Another way to say reactants is Question 4 options:

PLZ HELP THERE IS A FILE ATTACHED WILL GIVE U BRAINLIEST IF U HELP

erastovalidia [21]

Answer:

I think C (Might not be true)

Explanation:

6 0

3 years ago

Read 2 more answers

What is minimum volume of oxygen required to react with 42.5 g of aluminum in the synthesis of aluminum oxide at stp?

natka813 [3]

26.833 liters

Aluminum oxide has a formula of Al₂O₃, which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).

Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.

Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.

Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815

Finally, you need to calculate the volume of 1.1815 moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,

1.1815 * 22.7 = 26.8 liters of oxygen gas.

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

extremely hot.When a student went to open the car doo , he burned his finger. What two forms of energy were responsible for the

charle [14.2K]

Answer:

radiation and conduction

Explanation:

During a warm summer day, a car became extremely hot. When a student went to open the car door, he burned his fingers. What two forms of energy were responsible for the student burning his fingers?

Solution:

Heat is the transfer of energy from a warmer object to a cooler object. For heat transfer to occur, there have to be a difference in temperature between two objects.

Heat can be transferred in three ways: by conduction, by convection, and by radiation.

Conduction is the transfer of thermal energy between bodies through direct contact. Convection is the transfer of thermal energy through the movement of heat in a liquid or gas. Radiation is the transfer of thermal energy through thermal emission by electromagnetic waves.

During a warm summer day, The sun makes the car to become hot through energy transfer from the sun to the car. When the student touch the car, there is heat transfer as a result of conduction.

3 0

3 years ago

Can you check this for me?

Aliun [14]

Hi, your answer is correct.

3 0

3 years ago