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3 years ago

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How might you make sure you have the correct amount of nitrogen and hydrogen so that you have enough of each with not wasted sur

plus?

1 answer:

Grace [21]3 years ago

4 0

Answer:

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Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s

slava [35]

Answer:

Potassium carbonate (K₂CO₃)

Explanation:

The compounds dissociate into ions in water, as follows:

K₂CO₃ → 2 K⁺ + CO₃⁻ ⇒ 3 dissolved particles per mole

NaI → Na⁺ + I⁻ ⇒ 2 dissolved particles per mole

KBr → K⁺ + Br⁻ ⇒ 2 dissolved particles per mole

CH₃OH → CH₃O⁻ + H⁺ ⇒ 2 dissolved particles per mole

NH₄Cl → NH₄⁺ + Cl⁻ ⇒ 2 dissolved particles per mole

Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).

3 0

3 years ago

Please i need this!!!! The answer correct please

DIA [1.3K]

The table tells us that the larger the diameter the shorter the period of rotation is

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3 years ago

Identify which do not belong to the group?

nevsk [136]

Answer:

bronchi

epiglottis

plasma

Mittal

alveoli

5 0

2 years ago

Read 2 more answers

A certain organic compound is made up of the amino acids methionine, arginine, and leucine. Which type of organic compound is th

Andrews [41]

<em>Answer:</em>

The protein is an organic compound which is made up of the amino acids methionine, arginine, and leucine etc.

<em>Explanation:</em>

Most of the amino acid are linked with others amino acid through peptide linkage to form proteins. The peptide bond is formed between the two amino acid involving carboxylic and amino group.

The proteins produce by this way may be simple, conjugated and derived proteins.

4 0

3 years ago

Read 2 more answers

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago