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2 years ago

What is the number of electrons in an atom of scandium

1 answer:

4 0

The nuclear composition and electron configuration of an atom of Scandium 45 and the atomic number is 21 the most common isotype of this element is that nuclear consists of 21 protons and 24 neutrons Scandium is a transition metal group 3 period and the d-block of a periodic table

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A sample of aluminum is placed in a 25-ml graduated cylinder containing 10.0 ml of water. the level of the water rises to 18.0 m

OLga [1]

To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.

Then knowing volume of aluminum and the density of it, we can solve for the mass.

D = m/v

Dv = m

2.7 g/ml • 8 ml = 21.6 grams.

5 0

3 years ago

Read 2 more answers

Does the mass of a ball affect how far it rolls

dybincka [34]

Yes, it depends on the size of the ball and the weight. If it's heavy and big then it'll affect how far it rolls, if it's small and has no weight then it won't really move as much

3 0

3 years ago

Read 2 more answers

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

4 years ago

There is a gas at 780 mm of Hg, in a volume of 5 liters and a temperature of 37 C, the volume is changed to 5.5 liters and the

HACTEHA [7]

Answer:

32.8 C

Explanation:

- Use combined gas law formula and rearrange.

- Hope that helped! Please let me know if you need further explanation.

6 0

4 years ago

Law of Conservation of Mass in terms of atoms

VARVARA [1.3K]

No atoms are lost or made during the chemical reaction so the total mass of the products is equal to the total mass of the reactants. In an atom, protons and neutrons contribute to the mass and since the number of them doesn’t change, the mass doesn’t either.

4 0

3 years ago