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Leno4ka [110]
2 years ago
10

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

1.0°C, what is the specific heat of the substance?
Chemistry
1 answer:
Alexxandr [17]2 years ago
5 0
Specific heat capacity is the required amount of heat per unit of mass in order to raise teh temperature by one degree Celsius. It can be calculated from this equation: H = mCΔT where the H is heat required, m is mass of the substance, ΔT is the change in temperature, and C is the specific heat capacity.

H = m<span>CΔT
2501.0 = 0.158 (C) (61.0 - 32.0)

C = 545.8 J/kg</span>·°C
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Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to
Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0
2 years ago
what is 31/50000 If the toxic quantity is 1.5 g of ethylene glycol per 1000 g of body mass, what percentage of ethylene glycol i
Sveta_85 [38]

Answer:

\%m=0.15\%

Explanation:

Hello,

In this case, we are asked to compute the by mass percent representing the toxicity of ethylene glycol in the body mass. In such a way, since the by mass percent is computed with the shown below formula:

\%m=\frac{m_{ethylene \ glycol}}{m_{ethylene \ glycol}+m_{body\ mass}}*100\%

We can use the given masses to obtain:

\%m=\frac{1.5g}{1.5g+1000g}*100\%\\ \\\%m=0.15\%

Best regards.

7 0
2 years ago
Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

KH_2PO_4 acts as a weak acid and Na_2HPO_4 as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

7.00=6.86+log(\frac{base}{0.10})

7.00-6.86=log(\frac{base}{0.10})

0.14=log(\frac{base}{0.10})

Taking antilog:

10^0^.^1^4=\frac{base}{0.10}

1.38=\frac{base}{0.10}

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is Na_2HPO_4 required to make the buffer is 0.138M.

5 0
2 years ago
A precipitate of zinc hydroxide can be formed using the reaction below.
worty [1.4K]

Answer:

Option B is correct. KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

Explanation:

Step 1: Data given

Number of moles ZnCl2 = 0.36 moles

Number of moles KOH  = 0.54 moles

Step 2: The balanced equations

ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

Step 3: Calculate the limiting reactant.

KOH is the limiting reactant. It will completely be consumed (0.54 moles). ZnCl2 is in excess. There will react 0.54/2 = 0.27 moles

There will remain 0.36 - 0.27 = 0.09 moles.

Step 4: The products

There will be produced 2 moles KCl and 1 mol Zn(OH)2. Zn(OH)2 is the precipitate produced.

For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl

For 0.54 moles KOH, we will produce 0.27 moles precipitate (Zn(OH)2)

Option A is not correct because 0.27 mol of Zn(OH)2 will precipitate, not 0.54 mol

Option B is correct

Option C is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

Option D is not correct because ZnCl2 is not the limiting reactant, but the excess reactant

7 0
2 years ago
When a pH probe is inserted into a solution containing the chloride ion it is neutral. What is the pH of a solution containing t
vivado [14]

Answer:

a. the solution will be weakly basic.

b. Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.

Explanation:

a. The fluoride ion (F⁻) reacts with water thus:

F⁻ + H₂O → HF + OH⁻

That means that fluoride ions produce OH⁻ ions in solution doing <em>the solution will be weakly basic.</em>

b. The acidic equilibrium of NH₄⁺ is:

NH₄⁺ ⇄ NH₃ + H⁺ with a ka of 5,6x10⁻¹⁰.

The basic equilibrium of CN⁻ is:

CN⁻ + H₂O → HCN + OH⁻ with a kb of 2x10⁻⁵

That means that the production of OH⁻ from CN⁻ is higher than production of H⁺ from NH₄⁺. The CN⁻ is a stronger base than NH₄⁺ is an acid.

Thus, the pH of a salt solution of NH₄CN would be <em>Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.</em>

<em></em>

I hope ot helps!

3 0
2 years ago
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