What evidence is there that energy is conserved?

Does this sound catchy or invite you to read more? "GMOs and pesticides are general amongst farmers for generations and should n

siniylev [52]

Answer:

It sounds fine, but it may be a bit too long. It's difficult to shorten things like this, but getting more straight to the point would give it that "catchy" feel.

Explanation:

3 0

3 years ago

A solution contains an unknown amount of dissolved magnesium. Addition of

Scrat [10]

Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Mg²⁺: 1 mole
Na₂CO₃: 1 mole
MgCO₃: 1 mole
Na⁺: 2 moles

The molar mass of the compounds is:

Mg²⁺: 24.3 g/mole
Na₂CO₃: 106 g/mole
MgCO₃: 84.3 g/mole
Na⁺: 23 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
Na₂CO₃: 1 mole ×106 g/mole= 106 grams
MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
Na⁺: 2 moles ×23 g/mole= 46 grams

<h3>Mass of magnesium dissolved</h3>

The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?

$mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}$

<u><em>mass of magnesium= 2.13 grams</em></u>

Finally, 2.13 grams of magnesium was dissolved in the solution.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

4 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C

Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = $\frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\$

Now use the molar ratio to find how many moles of NaOH would be required to neutralize $5.5537*10^{-3} mol\\$ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

1 : 2 (we get this from the balanced equation)

$5.5537*10^{-3} mol\\$ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

$c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L$

5 0

4 years ago

Lucia wants to develop a process for dyeing shirts so that the color will not fade when the shirt is washed. She proceeds with t

Firlakuza [10]

I cant because it not full

3 0

4 years ago

Read 2 more answers