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3 years ago

How many grams of diborane will react with 7.5 mol of O2

Chemistry

1 answer:

lorasvet [3.4K]3 years ago

8 0

Answer: Mass of diborane that would react with 7.5 mole of O2 is 70g

Explanation: The explanation is contained in the picture attached.

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What is hydroponics

Aliun [14]

Answer:

Explanation:

Hydroponics is the process of growing crops using only water and liquid fertilizer. This process is great when your in the big city.

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3 years ago

HELP THIS IS TIMED!!!

Naddika [18.5K]

Answer:

See to them if they are balanced, i.e., the no. of atoms of each element should be equal on both the sides. If they are so, then its adhered to law of conservation of matter.

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3 years ago

C6H5NH3Cl is a chloride salt with an acidic cation. If 46.3 g of C6H5NH3Cl is dissolved in water to make 150 mL of solution, wha

vaieri [72.5K]

<u>Answer:</u> The initial molarity of cation is 2.38 M

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of $C_6H_5NH_3Cl$ = 46.3 g

Molar mass of $C_6H_5NH_3Cl$ = 129.6 g/mol

Volume of solution = 150 mL

Putting values in above equation, we get:

$\text{Molarity of }C_6H_5NH_3Cl=\frac{46.3\times 1000}{129.6\times 150}\\\\\text{Molarity of }C_6H_5NH_3Cl=2.38M$

1 mole of $C_6H_5NH_3Cl$ produces 1 mole of $C_6H_5NH_3^+$ cation and 1 mole of $Cl^-$ anion

So, molarity of $C_6H_5NH_3^+$ cation = (1 × 2.38) = 2.38 M

Hence, the initial molarity of cation is 2.38 M

8 0

3 years ago

PLZ HELP For the reaction: 2NO2(g) → N2O4(l),

Furkat [3]

Answer: $\Delta H_{rxn}=-20kJ/mol-(+66kJ/mol)$

Explanation:

Heat of reaction or enthalpy change is the energy released or absorbed during the course of the reaction.

It is calculated by subtracting the enthalpy of reactants from the enthalpy of products.

$\Delta H=H_{products}-H_{reactants}$

$\Delta H$ = enthalpy change = ?

$H_{products}$ = enthalpy of products

$H_{reactants}$ = enthalpy of reactants

For the given reaction :

$2NO_2(g)\rightarrow N_2O_4(l)$

$\Delta H=H_{N_2O_4}-2\times H_{NO_2}$

$\Delta H=-20kJ/mol-(+66kJ/mol)$

6 0

3 years ago

Plz answer correctly and make sure u put it in order

iren2701 [21]

Answer: look on my page ill help you with this one but im not sure if it would be right because i dont really understand myself but look on mine

Explanation:

8 0

3 years ago

Read 2 more answers