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Vikki [24]
4 years ago
6

What is the first step of the scientific method?

Chemistry
1 answer:
AleksAgata [21]4 years ago
5 0
The first step of the scientific method would be Observations. 
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If a gas has a volume of 3.67 L and a pressure of 790 mm Hg, what will the pressure be if the volume is compressed to 2.12 L? Wh
oksian1 [2.3K]

If each gas sample has the same temperature and pressure, which has the greatest volume? Since hydrogen gas has the lowest molar mass of the set, 1 g will have the greatest number of moles and therefore the greatest volume. What is the Ideal Gas Law?

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3 years ago
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If the strength of the magnetic field at B is 20 units, the strength of the magnetic field at A is
xxMikexx [17]

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A

Explanation:

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2. Draw a picture of your home. Where does the sun rise relative to your house? Where does the sunset? On the picture, indicate
Semmy [17]

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sun rises in the east so to the right of the front of your house and sets to the left which is south.

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3 years ago
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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
4 years ago
I just started learning about kinetic molecular theory, and I’m not sure how to answer the question circled below
lions [1.4K]

Answer : The value of 'R' is 0.0821\text{ L atm }mol^{-1}K^{-1}

Solution : Given,

At STP conditions,

Pressure = 1 atm

Temperature = 273 K

Number of moles = 1 mole

Volume = 22.4 L

Formula used :     R=\frac{PV}{nT}

where,

R = Gas constant

P = pressure of gas

T = temperature of gas

V = volume of gas

n = number of moles of gas

Now put all the given values in this formula, we get the values of 'R'.

R=\frac{(1atm)\times (22.4L)}{(1mole)\times (273K)}

R=0.0821\text{ L atm }mol^{-1}K^{-1}

Therefore, the value of 'R' is 0.0821\text{ L atm }mol^{-1}K^{-1}.

7 0
3 years ago
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