What 3 sub types of covergent plate boundaries?

you fill a rigid steel container that has a volume of 20 L with nitrogen gas to a final pressure of 2 x 10^4 kpa at 23 Celsius.

garik1379 [7]

Answer:

4.549 kg.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 2 x 10⁴ kPa/101.325 = 197.4 atm).

V is the volume of the gas in L (V = 20.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 23° C + 273 = 296 K).

∴ n = PV/RT = (197.4 atm)(20.0 L)/(0.0821 L.atm/mol.K)(296 K) = 162.5 mol.

To find the mass of N₂ in the cylinder, we can use the relation:

mass of N₂ = (no. of moles of N₂)*(molar mass of N₂) = (162.5 mol)*(28.0 g/mol) = 4549 g = 4.549 kg.

3 0

3 years ago

You are given the reaction Cu + HNO3 → Cu(NO3)2 + NO + H2O. Half-reactions: First: 3 upper C u right arrow 3 upper C u superscri

nexus9112 [7]

Answer: It is the first one. First: 3 upper C u right arrow 3 Upper C u superscript 2 plus, plus 6 e superscript minus. Second: 2 upper N superscript 5 plus, plus 6 e superscript minus right arrow 2 upper N superscript 2 plus.

Explanation:

Correct

4 0

2 years ago

Read 2 more answers

What is the range of accuracy for food temperature?.

telo118 [61]

Answer:

+ 2 degrees Fahrenheit

Explanation:

7 0

2 years ago

Which of the following you is true for a limiting reactant

skad [1K]

Answer:

C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

Explanation:

Please, find attached a complete question to determine which of the statements is or are true for a limiting reactant in a chemical equation.

First, remember that the limiting reactant is the substance that is consumed completely while the excess reactant is the substance that does not react completely.

The limiting reactant is found comparing the stoichiometry ratio and the actual ratio between the reactants.

The stoichiometry ratio is found using the coefficientes of the chemical equation.

For illustration, assume the general chemical equation:

$aA+bB\rightarrow cC+dD$

The stoichiometric ratio of the reactants is:

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

If the ratio of the available moles of substance A to the available moles of substance B is greater than the stoichiometric ratio, it means that there are more moles of the substance A than what is needed to react with the available moles of substance B, then A will be in excess and B will B the limiting reactant.

If, on the contrary, the ratio of the available moles of substance A to the available moles of substance B is is less than the stoichiometric ratio, then substance A is less than the necessary to make the all the moles of substance B react, meaning that the substance A will limit the reaction (it will be consumed completely), while the substance B will be in excess.

As for the options:

A) The limiting reactant is has the lowest coefficient in a balanced equation.

This is false, since it is not the magnitude of the coefficiente what determines the limiting reactant, but the comparison of the ratios.

B) The limiting reactant is the reactant for which you have the fewest number of moles.

This is false because it is not the number of moles what determines the limiting reactant , but the comparison of the ratios.

C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

This is true as proved below.

The stoichiometric ratio of the reactants is:

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

The actual ratio is:

$available\text{ }moles\text{ }of\text{ }A/available\text{ }moles\text{ }of\text{ }B$

Assume the first ratio is less than the second (which describes when the substance A is in excess and the limiting reactant is the substance B).

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

Change the relation to show the ratios of moles available of each substance to the cofficient in the chemical equation:

$available\text{ }moles\text{ }of\text{ }B/b\text{ }moles\text{ }of\text{ }B$

Then, in the scenary that the limiting reactant is the substance B, the ratio of the left is lower than the ratio of the right, which is the same that limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

D) The limiting reactant has the lowest ratio of coefficients in the balanced eqution/moles available.

This ratio is the inverse of the ratio of the previous statement, thus the relation is inverse, and, since the previous statement was true, this statement is false.

3 0

2 years ago

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl

Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

$\Delta S = S(products)-S(reactants)----(1)$

a) $PCl3(l) + Cl2(g) \rightarrow PCl5(s)$

Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) $2HgO(s) \rightarrow 2Hg(l) + O2(g)$

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) $H2(g) \rightarrow 2H(g)$

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) $U(s) + 3F2(g) \rightarrow UF6(s)$

Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

8 0

3 years ago