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malfutka [58]
3 years ago
12

How manyy grams of h2 are needed to react with 20 grams of o2. please show work

Chemistry
1 answer:
tamaranim1 [39]3 years ago
6 0

Would it be 10, you know 1h for every 2o, H2O


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3. A student measured 15.0 grams of ice in a beaker. The beaker was then
gregori [183]

Answer:

The heat that was used to melt the 15.0 grams of ice at 0°C is 4,950 Joules

Explanation:

The mass of ice in the beaker = 15.0 grams

The initial temperature of the ice = 0°C

The final temperature of the ice = 0°C

The latent heat of fusion of ice = 330 J/g

The heat required to melt a given mass of ice = The mass of the ice to be melted × The latent heat of fusion of ice

Therefore, the heat, Q, required to melt 15.0 g of ice = 15.0 g × 330 J/g = 4,950 J

The heat that was used to melt the 15.0 grams of ice = 4,950 Joules.

3 0
3 years ago
I need to know the measurements of this to the appropriate amount of significant figures
igomit [66]

Answer:

[See Below]

Explanation:

I'd say 44 something. It's probably ml but I can't see what it says on the tube.

3 0
3 years ago
How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant
schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

  • 1 g = 1,000 mg
  • The molar mass of Cu is 63.55 g/mol
  • When 1 mole of Cu is deposited, 2 moles of electrons circulate.
  • The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
  • 1 A = 1 C/s

The time  that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s

3 0
3 years ago
When are igneous rocks formed?
balandron [24]

Answer:

C. When melted rock solidifies.

3 0
2 years ago
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate
AlladinOne [14]

The given question is incomplete. The complete question is as follows.

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Explanation:

When Ba^{2+} and Na_{2}SO_{4} are added  then white precipitate forms. And, reaction equation for this is as follows.

       Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}

It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of BaSO_{4} is 233.43.

Now, we will calculate the number of moles as follows.

  No. of moles = mass × M.W

                        = \frac{0.212}{233.43}

                        = 0.00091 mol of BaSO_{4}

Hence, it means that 0.00091 mol of Ba^{2+}. Now, we will calculate the mass as follows.

       Mass = moles × MW

                 = 0.00091 \times 137.327

                 = 0.124 grams or 124 mg of barium

Thus, we can conclude that mass of barium into the original solution is 124 mg.

8 0
3 years ago
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