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Aneli [31]
2 years ago
7

Calculate the mass of liquid with density of 3.3 g/mL and a volume of 25 oz

Chemistry
1 answer:
love history [14]2 years ago
3 0

Answer: 215

Explanation:

Volume doesn’t add up into ounces that’s mass only

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Which formula shows a substance that is not molecular?
Scrat [10]
H ( hydrogen ) is the answer I believe.
8 0
3 years ago
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Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is
lianna [129]

Answer:

21.6 g

Explanation:

The reaction that takes place is:

  • CH₄ + 2O₂ → CO₂ + 2H₂O

First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:

  • 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
  • 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:

  • 0.6 mol CH₄ * \frac{2molH_2O}{1molCH_4} = 1.2 mol H₂O

Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:

  • 1.2 mol * 18 g/mol = 21.6 g
4 0
2 years ago
What is the name of the process that plants use to remove carbon from the atmosphere
Studentka2010 [4]
Photosynthesis is the process by which plants take in carbon and use it for energy and produce oxygen.
8 0
3 years ago
N2H4 + N2O4 --&gt; N2 + H2O
ahrayia [7]

Answer:

  1. The limiting reagent is N2O4
  2. 14,09g

Explanation:

  • First, we adjust the reaction.

2N_{2} H_{4} + N_{2} O_{4} ⇄6N_{2} +  4H_{2}O

  • Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using N_{2} H_{4} to form H_{2}O

               molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4}  }{32,04\frac{g}{mol}  N_{2} H_{4} }

                                           molH_{2} O = 1, 125 mol

Using N_{2} O_{4} to form H_{2} O

              molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol}  H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4}  }{92\frac{g}{mol}  N_{2} O_{4} }

                                           molH_{2} O = 0,783 mol

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴                          MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}

                                      MassOfH_{2}O = 14,09g

5 0
3 years ago
A sample of carbon of mass 250 mg from wood found in a tomb underwent 2480 carbon- 14 disintegrations in 20 h. Estimate the time
Ksenya-84 [330]

Explanation:

Let the age to be found in years is y.

Hence,       (2480 disintegrations) \times \frac{\frac{1.0g}{0.250g}}{1.84 \times 10^{4} disintegrations}

                        = (\frac{1}{2})^{\frac{y}{5730yr}}

Solve for y as follows.

                    0.53913 = (\frac{1}{2})^{\frac{y}{5730yr}}

Now, taking log on both the sides as follows

               log 0.53913 = (\frac{z}{5730}) log \frac{1}{2}&#10;

               \frac{log 0.53913}{log (1/2)} = \frac{z}{5730}

                       z = \frac{5730 \times log 0.53913}{log (1/2)}

                         = 5107 years

Thus, we can conclude that the time since death is 5107 years.

5 0
3 years ago
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