Two products of stars

snow_lady [41]

Answer:

1st one is...

Explanation:

The 1st one is change in size or shape. 2nd one is formation of precipitate. 3rd one is physical change. 4th one is formation of gas.

5 0

3 years ago

Identify the following elements based on their chemical symbols:

abruzzese [7]

Answer:C:Carbon

H:Hydrogen

O:Oxygen

Explanation:Every element is represented by a letter that is easy to tell which element it is so you don't have to get confused.

3 0

3 years ago

Describe how the molecular models you assembled are similar to real molecules.

mario62 [17]

Answer:

Atoms models are bigger but they don not match the true or positive way of atoms. to make them visible scientists make their models that resembled with the real ones.

Explanation:

Molecules are microscopic entities and only can be seen under a microscope.

4 0

3 years ago

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311

LuckyWell [14K]

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

5 0

4 years ago

Problem page gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o

iVinArrow [24]

$m(\text{CO}_2) = 2.24 \; \text{g}$

Ethene react with oxygen at a $2 : 7$ molar ratio:

$2\; \text{C}_2 \text{H}_6 (g) + 7\; \text{O}_2 (g) \to 6\; \text{H}_2{O} (g) + 4\; \text{CO}_2 (g)$

Convert the quantity of each reactant supplied to number of moles of particles:

$n(\text{C}_2\text{H}_6) = 0.60 \; \text{g} / 28.05 \; \text{g} \cdot \text{mol}^{-1} = 0.0214 \; \text{mol}$
$n(\text{O}_2) = 3.27 \; \text{g} / 32.00 \; \text{g} \cdot \text{mol}^{-1} = 0.102 \; \text{mol}$

The question stated not whether both reactants were used up in this process. Thus start by testing the assumption that e.g., ethene was used up while some oxygen gas were left unreacted (ethene as the <em>limiting </em>reagent.) Under this assumption, the relative availability of the two species, $n(\text{C}_2 \text{H}_6) /2$ and $n(\text{O}_2) /7$ (as seen in the balanced chemical equation) shall satisfy the relationship

$n(\text{O}_2) / 7 - n(\text{C}_2 \text{H}_6) / 2 > 0$

In other words,

$n(\text{O}_2)/7 > n(\text{C}_2 \text{H}_6)/2$

$n(\text{C}_2 \text{H}_6) / n(\text{O}_2) < 2/7 \approx 0.286$

Evaluating the expression $n(\text{C}_2 \text{H}_6) / n(\text{O}_2)$ with data given in the question yields approximately $0.210 < 0.286$ , which does satisfy the relationship. Hence the assumption holds and ethene is the limiting reactant.

The quantity of a reactant produced in a chemical reaction is related to its stoichiometric (of relating to proportions) relationship with the limiting reactant (or any of the reactants in case of more than one limiting reactant.) For this scenario, given the molar ratio $n(\text{C}_2\text{H}_6) : n( \text{CO}_2) = 2:4$ ,

$n(\text{CO}_2) = n(\text{C}_2\text{H}_6) \cdot (2 / 4) = 0.0510 \; \text{mol}$

$m(\text{CO}_2) = 0.0510 \; \text{mol} \times 44.01 \; \text{g} \cdot \text{mol}^{-1} = 2.24 \; \text{g}$

4 0

3 years ago