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3 years ago

Which of the following statement regarding orbits is true?

Physics

2 answers:

BigorU [14]3 years ago

4 0

Answer:

A is correct

Explanation:

Tatiana [17]3 years ago

3 0

Answer:

sorry i mean c

Explanation:

apex i got it right

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

3 years ago

What parts of an atom can change during a nuclear reaction that cannot change during a chemical reaction? Question 15 options: A

viva [34]

Neutrons and protons cannot be removed from nucleus from chemical reactions because they are strongly held together but nuclear reactions are strong enough to separate them. Hence option A is correct.

6 0

3 years ago

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Ella has a mass of 56 kg, and Tyrone has a mass of 68 kg. Ella is standing at the top of a skateboard ramp that is 1.5 meters ta

Ivahew [28]

Answer:Tyrone is 12 more kg more than Ella.

Explanation:

7 0

4 years ago

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What is the weight in newtons of a 10 kg mass on the earth's surface?

juin [17]

Answer:

98 kg

Mass is given as 10 kg. Therefore, Weight = 10 kg * 9.8 m/s^2. Weight = 98 kg.m/s^2. = 98 Newtons.

Explanation:

plz mark me brainleast

5 0

4 years ago

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Applications of pressure

Sunny_sXe [5.5K]

hydraulic press
hydraulic lift
hydraulic jack
hydraulic brake

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2 years ago