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2 years ago

For the following reaction, 5.21 grams of nitrogen gas are allowed to react with 5.91 grams of hydrogen gas.

Chemistry

1 answer:

xz_007 [3.2K]2 years ago

7 0

Answer:

Maximum amount of ammonia that can be formed is 6.32 g

Explanation:

The balanced equation is:

N2 + 3H2\rightarrow 2NH3

$Moles\ N2 = \frac{5.21g}{28g/mol} =0.186\\\\Moles H2 = \frac{5.21g}{2g/mol} =2.61$

N2 is the limiting reactant

Based on the reaction stoichiometry:

1 mole N2 produced 2 moles NH3

therefore, 0.186 mol N2 will yield: 2(0.186) = 0.372 moles NH3

Molar mass NH3 = 17 g/mol

$Maximum\ Mass\ NH3 = 0.372 mols*17g/mol=6.32g$

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How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?

Karo-lina-s [1.5K]

Answer:

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = <u>58902.9 joules</u>

Explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

Q = 129 g . 2.10 J/g°C . 23°C → 6230.7 joules

Then, the ice has melted. To be melted and change the state it required:

Q = C lat . m

Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

5 0

2 years ago

Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because __________.

olga55 [171]

Answer:

Present in both catabolic and anabolic pathways

Explanation:

Glyceraldehyde-3-phosphate abbreviated as G3P occurs as intermediate in glycolysis and gluconeogenesis.

In photosynthesis, it is produced by the light independent reaction and acts as carrier for returning ADP, phosphate ions Pi, and NADP+ to the light independent pathway. Photosynthesis is a anbolic pathway.

In glycolysis, Glyceraldehyde-3-phosphate is produced by breakdown of fructose-1,6 -bisphosphate. Further Glyceraldehyde-3-phosphate converted to pyruvate and pyruvate is further used in citric acid cycle for energy production. Therefore, it is used in catabolic pathway too.

Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because it is present in both catabolic and anabolic pathways.

4 0

2 years ago

Consider the following ionic compounds: CdCO3, Na2S, PbSO4, (NH4)3PO4, and Hg2Cl2. Which compounds will be soluble when added to

Grace [21]

Answer:

a. Na₂S and (NH₄)₃PO₄ .

Explanation:

To know which salts are soluble we need to remember some <em>solubility rules</em>. And apply these as follows:

* All compounds of group 1 (Li⁺, Na⁺, K⁺, etc.) are soluble, Na₂S is soluble.

* All ammonium salts are soluble,

<h3>(NH₄)₃PO₄ is soluble.</h3>

* All Cl⁻, I⁻, Br⁻ are soluble except when are with Ag⁺, Hg₂²⁺, Pb²⁺, Hg₂Cl₂ is <em>insoluble</em>

* Carbonates are insoluble except those with group 1 or ammonium, CdCO₃ is <em>insoluble</em>

* All sulfates are soluble except those with Ca²⁺, Sr²⁺, Ba²⁺, Pb²⁺, Ag⁺. PbSO₄ is <em>insoluble. </em>

Right solution is:

8 0

2 years ago

A neutral particle formed when atoms share electrons.

Yakvenalex [24]

Answer:

ion

ionic bond

compound

convalent bond

electrons

molecule

Explanation:

6 0

2 years ago

The normal boiling point of a liquid is 282 °C. At what temperature (in

ElenaW [278]

Answer:

The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C

Explanation:

Here we make use of the Clausius-Clapeyron equation;

$ln\left (\frac{p_{2}}{p_{1}} \right )=-\frac{\Delta H_{vap}}{R}\cdot \left (\frac{1}{T_{2}}-\frac{1}{T_{1}} \right )$

Where:

P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K

P₂ = 0.2 atm = The substance vapor pressure at temperature T₂

$\Delta H_{vap}$ = The heat of vaporization = 28.5 kJ/mol

R = The universal gas constant = 8.314 J/K·mol

Plugging in the above values in the Clausius-Clapeyron equation, we have;

$ln\left (\frac{0.2}{1} \right )=-\frac{28.5 \times 10^3}{8.3145}\cdot \left (\frac{1}{T_{2}}-\frac{1}{555.15} \right )$

$\therefore T_2 = \frac{-3427.95}{ln(0.2)-6.175}$

T₂ = 440.37 K

To convert to Celsius degree temperature, we subtract 273.15 as follows

T₂ in °C = 440.37 - 273.15 = 167.22 °C

Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.

5 0

3 years ago