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7nadin3 [17]
3 years ago
6

How can you tell a density of a reaction?

Chemistry
1 answer:
Lemur [1.5K]3 years ago
3 0
I don't really understand your question but I hope this helps. <span>density = mass divided my volume. 
</span>

You might be interested in
How many molecules of oxygen are needed to make 3.5 oxygens of water?​
max2010maxim [7]

Answer:

57.6g

Explanation:

So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.

8 0
2 years ago
g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0
2 years ago
For the reaction: 3 H2(g) + N2(g) &lt;--&gt; 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

Read more on equilibrium constant;

brainly.com/question/1619133

#SPJ1

5 0
2 years ago
given the following reaction: CuO (s) + H2 (g) to Cu(s) + H2O (g). If 250 L of hydrogen gas are used to reduce copper (II) oxide
QveST [7]

24g is needed no explanation

4 0
2 years ago
Hey hiihhbdndkdkdjdjddjdj
grigory [225]

Answer:

4 cups are about 0.25 gallons

Explanation:

To convert a cup measurement to a gallon measurement, divide the volume by the conversion ratio. The volume in gallons is equal to the cups divided by 16.

7 0
2 years ago
Read 2 more answers
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