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s2008m [1.1K]
1 year ago
5

Chlorine + sulfur dioxide + water = hydrochloric acid + sulfuric acid​

Chemistry
1 answer:
scZoUnD [109]1 year ago
7 0

The given example is a chemical reaction.

The contents (separated as reactants and products) :

\begin{tabular}{c | l}Reactants & Products \\\cline{1-2}Chlorine & Sulfuric Acid \\Water & Hydrochloric Acid \\Sulfur Dioxide & \\\end{tabular}

The written reaction is :

\boxed {Cl + SO_{2} + H_{2}O \implies HCl + H_{2}SO{4}}

<em>I hope it helped you solve the problem.</em>

<em>Good luck on your studies!</em>

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How many grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water?
dlinn [17]

Answer: Option A) 83.9g

Explanation:

KCl is the chemical formula of potassium chloride.

Given that,

Amount of moles of KCl (n) = ?

Volume of KCl solution (v) = 0.75L

Concentration of KCl solution (c) = 1.5M

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 1.5M x 0.75L

n = 1.125 mole

Now given that,

Amount of moles of KCl (n) = 1.125

Mass of KCl in grams = ?

For molar mass of KCl, use the molar masses of:

Potassium, K = 39g;

Chlorine,Cl = 35.5g

KCl = (39g + 35.5g)

= 74.5g/mol

Since, amount of moles = mass in grams / molar mass

1.125 mole = m / 74.5g/mol

m = 1.125 mole x 74.5g/mol

m = 83.81g

Thus, 83.9 grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water

3 0
3 years ago
A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid. the solid strontium sulfate formed is separated, d
seraphim [82]

Answer:

The formula of the original halide is SrCl₂.

Explanation:

  • The balanced equation of this reaction is:

SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.

  • From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
  • The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
  • The number of moles of SrX are  4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
  • n = mass / molar mass, n =  4.11 x 10⁻³ moles and mass = 0.652 g.
  • The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
  • The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
  • The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2  g/mole = 35.5 g/mole.
  • This is the atomic mass of Cl.
  • <em>So, the formula of the original halide is SrCl₂</em>.
4 0
3 years ago
Help ASAP!!! I need it now i have to write similar test tmrw
dusya [7]
A is Ea, which stands for activating energy. Energy is needed to get the reaction underway and Ea is the energy needed to “start” the reaction.

B is the temperature either released or absorbed.

The diagram shows that the reaction is exothermic based on the fact that the products energy is lower than the reactants. That is because energy (which is temperature in this case) is released during the process. If the reactants would have been lower than the products, the reaction would be endothermic.
3 0
2 years ago
Read 2 more answers
Which conclusion was drawn from the results of
torisob [31]

Answer;

-(2) An atom is mostly empty space.

Experiment

-Rutherford conducted the "gold foil" experiment where he shot alpha particles at a thin sheet of gold. The conclusion that can be drawn from these experiment is that an atom is mostly empty space.

-Rutherford found that a small percentage of the particles were deflected, while a majority passed through the sheet. This caused Rutherford to conclude that the mass of an atom was concentrated at its center, as the tiny, dense nucleus was causing the deflections.

3 0
3 years ago
A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo
Liula [17]

<u>Answer:</u> The molar mass of the insulin is 6087.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = 62.364\text{ L.mmHg }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol

Hence, the molar mass of the insulin is 6087.2 g/mol

8 0
3 years ago
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