What is the name of the compound Hg2O7

The tables show the mass, volume, and appearance of four unknown substances and the densities of four known substances.

Kazeer [188]

Answer:

Substance S is GLYCEROL.

Explanation:

To know what substance S represent, let us determine the density of substance S.

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

Mass of S = 92.09 g

Volume of S = 73.09 mL

Density of S =..?

Density = mass /volume

Density of S = 92.09/73.09

Density of S = 1.26 g/mL

Finally, we shall determine what substance S represent as follow:

Comparing the density of substance S (i.e 1.26 g/mL) with those given in question above, substance S is GLYCEROL.

4 0

3 years ago

What is S for the reaction N2(g) +O2(g) 2NO? Use G = 173.3 kJ at 303.0 K and H = 180.7 kJ. Use G = H – TS.

Rashid [163]

Given information :

G = 173.3 KJ

H = 180.7 KJ

T = 303.0 K

S = unknown (?)

By using the given formula : G = H - TS , we can calculate the value of 'S'

On rearranging the formula we get : S = $\frac{H-G}{T}$

Plug in the value of G , H and T in the above formula :

S = $\frac{(180.7 KJ - 173.3KJ)}{303.0K}$

S = 0.02442 $\frac{KJ}{K}$

4 0

3 years ago

Read 2 more answers

If 5.0 grams of sucrose, C12H22O11, are dissolved in 10.0 grams of water, what will be the boiling point of the resulting soluti

GaryK [48]

Answer : The boiling point of the resulting solution is, $100.6^oC$

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

or,

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^oC$

$k_b$ = boiling point constant = $0.52^oC/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute (sucrose) = 5.0 g

$w_1$ = mass of solvent (water) = 10.0 g

$M_2$ = molar mass of solute (sucrose) = 342.3 g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}$

$T_b=100.6^oC$

Therefore, the boiling point of the resulting solution is, $100.6^oC$

5 0

3 years ago

Balance the following reaction and use the equation to calculate the ΔHrxn. C3H8(g) + O2(g) → CO2(g) + H2O(g) Round your answer

gladu [14]

The reaction corresponds to the combustion of propane (C3H8). The balanced reaction is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

The reaction enthalpy is given as:

ΔHrxn = ∑nΔH°f(products) - ∑ nΔH°f(reactants)

= [3ΔH°f(CO2(g)) + 4ΔH°f(H2O(g)] - [1ΔH°f(C3H8(g)) + 5ΔH°f(O2(g)]

= [3(-393.5) + 4(-241.8)] - [-103.9 + 5(0)] = -2043.8 kJ

The enthalpy for the combustion of propane is -2044 kJ

4 0

3 years ago

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

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pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

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The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

3 0

3 years ago