What happens to a solution when you add an acid to

Force and distance are used to calculate work, work is measured in

Natali [406]

Work is measured in joules. hope this helps!

7 0

3 years ago

\The specific heat of aluminum is 0.21 cal g°C . How much heat is released when a 10 gram piece of aluminum foil is taken out of

atroni [7]

Answer: The amount of heat released is 84 calories.

Explanation:

The equation used to calculate the amount of heat released or absorbed, we use the equation:

$Q= m\times c\times \Delta T$

where,

Q = heat gained or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

$\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC$

$Q=10g\times 0.21Cal/g^oC\times-40^oC$

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

8 0

3 years ago

What is the formula of sulfurous acid? H2S HSO4 H2SO4 H2SO3

Bas_tet [7]

The correct answer is H2SO3 I took the assignment already

6 0

3 years ago

Nonmetals gain electrons under certain conditions to attain a noble gas electron configuration. How many electrons must be gaine

gogolik [260]

Answer:

1

Explanation:

For non metals to attain a noble gas configuration, they gain the number of electrons needed to attain the noble gas configuration of the noble gas at the end of their periods. This means that these non metals would only take up the configuration of the last element on their periods which of course is always a noble gas.

The last element on the hydrogen period or more conservatively the only other element on the hydrogen period is helium, with an atomic number of 2. The atomic number is the number of protons in he nucleus of an atom. For an electrically neutral atom, the number of electrons equal the number of protons.

Hence we can deduce that helium has 2 electrons while hydrogen has one electron. Thus for it to attain the configuration of helium, it just needs to gain one more electron

6 0

3 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

For $N_2$ :

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

For $H_2$ :

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

3 years ago