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Marianna [84]
3 years ago
9

At STP, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water?

Chemistry
1 answer:
Delvig [45]3 years ago
7 0
Should be 1.8L.
2 moles of hydrogen react with 1 mole of oxygen. If 2 moles of hydrogen is 3.6L, 1 mole of oxygen should be 1.8L.
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A scientist obtains a set of measurements that are very close to one another. These can be said to be:
pickupchik [31]

Answer:

I believe this is a K-12 test question. If the answers below are what you have on your test . . .

- Precise

- Accurate

- Identical

- None of the above

Then the answer is <u>precise</u>.

3 0
3 years ago
What is the summary at the end of an experiment that explains the results
jasenka [17]

Answer:

The conclusion

Explanation:

6 0
3 years ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2
Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

5 0
2 years ago
What is the overall cell potential for this redox reaction? Ca2+ + 2Li = Ca + 2Li+ Ca2+ + 2e- = Ca -2.87 V Li = Li+ + e- -3.05 V
Jet001 [13]

The redox reaction is

Ca^{+2}  + 2Li(s)  = Ca(s) + 2Li^{+}

Here

Calcium undergoes reduction, and acts as cathode

Lithium undergoes oxidation and acts as anode

The reduction potential of calcium is -2.87 V

The reduction potential of lithium is - -3.05 V

We know that

Ecell = Ecathode - Eanode

Ecell = -2.87 - (-3.05)  = 0.18 V


4 0
4 years ago
What is the weight of a 90kg man standing on the moon, where gmoon = 1.64 m/s2?
Doss [256]

Answer:

The weigth of a 90kg man standing on the moon is <u><em>147.6 N (option C)</em></u>

Explanation:

Weight is called the action exerted by the force of gravity on the body.

The mass (amount of matter that a body contains) of an object will always be the same, regardless of where it is located. Instead, the weight of the object will vary according to the force of gravity acting on it.

The formula that allows you to calculate the weight of any body is:

W = m*g

where:

  • W = weight measured in N.
  • m = mass measured in kg.
  • g = acceleration of gravity measured in m/s². The acceleration of gravity g is the same for all objects that fall due to gravitational attraction, whatever their size or composition. For example, as an approximate value on Earth, g = 9.8 m/s².

In this case,  the mass m has a value of 90 kg and the gravity g has a value of 1.64 m/s², which is the value of the acceleration of gravity of the moon. Then:

W=90 kg* 1.64 m/s²

<u><em>W= 147.6 N</em></u>

Finally, <u><em>the weigth of a 90kg man standing on the moon is 147.6 N (option C)</em></u>

3 0
3 years ago
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