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3 years ago

Why can hydrogen only form one covalent bond?

1 answer:

7 0

Because its in group one....which only lose or gain electrons.....thus it will only have oneelectron to fulfill its octect rule

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H+(aq) + OH–(aq) → H2O(l) + 55.8 kJ In this reaction there is conservation of

pentagon [3]

In the equation given above, there is conservation of MASS, CHARGE AND ENERGY.
These three parameters are usually conserved during the course of chemical reactions. When any of these parameter experience a reduction during the course of chemical reaction, such loss is always gained by other elements involved in the same reaction, so that at the end of the day, they are not considered as lost.

4 0

3 years ago

Which formula equation represents the burning of sulfur to produce sulfur dioxide?

Dmitriy789 [7]

Answer:

option A = S(s) + O₂(g) → SO₂ (s)

Explanation:

Chemical equation:

S(s) + O₂(g) → SO₂ (s)

when sulfur burned in the presence of oxygen it produce sulfur dioxide. The sulfur dioxide can further react with oxygen to produce sulfur trioxide and then react with water to form sulfuric acid.

Uses of sulfur dioxde:

It is used as a solvent and reagent in laboratory.

Sulfur dioxide is used to produce sulfuric acid.

It is used as a disinfectant

It is also used as a reducing agent.

It is used to preserve the dry food.

7 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

Question Megan made a model of the respiratory system using a water bottle and balloons. When she pulled down on the balloon str

ruslelena [56]

Answer:

The lung

Explanation:

The model of the respiratory system made by Megan consists of two balloons. The first balloon stretched across the bottom of the bottle represents the diaphragm which contracts and relaxes to allow air in and out of the lungs. The balloon inside the bottle represents one lung.

Breathing in causes the balloon inside the bottle to be filled with air. This is preceded by the expansion of the diaphragm which makes the lungs to be filled with air. Breathing out causes a contraction of the diaphragm thus making the lungs to let out air.

4 0

2 years ago

Read 2 more answers

4. Know the location of metals, non-metals and metalloids

erastovalidia [21]

Answer:

The metals are to the left of the line (except for hydrogen, which is a nonmetal), the nonmetals are to the right of the line, and the elements immediately adjacent to the line are the metalloids.

Hope this helped

7 0

3 years ago