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3 years ago

Na2S2O3 (aq) + 2HNO3 (aq) →

Chemistry

1 answer:

Bezzdna [24]3 years ago

3 0

Answer: The correct answer is Option a.

Explanation:

The reaction between sodium thiosulfate and nitric acid produces a number of products. The balanced chemical equation for the reaction of two follows:

$Na_2S_2O_3(aq.)+2HNO_3(aq.)\rightarrow 2NaNO_3(aq.)+H_2O(l)+SO_2(g)+S(s)$

By Stoichiometry of the reaction:

1 mole of sodium thiosulfate reacts with 2 moles of nitric acid to produce 2 moles of sodium nitrate, 1 mole of water, 1 moles of sulfur dioxide and 1 mole of solid sulfur.

Hence, the correct answer is Option a.

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"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

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4 years ago

At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume

Natalija [7]

Answer: Volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

Explanation:

Given: $T_{1} = 35^{o}C = (35 + 273) K = 308 K$ , $V_{1}$ = 256 mL,

$P_{1}$ = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

$T_{1} = 22^{o}C = (22 + 273) K = 295 K$ , $P_{2} = 1.25 atm$

Formula used to calculate volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL$

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

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3 years ago

Why does a 4 carbon linear alkane, butane c4h10, have two more hydrogens ?

Scilla [17]

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3 years ago