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DiKsa [7]
2 years ago
10

5. The heat of fusion of lead is 25 J/g and its melting point is 601 K. How much heat is given off as 3.0 g

Chemistry
1 answer:
PIT_PIT [208]2 years ago
5 0

Answer:

the heat given off is 75 J.

Explanation:

Given;

latent heat of fusion of lead, L= 25 J/g

mass of liquid lead, m = 3.0g

The heat given off is calculated as;

H = Lm

Where;

H is the quantity of heat given off

H = 25 x 3

H = 75 J.

Therefore, the heat given off is 75 J.

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A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num
-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}

<em>B) Radius</em>

\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}

V = \frac{ 4}{3 }\pi r^{3}

r^{3} = \frac{3V }{4 \pi }\

r= \sqrt [3]{ \frac{3V }{4 \pi } }

r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}

3 0
3 years ago
What is the amount of moles for 12.0 g Ar?
Digiron [165]

if i am correct it shall be 12. because i am thinking, 1 mole = 1 ar.

8 0
3 years ago
Read 2 more answers
The limestone used in building the Egyptian pyramids was used because it is _____________
kvasek [131]

Answer:

hard and brittle

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3 years ago
Read 2 more answers
1.
Makovka662 [10]

Answer:

1) 70

2) conduction

3) The ground absorbs the solar radiation and

releases it throughout the later afternoon.

4) transpiration

5) convection

6) cumulonimbus

7) reflectivity of Earth’s surface

8) As temperature increases, air pressure

increases.

9) thermosphere

10) convection cells, pressure belts, and the

Coriolis effect

11) radiation

12) pollen

13) scattering

14) thermosphere

15) The different types of clouds are:

<u>cumulus:</u><em><u> </u></em><em>Cumulus clouds are clouds which have flat bases and are often described as "puffy", "cotton-like" or "fluffy" in appearance. Their name derives from the Latin cumulo-, meaning heap or pile. </em>

<u>cirrus</u><em>:</em><em> </em><em>Cirrus is a genus of atmospheric cloud generally characterized by thin, wispy strands, that typically appear white or light grey. The name is derived from the Latin word cirrus, meaning 'ringlet' or 'curling lock of hair'. Such a cloud can form at any altitude between 5,000 and 13,700 m above sea level.</em>

<u>stratus:</u><em> </em><em>Stratus clouds are low-level clouds characterized by horizontal layering with a uniform base, as opposed to convective or cumuliform clouds that are formed by rising thermals.</em>

<u>nimbus</u>: <em>Nimbostratus clouds are dark, grey, featureless layers of cloud, thick enough to block out the Sun and produce persistent rain. Height of base: 2,000 - 10,000 ft. Shape: Bands or areas of individual cells. Latin: nimbus - rainy cloud; stratus - flattened or spread out. Precipitation: Continuous rain or snow likely.</em>

16) the tradewinds

17) the location’s latitude

18) Cirrus clouds

19) the ozone layer

20) Global winds are created by both the spin of

the Earth (Coriolis effect) and the differences

in temperature between the equator and

the polar areas. These winds are often

grouped together as trade winds, easterlies,

and westerlies.

21) nitrogen

22) the Miller-Urey hypothesis

the Oparin-Haldane hypothesis

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25) It determines the mass of water in a volume of air

Explanation:

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7 0
2 years ago
A total of 20.0 mL of sodium hydroxide (NaOH) was neutralized by 30.0 mL of 0.250 M hydrogen bromide (HBr). What was the concent
Karolina [17]

Answer:

0.375 M

Explanation:

NaOH(aq) + HBr(aq) ------------> NaBr(aq) + H2O(l)

Concetration of acid CA= 0.250M

Concentration of base CB= ????

Volume of acid VA= 30.0mL

Volume of base VB= 20.0mL

Number of moles of acid nA= 1

Number of moles of base nB= 1

CA VA/CB VB= nA/nB

CB= CAVAnB/VB nA

= 0.25× 30×1/20×1= 0.375 M

8 0
3 years ago
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