Magnesium metal a gray solid, is heated in a crucble in the presence of oxygen

What is the angle between two of the nitrogen-hydrogen bonds in the ammonium (NH4+) ion?

Arlecino [84]

Answer:

109° 27'

Explanation:

The ammonium ion is tetrahedral in shape, all the HNH bonds are exactly at the tetrahedral bond angle since there are only bond pairs in the structure and no lone pairs. Recall that lone pairs decrease the bond angke from the ideal value in a tetrahedron due to higher repulsion.

8 0

3 years ago

Please help

Alika [10]

Answer:

there is no question to answer :(

4 0

2 years ago

You are a researcher with an interest in understanding the mechanism of serine protease enzymes. To test the effect of the role

yawa3891 [41]

Ngl didn’t read it so I’m not gonna answer it. Don’t report please just want to ask a question but have to answer two

5 0

3 years ago

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC

Salsk061 [2.6K]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.6 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:

$p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm$

Putting values in above equation, we get:

$K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92$

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 = 22.92

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.6 kJ/mol

7 0

3 years ago

what blood vessels and chambers does a molecule of glucose absorbed through the gut passes on its way to an active muscle in the

posledela

Answer:

Arteries are blood vessels that generally carry oxygen rich blood AWAY from the heart and lungs to the capillaries. The walls of ... and glucose, are absorbed from the digestive system and transported to the cells. Plasma ...

4 0

3 years ago