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3 years ago

What does (w/w)% mean?

1 answer:

3 0

Answer: 2% w / w solution means grams of solute is dissolved in 100 grams of solution. ... 5% v / v solution means 5 ml of solute is dissolved 100 ml of solution.

Explanation:

so (w/w)% u would need a number to use it.

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What is h2o? I’m in 1st grade

Crazy boy [7]

Answer:

Water

Explanation:

H20 means

H = Hydrogen

2 = 2 Hydrogen atoms

O = Oxygen (1 atom)

Thanks!

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3 years ago

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Mass (g) 4.25 g 7.48 g Initial Volume of Graduated Cylinder (mL) 7 mL 6 mL Final Volume of Graduated Cylinder (mL) 7.5 mL 7 Obje

nata0808 [166]

Answer:

A table was attached to the question

Explanation:

The step by step calculation is as shown

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4 years ago

In a homogeneous mixture, the soluble ingredient that dissolves is the

murzikaleks [220]

The answer would be A. Solute because it is a substance in which the solute is dissolved.

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3 years ago

Sulfur dioxide gas (SO2) reacts with excess oxygen gas (O2) and excess liquid water (H2O) to form liquid sulfuric acid (H2SO4).

inessss [21]

A.2SO₂ + O₂ + 2H₂O → 2H₂SO₄
B.Moles of SO₂ = 67.2 / 22.4Moles of SO₂ = 3 molesMoles of H₂SO₄ = 3 molesMass of H₂SO₄ = 3 x 98Mass of H₂SO₄ = 294 grams
Assuming sulfuric acid to have the same density as water,density = 1000 g / LVolume = mass / densityVolume = 294 / 1000Volume = 0.29 liters of sulfuric acid

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3 years ago

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A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass

valkas [14]

Answer:

$\large \boxed{\text{0.0035 mol/L}}$

Explanation:

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

We know that we will need moles, so, lets assemble all the data in one place.

Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

V/mL: 3.00 7.00

c/mol·L⁻¹: 0.050 0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

(ii) From NH₃

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

$\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}$

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3 years ago