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1 month ago

Calculate the heat change in calories for melting 65 g of ice at 0 ∘c.

1 answer:

5 0

When ice transitions to water, the phase changes from solid to liquid. The energy or heat (q) necessary to convert a unit mass (m) of a substance from solid to liquid is called the enthalpy or heat of fusion (ΔHf). The relationship between q, m, and ΔHf is expressed as:

q = m * ΔHf

The ice mass is given as m = 65 g

The heat of fusion of water at 0°C is ΔHf = 334 J/g

Thus, we obtain: q = 65 g * 334 J/g = 21710 J

4.184 J equals 1 cal

meaning: 21710 J = 1 cal * 21710 J/4.184 J = 5188.8 cal

Therefore, the heat required is approximately 5188.8 cal, or about 5.2 Kcal.

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What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

eduard [2782]

Explanation:

The following data has been provided:

Energy of radiation absorbed by the electron in the hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

According to the De-Broglie equation $\lambda = \frac{h}{p}$

with p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$ = $3.6 \times 10^{-26} J/m$

Squaring both sides gives us:

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

$12.96 \times 10^{-52}$ = $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where m = mass of the electron

Therefore, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

1 month ago

A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

Anarel [2989]

Response:

702 mL

To elaborate:

Given the following:

Mass of sodium hydroxide = 13.20 g

Molarity of H₂SO₄ = 0.235 M

We're tasked with finding the volume of acid necessary to neutralize the sodium hydroxide solution

Step 1: Write the balanced reaction equation

The reaction between H₂SO₄ and NaOH can be summarized as follows:

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Step 2: Calculate the moles of NaOH

Moles are calculated by dividing mass by molar mass

The molar mass of NaOH is 40.0 g/mol

Hence;

Number of moles of NaOH = 13.20 g ÷ 40 g/mol

= 0.33 moles of NaOH

Step 3: Determine moles of H₂SO₄ that react

According to the balanced equation, 2 moles of NaOH react with 1 mole of H₂SO₄

Therefore, the ratio giving moles of H₂SO₄ = Moles of NaOH ÷ 2

= 0.33 moles ÷ 2

= 0.165 moles

Step 4: Find the volume of H₂SO₄

Molarity indicates the concentration of the solution in moles per liter

Molarity = Moles ÷ Volume

By rearranging the formula, we find volume = Moles ÷ Molarity

= 0.165 moles ÷ 0.235 M

= 0.702 L

= 702 mL

Thus, the volume of the 0.235 M H₂SO₄ acid solution required equals 702 mL

8 0

24 days ago

doctor has ordered that a patient be given 20 g of glucose, which is available in a concentration of 70.00 g glucose/1000.0 mL o

lions [2927]

The patient needs to receive 285.71 ml.
1000 ml contains 70 gr of glucose.
x contains 20 gr of glucose.

x=1000*20/70

8 0

26 days ago

45.0 g of Ca(NO3)2 are used to create a 1.3 M solution. What is the volume of the solution

Tems11 [2777]

The formula for Molarity is given by:

M = moles / V
To isolate V,
V = moles / M ------------------(1)
Moles can also be calculated as:
moles = mass / M.mass -------------(2)
Substituting the value of moles from equation 2 into equation 1 yields:
V = (mass / M.mass) / M
Plugging in the numbers gives:
V = (45 g / 164 g/mol) / 1.3 mol/dm³

V = 0.21 dm³.

6 0

1 month ago

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Alekssandra [3086]

Answer:

Quantity of $H_{2}O$ generated will be reduced to fifty percent of its initial amount.

Explanation:

Equilibrium reaction: $HCl+NaOH\rightarrow NaCl+H_{2}O$

In accordance with the balanced equation, 1 mol of HCl interacts with 1 mol of NaOH leading to the formation of 1 mol of $H_{2}O$

0.5 mol of HCl interacting with 0.5 mol of NaOH yielding 0.5 mol of $H_{2}O$ .

Consequently, it is clear that the total of $H_{2}O$ produced will be halved if the quantities of the reactants are halved.

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5 0

2 months ago