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1 year ago

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A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is

the specific heat of this metal

1 answer:

Vlada [557]1 year ago

4 0

Answer:

$c=0.127\ J/g^{\circ} C$

Explanation:

Given that,

Mass of the sample, m = 275 g

It required 10.75 kJ of heat to change its temperature from 21.2 °C to its melting temperature, 327.5 °C.

We need to find the specific heat of the metal. The heat required by a metal sample is given by :

$Q=mc\Delta T$

c is specific heat of the metal

$c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{10.75\times 10^3\ J}{275\times (327.5 -21.2)}\\\\=0.127\ J/g^{\circ} C$

So, the specific heat of metal is $0.127\ J/g^{\circ} C$ .

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Calculate the energy of a photon with a frequency of 9.78 x 1021 Hz.

Alex_Xolod [135]

Answer:

9.98538 kilohertz

Explanation:

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2 years ago

0.446 g of hydrogen gas fills a 5.0 L bag determine the density of hydrogen

Oduvanchick [21]

The density of hydrogen : ρ = 0.0892 g/L

<h3>Further explanation</h3>

Given

mass of Hydrogen : 0.446 g

Volume = 5 L

Required

The density

Solution

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

The unit of density can be expressed in g/cm³, kg/m³, or g/L

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

Input the value :

ρ = m : V

ρ = 0.446 g : 5 L

ρ = 0.0892 g/L

5 0

2 years ago

Help and show work please

olya-2409 [2.1K]

ANWERS ~
We know that :
1 cal (th) = 4.184 J
1 J = 0.2390057361 cal (th) , so :

•55.2 j to cal > 13.193116635 cal
•110 call > 460.24 joule
•65 kj > divide the energy value by 4.184
= 15.535 kilocalories calorie (IT)
——————
Converting form C to F > (F-32)*5/9Understand it better if we have Fahrenheit just add to the equation mentioned to find Celsius.
+to find F to C> (9/5*C)+32

•425 Fahrenheit = (425- 32) × 5/9 =218.33333333 Celsius

•1935 C = 3515 F
———————————-
Converting Celsius to kelvin,We know that :
K = C + 273.15
C = K - 273.15
And from F to K=9/5(F+459.67)
And K to F =(9/5 *k)-459.67
•39.4 Celsius = 312.55 kelvin
•337 Fahrenheit = (337+ 459.67) × 5/9 =442.594 kelvin

4 0

2 years ago

A brick has a mass of 4.0 kg and the Earth has a mass of 6.0 × 1027 g. Use this information to answer the questions below. Be su

Vedmedyk [2.9K]

Answer:

a) $2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

b) $2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

Explanation:

a) Mass of brick = 4.0 kg

1 mole = $N_A=6.022\times 10^{23}$ particles/ atoms/molecules

Mass of $N_A$ bricks :

$=6.022\times 10^{23}\times 4.0 kg=2.4088\times 10^{24} kg\approx 2.4\times 10^{24} kg$

$2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

b)

Mass of the Earth = M = $6.0\times 10^{27} kg$

Mass of 1 mole of brick = m= $2.4\times 10^{24} kg$

Let the moles of brick with equal mass of the Earth be x.

$m\times x=M$

$x=\frac{M}{m}=\frac{6.0\times 10^{27}kg}{2.4\times 10^{24} kg}=2.5\times 10^3$

$2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

7 0

2 years ago

Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) => CH 3OH(l) Using the following information: C(graphite) +

Alika [10]

Answer:

$\Delta H$ for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

$C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)$ $\Delta H_{1}= -393.5 kJ$

$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)$ $\Delta H_{2}= (2\times -285.8)kJ$

$CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)$ $\Delta H_{3}= 726.4 kJ$

---------------------------------------------------------------------------------------------------

$C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)$

$\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ$

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2 years ago