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Igoryamba
1 year ago
7

Material A has a small latent heat of fusion. Material B has a large heat of fusion. Which of the following statements is true?

(assume equal masses of both materials) material A will take less energy to phase change to a liquid material A will take more energy to phase change to a liquid material A will take less energy to phase change to a gas material A will take more energy to phase change to a gas
Chemistry
2 answers:
neonofarm [45]1 year ago
8 0

Answer: Material A will take less energy to phase change to a liquid material.

Explanation:

The energy provided to change the enthalpy of a substance is known as latent heat of fusion. This energy actually helps in changing the state of a substance that is from solid to liquid or vice versa.

Hence, a substance with small latent heat of fusion will require less energy to change its phase. Whereas a substance with large heat of fusion will require large energy to change its phase.

There, it can be concluded that material A will take less energy to phase change to a liquid material

Olin [163]1 year ago
7 0
The correct answer would be the first option. Material A having a smaller latent heat of fusion would mean that it will take only less energy to phase change into the liquid phase. Latent of heat of fusion is the amount of energy needed of a substance to phase change from solid to liquid or liquid to solid.
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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
1 year ago
Use the specific heat capacity that you calculated for granite to determine how many grams of granite at the initial temperature
Valentin [98]

Answer:

Explanation:

specific heat of granite s = .79 J / g / k

let the mass of granite = m

heat lost by granite = heat gained by water

heat lost = mass x specific heat x drop in temperature

= m x .79 x (80 - 20.45)

heat gained by water

= 3000 x 4.186 x (20.45- 20)

heat lost by granite = heat gained by water

m x .79 x 59.55  =  3000 x 4.186 x .45

m = 120.12 g .

4 0
2 years ago
What is the mass of solute in 200.0 L of a 1.556-M solution of KBr
sergejj [24]
The molarity of KBr solution is 1.556 M
molarity is defined as the number of moles of solute in volume of 1 L solution.
the number of KBr moles in 1 L - 1.556 mol
Therefore in 200.0 L - 1.556 mol/L x 200.0 L = 311.2 mol
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4 0
1 year ago
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
1 year ago
A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista
baherus [9]

Explanation:

Formula for work done is as follows.

           W = -k \frac{q_{1}q_{2}}{d}    

where,  k = proportionality constant = 8.99 \times 10^{9} Jm/C^{2}

            q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C

            q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C

            d = separation distance = 0.45 nm = 0.45 \times 10^{-9} m

Now, we will put the given values into the above formula and calculate work done as follows.

         W = -k \frac{q_{1}q_{2}}{d}    

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           = 3.68 \times 10^{-18} J

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is 3.68 \times 10^{-18} J.

7 0
2 years ago
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