Which deflects winds to the right or left

An unbalanced equation is shown. In this reaction, 200.0 g of FeS2 is burned in 100.0 g of oxygen, and 55.00 g of Fe2O3 is produ

Crazy boy [7]

<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>

Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.

Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3

To find for the theoretical yield, we first determine the limiting reactant.

100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2

Therefore, the limiting reactant is O2.

Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3

Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%

3 0

3 years ago

Surrounding

umka2103 [35]

Based on the data given in this question, the statement that shows a correct interpretation of the chemical reactions is as follows: reaction A was exothermic and reaction B was endothermic.

<h3>What are endothermic and exothermic reactions?</h3>

Endothermic reaction is a chemical reaction that absorbs heat energy from its surroundings while exothermic reaction is a reaction that releases energy in the form of heat.

Endothermic reactions leave their surroundings cooler while exothermic reactions leave their surroundings hotter.

According to this question, the initial and final temperatures of two reactions are given as follows:

Reaction A: 25.1°C and 30.2°C
Reaction B: 25.1°C and 20.0°C

From the above data, reaction A was exothermic because it increased the surrounding temperature and reaction B was endothermic because it reduced the surrounding's temperature.

Learn more about endothermic and exothermic at: brainly.com/question/23184814

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4 0

2 years ago

A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot

kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{O_2}$ = partial pressure of $O_2$ = ?

$\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}$

$P_{T}$ = total pressure of mixture = 525 mmHg

${\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles$

${\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles$

Total moles = 1.94 + 1.35 = 3.29 moles

$\chi_{O_2}=\frac{1.94}{3.29}=0.59$

$p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg$

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0

3 years ago

How many moles of electrons must be transferred to plate out 110 g of manganese (MW ~ 55g/mol) from a solution of permanganate (

ololo11 [35]

Answer:

14 mol e⁻

Explanation:

Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese

8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)

Step 2: Calculate the moles corresponding to 110 g of manganese

The molar mass of Mn is 55 g/mol.

110 g × 1 mol/55 g = 2 mol

Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn

According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.

2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻

8 0

3 years ago

HNO3 and H2CO3 are examples of ?

Basile [38]

A) acids because they start with h

7 0

3 years ago