What is an examole of mass

yKpoI14uk [10]

Answer:

a weak bond between two molecules resulting from an electrostatic attraction between a proton in one molecule and an electronegative atom in the other.

Explanation:

For example, in water molecules (H2O), hydrogen is covalently bonded to the more electronegative oxygen atom. Therefore, hydrogen bonding arises in water molecules due to the dipole-dipole interactions between the hydrogen atom of one water molecule and the oxygen atom of another H2O molecule.

4 0

3 years ago

A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul

gayaneshka [121]

Answer:

10°C

Explanation:

Heat gain by water = Heat lost by the slice of pizza

Thus,

$m_{water}\times C_{water}\times \Delta T=Q$

<u>For water: </u>

Volume = 50.0 L

Density of water= 1 kg/L

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1 kg/L \times {50.0\ L}$

Mass of water = 50 kg

Specific heat of water = 1 kcal/kg°C

ΔT = ?

For slice of pizza:

Q = 500 kcal

So,

$50\times 1\times \Delta T=500$

ΔT = 10°C

Increase in temperature = 10°C

3 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

68 POINTS WILL GIVE BRAINliest!!!

AleksandrR [38]

Answer:

D

C

Explanation:

4 0

3 years ago

Read 2 more answers

Muszę napisać wzór sumatryczny wysyłam załącznik

ioda

Cześć, nie mówię po polsku, więc zrobiłem to w Tłumaczu Google, więc przepraszam, jeśli coś brzmi śmiesznie.

chlor (VII) i tlen - ten wzór to $CI_{2} O_{7}$ , a tlenek chloru jest bezwodnikiem kwasu nadchlorowego.

węgiel i wodór - wzór na węgiel i wodór to CnH2n + 2), jest to związek organiczny, a niższa klasyfikacja jest taka, że jest to również węglowodór aromatyczny.

Mam nadzieję, że to pomoże, błogosławionego i cudownego dnia! :-)

-Cutiepatutie

7 0

3 years ago