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3 years ago

Vrite skeleton equations for the following word equations.

Chemistry

2 answers:

almond37 [142]3 years ago

4 0

Answer:

Explanation:

Bad White [126]3 years ago

3 0

H + Br -> HBr
Criss cross the charges because hydrogen bromide is ionic

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What is 25 °C in °F?

kifflom [539]

Answer:

77°F

Explanation:

(25°C × 9/5) + 32 = 77°F

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3 years ago

The layers of earth are the crust, mantle, and core, with the core being divided into inner and outer layers. Which of the follo

Westkost [7]

Answer:

C- The core is made up of dense elements, such as iron and nickel.

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3 years ago

Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.

Bezzdna [24]

Answer:

$\large \boxed{\text{ 0.066 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ: 58.44

NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g: 0.245

V/mL: 50.

c/mmol·mL⁻¹: 0.0180

2. Calculate the moles of each reactant

$\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}$

3. Identify the limiting reactant

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:

The molar ratio of NaCl to AgCl is 1:1.

$\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}$

From AgNO₃:

The molar ratio of AgNO₃ to AgCl is 1:1.

$\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}$

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)

I/mmol: 0.900 4.192 0

C/mmol: -0.900 -0.900 +0.900

E/mmol: 0 3.292 0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

$\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}$

8 0

3 years ago

What would be an advantage and a disadvantage to using litmus as a titration indicator???

lesantik [10]

To indicate whether the elements is basic or acid by changing of colour.(advantage)
can give out bad gas which can cause suffocate to the people.(disadvantage).

4 0

3 years ago

1.00 mole of an ideal gas at STP is cooled to -41°C while the

qwelly [4]

Answer:

V₂ = 18.13 L

Explanation:

Given data:

Mole of gas = 1 mol

Initial temperature = 273 K

Initial pressure = 1 atm

Final volume = ?

Final temperature = -41°C (-41+273 = 232 K)

Final pressure = 805 mmHg (805/760 = 1.05 atm)

Solution:

First of all we will calculate the initial volume of gas.

PV = nRT

V = nRT/P

V = 1 mol × 0.0821 mol.L/atm.K × 273 K / 1 atm

V = 22.4 L/atm / 1 atm

V = 22.4 L ( initial volume)

Now we will determine the final volume by using equation,

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Now we will put the values.

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 1 atm × 22.4 L × 232 K / 273 K × 1.05 atm

V₂ = 5196.8 atm .L. K / 286.65 atm.K

V₂ = 18.13 L

5 0

3 years ago