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2 years ago

If 14.5 kJ of heat were added to 485 g of liquid water, How much would its temperature increase?

Chemistry

1 answer:

olga55 [171]2 years ago

3 0

<u>Answer:</u> The increase in temperature is 7.14°C

<u>Explanation:</u>

To calculate the change in temperature, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed = 14.5 kJ = 14500 J (Conversion factor: 1 kJ = 1000 J)

m = mass of water = 485 g g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = ?

Putting values in above equation, we get:

$14500J=485g\times 4.184J/g.^oC\times \Delta T\\\\\Delta T=7.14^oC$

Hence, the increase in temperature is 7.14°C

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How could installing new technology, such as scrubber machines, affect the factories required to install them? Name a positive a

Sindrei [870]

Answer:

Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.

Explanation:

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2 years ago

A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me

DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0

2 years ago

Read 2 more answers

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

2 years ago

A common way of initiating certain chemical reactions with light involves the generation of free halogen atoms in solution. if δ

STatiana [176]

Answer: The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

Explanation:

$Cl_2\overset{h\nu}\rightarrow Cl^-,Delta H_{rxn}=242.8kJ/mol$

Energy required to produce free chlorine atoms from one mole of chlorine gas :

= 242.8kJ = $242.8\times 1000=242800 Joules$ (1kJ=1000J)

1 mole = $6.022\times 10^{23}$ molecules

For $6.022\times 10^{23}$ molecules = 242,800 Joules

For one molecule of chlorine gas = $\frac{242800 Joules/mol}{6.022\times 10^{23} mol^{-1}}=40,318.83\times 10^{-23}Joules$

According to photoelectric equation:

$E=h\nu=\frac{hc}{\Lambda }$

E = Energy of the photon of light used to produce free chlorine atoms

$\nu$ = frequency of the light used to produce free chlorine atoms

h = Planck's constant = $6.626\times 10^{-34}J.s$ , c = speed of light= $3\times 10^8 m/s$

$\lambda$ = wavelength of the light used to produce free chlorine atoms

$40,318.83\times 10^{-23}J=\frac{hc}{\Lambda }=\frac{6.626\times 10^{-34} J.s\times 3\times 10^8 m/s}{\lambda }$

$\lambda=0.0004930203\times 10^{-3} m=493.0203\times 10^{-9} m=493 nm$

The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

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2 years ago

Products or solutions found at home or in store and their characteristics

AleksAgata [21]

Answer:

bruh, just go in your bathroom and look for cleaning products

Explanation:

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2 years ago