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2 years ago

What is the identity of a sample that has a mass of 2.44 g and a volume of 0.34 cm3?

Chemistry

2 answers:

kramer2 years ago

7 0

Answer:

Zinc is a chemical element with symbol Zn and atomic number 30. Zinc is a slightly brittle metal at room temperature and has a blue-silvery appearance when oxidation is removed. It is the first element in group 12 of the periodic table.

borishaifa [10]2 years ago

3 0

The answer would be zinc

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A 52.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 26.0 mL of KOH at 25 ∘C.

prohojiy [21]

The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7

7 0

2 years ago

How many moles of chromium(iii) nitrate are produced when chromium reacts with 0.85 moles of lead(iv) nitrate to produce chromiu

Alexxandr [17]

The moles of chromium (iii) nitrate produced is calculated as follows

write the equation for reaction

3 Pb(NO3)2 + 2 Cr = 2 Cr(NO3)3 + 3 Pb

by use of mole ratio between Pb(NO3)2 to Cr(NO3)3 which is 3 : 2 the moles of Cr(NO3)3 is therefore
= 0.85 x2 /3 = 0.57 moles

5 0

2 years ago

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

Answer: The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0

2 years ago

Sodium acetate spontaneously crystallizes out of a supersaturated solution upon standing or upon the addition of a seed crystal.

AlekseyPX

Answer:

Option (c) is correct.

Explanation:

According to 2nd law of thermodynamics, every spontaneous process is associated with increase in entropy ( $\Delta S>0$ ).

As sodium acetate crystallizes out spontaneously therefore it is associated with $\Delta S>0$ .

During crystallization, bond formation takes place which results evolution of heat. In other words, crystallization phenomenon is exothermic ( $\Delta H$ ).

So, the given crystallization process renders the following thermodynamic quantities- $\Delta S>0$ J/K and $\Delta H$ J

Option (c) is correct.

6 0

2 years ago