Where was hydrogen discovered

Calcium chloride solution reacts with sodium carbonate solution

Klio2033 [76]

Reactions can occur because of a precipitation reaction

<h3>Further explanation</h3>

Double-Replacement reactions, happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

To predict whether this reaction can occur or not is one of them, the precipitation reaction. A precipitation reaction occurs if two ionic compounds which are dissolved reacted to produce one of the products of the ion compound does not dissolve. Formation of these precipitating compounds that cause reactions can occur

Chloride solution reacts with Sodium carbonate solution

CaCl₂(aq)+Na₂CO₃(aq)⇒CaCO₃(s)+NaCl(aq)

Because Calcium carbonate, CaCO₃ an insoluble solid , it will precipitates so that a reaction can occur

8 0

3 years ago

A mixture of so2(g) and o2(g), formed by the complete decomposition of so3(g), is collected over water at 34°c at a total pressu

Sedbober [7]

The decomposition of so3 to so2 gas and o2 gas can be described in the balanced chemical equation:

2so3(g) = 2 so2 (g) + 02(g).

so assuming a complete reaction, the ratio of so2 gas to total products is 2/3 while that of 02 is 1/3.

Subtracting water's water vapor pressure, 760-40 mm hG = 720 mm Hg.

then the products partial pressures are

so2 = 2/3 * (720) = 480 mm Hg.
o2 = 720-480 = 240 mm Hg.

8 0

3 years ago

The osmotic pressure of an aqueoussolution of urea at 300 K is

den301095 [7]

<u>Answer:</u> The freezing point of solution is -0.09°C

<u>Explanation:</u>

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

$\pi=imRT$

where,

$\pi$ = osmotic pressure of the solution = 120 kPa

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

m = concentration of solute in terms of molality = ?

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the solution = 300 K

Putting values in above equation, we get:

$120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=i\times K_f\times m$

where,

i = Vant hoff factor = 1 (for non-electrolytes)

$K_f$ = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

$\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

The equation used to calculate freezing point of solution is:

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

where,

$\Delta T$ = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC$

Hence, the freezing point of solution is -0.09°C

4 0

3 years ago

For each of the following unbalanced chemical equations suppose that exactly 50.0 g of each reactant is taken. Determine which r

Helen [10]

Answer:

1) Br2 is the limiting reactant.

Mass NaBr produced = 64.4 grams

2) CuSO4 is the limiting reactant

Mass Cu = 19.89 grams

Mass ZnSO4 = 50.54 grams

3) NH4Cl is the limiting reactant

Mass NaCl = 54.6 grams

Mass NH3 =15.9 grams

Mass H2O =16.8 grams

4) Fe2O3 is the limiting reactant

Mass Fe = 35.0 grams

Mass CO2 = 41.3 grams

Explanation:

1) Na+br2 ------------->Nabr

Step 1: Data given

Mass Na = 50.0 grams

Mass Br2 = 50.0 grams

Molar mass Na = 22.99 g/mol

Molar mass Br2 = 159.81 g/mol

Step 2: The balanced equation

2Na + Br2 → 2NaBr

Step 3: Calculate moles

Moles = mass / molar mass

Moles Na = 50.0 grams / 22.99 g/mol = 2.17 moles

Moles Br2 = 50.0 grams / 159.81 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

Br2 is the limiting reactant. It will completely be consumed (0.313 moles).

Na is in excess. There will react 2*0.313 = 0.626 moles

There will remain 2.17 - 0.626 = 1.544 moles

Step 5: Calculate moles NaBr

For 1 mol Br2 we'll have 2 moles NaBr

For 0.313 moles we'll have 0.626 moles NaBr

Step 6: Calculate mass NaBr

Mass NaBr = 0.626 moles * 102.89 g/mol

Mass NaBr = 64.4 grams

2) Zn+cuso4 -------------->Znso4+Cu

Step 1: Data given

Mass Zn = 50.0 grams

Mass CuSO4 = 50.0 grams

Molar mass Zn = 65.38 g/mol

Molar mass CuSO4 = 159.61 g/mol

Step 2: The balanced equation

Zn + CuSO4 → Cu + ZnSO4

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol = 0.765 moles

Moles CuSO4 = 50.0 grams / 159.61 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

CuSO4 is the limiting reactant. It will completely be consumed (0.313 moles).

Zn is in excess. There will react 0.313 moles

There will remain 0.765 - 0.313 = 0.452 moles

Step 5: Calculate moles products

For 1 mol Zn we need 1 mol CuSO4 to produce 1 mol Cu and 1 mol ZnSO4

For 0.313 moles CuSO4 we'll have 0.313 moles Cu and 0.313 moles ZnSO4

Step 6: Calculate mass products

Mass Cu = 0.313 moles * 63.546 g/mol = 19.89 grams

Mass ZnSO4 = 0.313 moles * 161.47 g/mol = 50.54 grams

3) NH4cl+NaOH -------------->NH3+H2O+NaCl

Step 1: Data given

Mass NH4Cl = 50.0 grams

Mass NaOH = 50.0 grams

Molar mass NH4Cl = 53.49 g/mol

Molar mass NaOH = 40.0 g/mol

Step 2: The balanced equation

NH4Cl + NaOH → NaCl + NH3 + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles NH4Cl = 50.0 grams / 53.49 g/mol = 0.935 moles

Moles NaOH = 50.0 grams / 40.0 g/mol = 1.25 moles

Step 4: Calculate limiting reactant

NH4Cl is the limiting reactant. It will completely be consumed (0.935 moles).

NaOH is in excess. There will react 0.935 moles

There will remain 1.25 - 0.935 = 0.315 moles

Step 5: Calculate moles products

For 1 mol NH4Cl we need 1 mol NaOH to produce 1 mol NaCl, 1 mol NH3 and 1 mol H2O

For 0.935 moles NH4Cl we'll have 0.935 moles NaCl, 0.935 moles NH3 and 0.935 moles H2O

Step 6: Calculate mass products

Mass NaCl = 0.935 moles * 58.44 g/mol = 54.6 grams

Mass NH3 = 0.935 moles * 17.03 g/mol = 15.9 grams

Mass H2O = 0.935 moles * 18.02 g/mol = 16.8 grams

4) Fe2O3+CO ------------>Fe+CO2

Step 1: Data given

Mass Fe2O3 = 50.0 grams

Mass CO = 50.0 grams

Molar mass Fe2O3 = 159.69 g/mol

Molar mass CO = 28.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe2O3 = 50.0 grams / 159.69 g/mol = 0.313 moles

Moles CO = 50.0 grams / 28.01 g/mol = 1.785 moles

Step 4: Calculate limiting reactant

Fe2O3 is the limiting reactant. It will completely be consumed (0.313 moles).

CO is in excess. There will react 3* 0.313 = 0.939 moles

There will remain 1.785 - 0.939 = 0.846 moles

Step 5: Calculate moles products

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe, 3 moles CO2

For 0.313 moles Fe2O3 we'll have 0.626 moles Fe and 0.939 moles CO2

Step 6: Calculate mass products

Mass Fe = 0.626 moles * 55.845 g/mol = 35.0 grams

Mass CO2 = 0.939 moles * 44.01 g/mol = 41.3 grams

8 0

3 years ago

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Lisa [10]

Answer:

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3 0

3 years ago

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