = 58%

Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where, ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

${ee}=\frac{\%(+)-50\%}{2}$

$=\frac{79\%-50\%}{2}$

= 58%

7 0

2 years ago

Read 2 more answers

Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati

worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4 + Cl-

Firstly, we calculate the number of moles of the ammonia as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

NH4+ H2O ⇄ NH3 H3O+

I 0.242 0 0

C -X +x +X

E 0.242-X X X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b = K w

, where K w- the self-ionization constant of water, equal to

10 ^-14 at room temperature

This means that you have

K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈ 0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0

2 years ago

How many moles of h2o are produced when using 7 moles of h2

Olenka [21]

1) Write the balaced chemical equation:

H2 + 2O2 → 2H2O

2) Infere the molar ratios:

1 mol H2 : 2 mol of water

3) Make the calculus as the direct proportion relation:

[2 mol H2O] / [1 mol H2] * 7 mol H2 = 14 mol H2

As you see you produce the double number of moles of H2O than number of moles of H2 used.

Answer: 14 moles

6 0

3 years ago