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Margarita [4]
4 years ago
9

PLEASEEEE HELPPPP ASAP

Chemistry
1 answer:
Alexus [3.1K]4 years ago
3 0

Answer:

C

Explanation:

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29.5 g of mercury is heated from 32°C to 161°C, and absorbs 499.2 joules of heat in the process. Calculate the specific heat cap
Finger [1]

Answer:

c = 0.13 j/ g.°C

Explanation:

Given data:

Mass of mercury = 29.5 g

Initial temperature = 32°C

Final temperature = 161°C

Heat absorbed = 499.2 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 161°C - 32°C

ΔT  = 129 °C

Q = m.c. ΔT

c = Q / m. ΔT

c = 499.2 j / 29.5 g. 129 °C

c =  499.2 j / 3805.5 g. °C

c = 0.13 j/ g.°C

5 0
3 years ago
Which elements are<br> considered "Noble Metals"?
Lisa [10]

Answer:

ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).

Explanation:

7 0
3 years ago
A roller coaster track is 3,000 meters long. It takes 100 seconds totravel once around the roller coaster. What is the average s
postnew [5]
The rollercoaster went 200 miles.
3 0
3 years ago
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Explain how much risk you think your area has for water scarcity????
xz_007 [3.2K]
Not very much because half the time the water is purified but just normal water in my area probably has serious levels of scarcity.
3 0
4 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
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