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2 years ago

13

Which is an example of a vascular flowering seeded plant? \

1 answer:

olga55 [171]2 years ago

6 0

One plant would be the angiosperm. A
vascular, seed-producing plants whose seeds are enclosed in a fruit.

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Use Lewis diagrams to show how electrons are shared to form covalent bonds in compounds wit the following atoms. Write the formu

bixtya [17]

The Lewis structure/diagram for CH2O (aka Formaldehyde) can be written in either of the following ways shown in the picture.
The dots represent electrons in the valence shell of the atom (the outermost shell). The green dots are electrons that belong to the Oxygen atom, the blue belong to the Carbon atom, and the pink belong to the Hydrogen atoms.
Covalent bonds are bonds between atoms where atoms share electrons with each other. Atoms bond because they obey the octet rule ( the rule states that most atoms of main-group elements tend to want 8 electrons in their valence shells).
Oxygen has 6 valence electrons, Carbon has 4, and Hydrogen has 1. H does not follow the octet rule, but C and O do, so the atoms are arranged in this way so that the O and C atoms have a full octet of electrons in their valence.

8 0

2 years ago

Look at the diagram of the atom. What will it change to during beta decay? A)An atom with 3 protons, 4 neutrons, and 3 electrons

GREYUIT [131]

D) An atom with 4 protons, 3 neutrons, and 3 electrons

5 0

3 years ago

What enzyme is coded for in the chromosome of bacteria?

Mariulka [41]

The correct answer would be, "<span>D. beta-galactosidase".
As all the others are made in different organisms.

Hoped I helped.</span>

3 0

2 years ago

Read 2 more answers

In addition, they are poor conductors of

MA_775_DIABLO [31]

Answer: electricity

Explanation:

6 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