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ddd [48]
2 years ago
10

Calculate the volume of 0.10 g of titanium (4.51 g/cm³).

Chemistry
1 answer:
Juli2301 [7.4K]2 years ago
4 0

The volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

<h3>What is volume?</h3>

Volume is known to be equal to the mass divided by the density.

It is written thus:

Volume = Mass / density

<h3>How to calculate the volume</h3>

The volume is calculated using the formula:

Volume = mass ÷ density

Given the mass = 0. 10g

Density = 4.51 g/cm³

Substitute the values into the formula

Volume of titanium = 0. 10 ÷ 4.51 = 0. 02 cm³

Thus, the volume of titanium with mass of 0. 10g and density of 4. 51 g/cm³ is 0. 02 cm³

Learn more about volume here:

brainly.com/question/1762479

#SPJ1

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Calculate the pH of a buffer solution prepared by mixing 60.0 mL of 1.00 M lactic acid and 25.0 mL of 1.00 M sodium lactate.
marshall27 [118]
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:

pH = pKa + log[(salt/acid]

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Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
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4 0
3 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

7 0
3 years ago
Choose all the answers that apply.
Stolb23 [73]

Answer:

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Answer:

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\displaystyle \Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}

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3 years ago
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