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4 years ago

Indicate how the atomic number and the mass number of an isotope will change when it’s nucleus loses

1 answer:

7 0

Answer:The mass number of an isotope is the total number of protons and neutrons in an atomic nucleus. If you know that a nucleus has 6 protons and 6 neutrons, then its mass number is 12. If the nucleus has 6 protons and 7 neutrons, then its mass number is 13. The symbol for an isotope lists the mass number as a superscript at the upper left.

Explanation: SIRI

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What is the volume of 1.50 lb (pounds) of mercury? The density of mercury is 13.546 g/mL. Use the conversion that 1 kg = 2.20 lb

Arada [10]

Answer:

50.3mL of mercury are in 1.50lb

Explanation:

Punds are an unit of mass. To convert mass to volume we must use density (13.546g/mL). Now, As you can see, density is in grams but the mass of mercury is in pounds. That means we need first, to convert pounds to grams to use density and obtain volume of mercury.

<em>Mass mercury in grams:</em>

1.50lb * (1kg / 2.20lb) = 0.682kg = 682g of mercury.

<em>Volume of mercury:</em>

682g Mercury * (1mL / 13.546g) =

<h3>50.3mL of mercury are in 1.50lb</h3>

7 0

4 years ago

Which function of lysosomes is carried out by vacuoles in plant cells

user100 [1]

Answer:

Lysosomes are predominantly found in eukaryotic animal cells and are responsible for breaking down cellular debris

Explanation:

7 0

3 years ago

Liquid in a graduated cylinder has a volume of 70 mL. You put a small solid ball into the cylinder, and the height of the liquid

emmasim [6.3K]

Yes sir

Have a nice day

8 0

3 years ago

a student determined that it requires 106220 j of energy to vaporize 47g of water. is the student is right

shusha [124]

<h2>Answer:</h2>

He is right that the energy of vaporization of 47 g of water s 106222 j.

<h3>Explanation:</h3>

Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.

In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.

It means for vaporizing 18 g, 40.65 kJ energy is needed.

So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ

Hence the student is right about the energy of vaporization of 47 g of water.

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

4 years ago