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fredd [130]
3 years ago
13

If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr

avity of the product (to two decimal places)
Chemistry
1 answer:
xxMikexx [17]3 years ago
6 0

The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be \frac{49}{40}=1.225 g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

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Malleable ductile and sonorous
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Consider this initial-rate data at a certain temperature for the reaction described by N2O3(
Yuliya22 [10]

<span><span>N2</span><span>O3</span><span>(g)</span>→NO<span>(g)</span>+<span>NO2</span><span>(g)</span></span>

<span><span>[<span>N2</span><span>O3</span>]</span> Initial Rate</span>
<span>0.1 M     r<span>(t)</span>=0.66</span> M/s
<span>0.2 M     r<span>(t)</span>=1.32</span> M/s
<span>0.3 M     r<span>(t)</span>=1.98</span> M/s

We can have the relationship:

<span>(<span><span>[<span>N2</span><span>O3</span>]/</span><span><span>[<span>N2</span><span>O3</span>]</span>0</span></span>)^m</span>=<span><span>r<span>(t)/</span></span><span><span>r0</span><span>(t)
However,
</span></span></span>([N2O3]/[N2O3]0) = 2

Also, we assume m=1 which is the order of the reaction.

Thus, the relationship is simplified to,

r(t)/r0(t) = 2

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6 0
3 years ago
Read 2 more answers
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
Anna71 [15]

Answer:

31.652g of Na3PO4

Explanation:

We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:

Na3PO4 will dessicate in solution as follow:

Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.

Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e

xM Na3PO4 = (1.10 x 1)/3

xM Na3PO4 = 0.367M

Therefore, the molarity of Na3PO4 is 0.367M.

Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:

Molarity of Na3PO4 = 0.367M

Volume = 525mL = 525/1000 = 0.525L

Mole of Na3PO4 =..?

Molarity = mole /Volume

0.367 = mole /0.525

Cross multiply

Mole of Na3PO4 = 0.367 x 0.525

Mole of Na3PO4 = 0.193 mole.

Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:

Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol

Mole of Na3PO4 = 0.193 mole

Mass of Na3PO4 =.?

Mass = mole x molar mass

Mass of Na3PO4 = 0.193 x 164

Mass of Na3PO4 = 31.652g

Therefore, 31.652g of Na3PO4 is needed to prepare the solution.

6 0
3 years ago
What charge would you expect for an formed by S?
sp2606 [1]

Answer:

in this the correct answer is option 2.

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marin [14]

Answer: 63.26%

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Which is equal to <u>63.26%</u>

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