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Sunny_sXe
13 days ago
11

If the same amount of heat is added to 50.0 g samples of each of the metals which are all at the same temperature, which metal w

ill reach the highest temperature?
Copper 0.385 J/gºC
Magnesium 1.02 J/gºC
Mercury 0.138 J/g °C
Silver 0.237 J/g °C
Lead 0.129 J/gºC
a. Copper
b. Magnesium
c. Mercury
d. Silver
e. Lead
Chemistry
1 answer:
Tems11 [2.7K]13 days ago
4 0
e. Lead. In this scenario, since the computation of heat in a heating or cooling process utilizes specific heat, it is evident that a lower specific heat results in a higher achieved temperature, given that they share an inversely proportional relationship. Thus, Lead will achieve the highest temperature among the mentioned metals when subjected to the same amount of heat.
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A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma
alisha [2963]
a
8 0
12 days ago
How many micrograms of iron were in the 8.0 mL sample of Greg's blood?
castortr0y [3046]
The result is: 3.36 micrograms of iron in<span> Greg's blood sample.
</span>m(Fe) = 42 mcg(micrograms).
V(Fe) = 1 dL = 1 dL · 100 mL/1dL.
V(Fe) = 100 mL.
Using proportions: m(Fe): 8 mL = 42 mcg: 100 mL.
Thus, 100 mL · m(Fe) = 8 mL · 42 mcg.
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4 0
1 month ago
Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.
Tems11 [2777]

Answer:

vHe / vNe = 2.24

Explanation:

To determine the velocity of an ideal gas, one should apply the formula:

v = √3RT / √M

In this equation, R represents the gas constant (8.314 kgm²/s²molK); T refers to temperature, and M indicates the molar mass of the gas (4x10⁻³kg/mol for helium and 20.18x10⁻³ kg/mol for neon). Hence:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio simplifies to:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

vHe / vNe = 2.24

I hope it assists you!

8 0
1 month ago
Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d
KiRa [2933]

Response: k = 23045 N/m

Clarification:

To determine the spring constant, one must consider the maximum elastic potential energy that the spring can withstand. The kinetic energy of the vehicle should equal at minimum the elastic potential energy of the spring when it is fully compressed. Hence, we express it as:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the vehicle = 1050 kg

k: spring constant =?

v: car speed = 8 km/h

x: maximum spring compression = 1.5 cm = 0.015m

You need to resolve equation (1) for k. Beforehand, convert the speed v to meters per second:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant calculates to 23045 N/m

3 0
1 month ago
In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a
KiRa [2933]

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

\Delta H =\frac{Q}{n}

To calculate the heat (Q):

Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J

Next, we need to calculate the number of moles involved in the reaction (n):

n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles

With these two values, we can substitute them into the first equation:

\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole

Have a great day!

5 0
1 month ago
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