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nata0808 [166]
3 years ago
7

What does it mean for an element to be oxidized?

Chemistry
1 answer:
Delicious77 [7]3 years ago
7 0
Oxgen has been added to the element
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What is the maximum amount of HCl, in grams, that can be produced if 37.5 g of BCl3 and 60.0 g of H2O are reacted according to t
Rus_ich [418]
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.

First, how many moles of each substance are there

the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.

Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.

But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.

.96 x 36.46 = ~35 g</span>
6 0
3 years ago
The combustion of hydrogen and oxygen is commonly used to
posledela

Answer:

6.66 mol

Explanation:

(atm x L) ÷  (0.0821 x K)

(0.875 x 250) ÷  (0.0821 x 400)

=6.66108

5 0
3 years ago
If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
An experiment was conducted to determine the density of a rock sample. The table shows a partial record of the experiment.
mart [117]

Answer: the density of the rock

55.91g/ml

Explanation:

The density of the rock after it had being placed in the cylinder of water so the calculation should look like this:

Volume of water substract the mass of the rock:

And that is 142.5 ml - 86.59g =

Answer 55.91 g/ml

So the density of the rock is 55.91g/ml

5 0
3 years ago
What might you expect for the value of S∘ (entropy) for butane, C4H10?
Norma-Jean [14]

Answer:

340 J·K⁻¹mol⁻¹  

Explanation:

I looks like the standard entropy increases approximately linearly with the number of C atoms.

I plotted S° vs the number of C atoms and got the graph shown below.

It appears that S° for four carbon atoms should be about 340 J·K⁻¹mol⁻¹.·

4 0
3 years ago
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