Nonliving and living things are made of matter.which of the following statements is true about matter

`You have to be careful about pouring drano down your pipes since it is mainly hydrochloric acid--you can't do it if they are ma

Zanzabum

Answer:

6.67 moles

Explanation:

Given that:-

Moles of hydrogen gas produced = 10.0 moles

According the reaction shown below:-

$2Al + 6HCl\rightarrow 2AlCl_3 +3H_2$

3 moles of hydrogen gas are produced when 2 moles of aluminium undergoes reaction.

Also,

1 mole of hydrogen gas are produced when $\frac{2}{3}$ moles of aluminium undergoes reaction.

So,

10.0 moles of hydrogen gas are produced when $\frac{2}{3}\times 10.0$ moles of aluminium undergoes reaction.

Moles of Al needed = $\frac{2}{3}\times 10.0$ moles = 6.67 moles

6 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0

3 years ago

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.0 g of lead(II) oxide. Calculate the percent yi

user100 [1]

Let MM(x) be the molar mass of x.

MM(Pb) : MM(PbO)
=207.21 : 223.20 = 451.4 g : x g

cross multiply and solve for x
x=223.2/207.21*451.4
= 486.23 g

Percentage yield = 365.0/486.23= 0.75067 = 75.07% (rounded to 4 sign. fig.)

4 0

3 years ago

A solution that has a large amount of dissolved solute is called

kvasek [131]

Answer: It is an unsaturated solution

Explanation: This is because it has more solute than a normal solution.

6 0

3 years ago

What mass of Sodium Chloride is required to make 100.0 mL of 3.0 M solution?

Julli [10]

Answer:

17.55 g of NaCl

Explanation:

The following data were obtained from the question:

Molarity = 3 M

Volume = 100.0 mL

Mass of NaCl =..?

Next, we shall convert 100.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

100 mL = 100/1000

100 mL = 0.1 L

Therefore, 100 mL is equivalent to 0.1 L.

Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:

Molarity = 3 M

Volume = 0.1 L

Mole of NaCl =?

Molarity = mole /Volume

3 = mole of NaCl /0.1

Cross multiply

Mole of NaCl = 3 × 0.1

Mole of NaCl = 0.3 mole

Finally, we determine the mass of NaCl required to prepare the solution as follow:

Mole of NaCl = 0.3 mole

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.3 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 0.3 × 58.5

Mass of NaCl = 17.55 g

Therefore, 17.55 g of NaCl is needed to prepare the solution.

5 0

3 years ago