Lil Donald went to his friend's Fat James's house. Fat James house is

The reaction of aluminum with bromine is shown here. The equation for the reaction is

m_a_m_a [10]

Answer is (b)

2,3,1 is the stoichiometry

The values in front of the elements are the stoichiometric values
(Since Al2Br6 has no value in front, it's considered 1)

5 0

2 years ago

A scientist measures out equal volumes and concentrations of hydrochloric acid and sodium hydroxide solutions. When mixed, these

olasank [31]

Answer:

Neutral solution is formed.

Explanation:

When the hydrochloric acid and sodium hydroxide which is a strong base are combined together, it produces sodium chloride which is a salt and water. This solution is known as Neutral solution because the solution do not have the characteristics or properties of either an acid or a base. If the concentration of one of the reactant is higher as compared to another reactant so the product has the characteristics of that reactant.

8 0

2 years ago

Read 2 more answers

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

$n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}$

$n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{Fe} = 0.5\:mol}$

* to n(S)

$n_{S} = \dfrac{m_{S}}{MM_{S}}$

$n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{S} = 0.5\:mol}$

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

$Fe + S \Longrightarrow FeS$

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

$n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}$

$m_{FeS} = n_{FeS}*MM_{FeS}$

$m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

44 grams of iron sulfide

___________________________

$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

4 0

2 years ago

A steel cylinder of oxygen is being store in a room at 25.0 °C under a pressure of 1250 atm. What pressure would be exerted by t

Rudik [331]

Answer:

P₂ = 1312.88 atm

Explanation:

Given data:

Initial temperature = 25°C

Initial pressure = 1250 atm

Final temperature = 40°C

Final pressure = ?

Solution:

Initial temperature = 25°C (25+273.15 = 298.15 K)

Final temperature = 40°C ( 40+273.15 = 313.15 k)

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1250 atm / 298.15 K = P₂/313.15 K

P₂ = 1250 atm × 313.15 K / 298.15 K

P₂ = 391437.5 atm. K /298.15 K

P₂ = 1312.88 atm

6 0

3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

5 0

3 years ago