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3 years ago

Según la ecuación, HNO3 + H2S ----------- NO + S+ H2O el elemento que se oxida es

Chemistry

1 answer:

Serhud [2]3 years ago

5 0

Answer:

HNO3 + H2S = H2O + NO + S - Chemical Equation Balancer. Balanced Chemical Equation. 2 HNO 3 + 3 H 2 S → 4 H 2 O + 2 NO + 3 S. Reaction Information. Nitric Acid + Hydrogen Sulfide = Water + Nitric Oxide (radical) + Sulfur . Reactants.

Explanation:

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Which of the following is the largest volume? 1.2 x 101 m3 1.2 x 108 cm3 2.0 x 104 dm3 1.2 x 108 mm3

alexandr1967 [171]

In order to answer this question, the units of volume must be consistent. In this problem, we decide the unit m3 to be uniform. Option A is equal to 12 m3, option b is equal to 1.2x10^8/100^3 or 120 m3. Option C is 2.0 x10^4/ 10^3 or 20 m3. Option D is 1.2x10^8/ 1000^3 or 0.12 m3. The greatest volume is option b. 120 m3.

7 0

3 years ago

Analysis of an athletes urine found the presence of a compound with a molar mass of 312 g/mol. How many moles of this compound a

rewona [7]

<h3>Answer:</h3>

= 5.79 × 10^19 molecules

<h3>Explanation:</h3>

The molar mass of the compound is 312 g/mol

Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)

We are required to calculate the number of molecules present

We will use the following steps;

<h3>Step 1: Calculate the number of moles of the compound </h3>

$Moles=\frac{mass}{molar mass}$

Therefore;

Moles of the compound will be;

$=\frac{0.030}{312g/mol}$

= 9.615 × 10⁻5 mole

<h3>Step 2: Calculate the number of molecules present </h3>

Using the Avogadro's constant, 6.022 × 10^23

1 mole of a compound contains 6.022 × 10^23 molecules

Therefore;

9.615 × 10⁻5 moles of the compound will have ;

= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules

= 5.79 × 10^19 molecules

Therefore the compound contains 5.79 × 10^19 molecules

5 0

3 years ago

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req

tigry1 [53]

Answer:

the concentration of $Cd^{2+}$ in the original solution= 0.0088 M

the concentration of $Mn^{2+}$ in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample containing Cd2+ and Mn2+ = 50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = $\frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}$

= $\frac{14.2*0.0310}{0.0600}$

= 7.3367 mL

concentration of $Cd^{2+}$ = $\frac{volume \ of \ newly \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}$

= $\frac{7.3367*0.0600}{50}$

= 0.0088 M

Thus the concentration of $Cd^{2+}$ in the original solution = 0.0088 M

Volume of excess unreacted EDTA = $\frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}$

= $\frac{16.1*0.0310}{0.0600}$

= 8.318 mL

Volume of EDTA required for sample containing $Cd^{2+}$ and $Mn^{2+}$ = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for $Mn^{2+}$ = Volume of EDTA required for

sample containing $Cd^{2+}$ and

$Mn^{2+}$ -- Volume of newly freed EDTA

Volume of EDTA required for $Mn^{2+}$ = 55.682 - 7.3367

= 48.3453 mL

Concentration of $Mn^{2+}$ = $\frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}$

Concentration of $Mn^{2+}$ = $\frac{48.3453*0.0600}{50}$

Concentration of $Mn^{2+}$ in the original solution= 0.058 M

Thus the concentration of $Mn^{2+}$ = 0.058 M

6 0

3 years ago

A sample of nitrogen gas had a volume of 500. mL, a pressure in its closed container of 740 torr, and a temperature of 25 °C. Wh

iren2701 [21]

Answer: Thus the new volume of the gas is 530 ml

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 740 torr

$P_2$ = final pressure of gas = 760 torr

$V_1$ = initial volume of gas = 500 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $25^oC=273+25=298K$

$T_2$ = final temperature of gas = $50^oC=273+50=323K$

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

$\frac{740\times 500}{298}=\frac{760\times V_2}{323}$

$V_2=530ml$

Thus the new volume of the gas is 530 ml

6 0

3 years ago

True or False: In every gram of sucrose (table sugar) there is 0.513 g of oxygen

Ber [7]

Answer:

True

Explanation:

Step 1: Find molecular formula of sucrose

C₁₂H₂₂O₁₁

Step 2: Convert moles of oxygen present to grams

1 mol O = 16 g O

11 mol O = 176 g O

Step 3: Find molar mass of sucrose

C - 12.01 g/mol

H - 1.01 g/mol

O - 16.00 g/mol

12.01(12) + 22(1.01) + 11(16.00) = 342.34 g/mol C₁₂H₂₂O₁₁

Step 4: Set up dimensional analysis

$1 gC_{12}H_{22}O_{11}(\frac{176 gO}{342.34gC_{12}H_{22}O_{11}} )$

Step 5: Multiply/Divide and cancel out units

Grams of C₁₂H₂₂O₁₁ and grams of C₁₂H₂₂O₁₁ cancel out.

We are left with grams Oxygen

176/342.34 = 0.514109 grams Oxygen

4 0

3 years ago