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Angelina_Jolie [31]
3 years ago
5

What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?

Chemistry
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

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An exacuted glass vessel weighs 50 g when empty, 148g when filled with an liquid of density o.989/cc and 50.5 g whenfilled with
Zanzabum

MW of gas : 124.12 g/mol

<h3>Further explanation  </h3>

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of density  

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

Density formula:  

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

ρ = density  

m = mass  

v = volume  

glass vessel wieight = 50 g

glass vessel + liquid = 148 ⇒ liquid = 148 - 50 =98 g

volume of glass vessel :

\tt V=\dfrac{m}{\rho}=\dfrac{98}{0.989}=99.1~ml

An ideal gas :

m = 50.5 - 50 = 0.5 g

P = 760 mmHg = 1 atm

T = 300 K

\tt PV=\dfrac{mass(m)}{MW}.RT\\\\MW=\dfrac{m.RT}{PV}\\\\Mw=\dfrac{0.5\times 0.082\times 300}{1\times 0.0991}=124.12~g/mol

3 0
3 years ago
Rank the solutions in order of decreasing [H3O ]. Rank solutions from largest to smallest hydronium ion concentration. To rank i
Usimov [2.4K]

Answer:

0.10M HCN  <  0.10 M HClO  <  0.10 M HNO₂  < 0.10 M HNO₃

Explanation:

We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.

HNO₃ is a strong acid and will have the largest hydronium concentration.

HCN  Ka = 6.2 x 10⁻¹⁰

HNO₂ Ka = 4.0 x 10⁻⁴

HClO  Ka = 3.0 x 10⁻⁸

The ranking from smallest to largest hydronium concentration will then be:

0.10M HCN  <  0.10 M HClO  <  0.10 M HNO₂  < 0.10 M HNO₃

5 0
3 years ago
Why does the moon spin around the earth and the earth spins around the sun
liberstina [14]

This easy bro there is only one answer Gravity

but here my advance word

Sometimes the gravity of big objects would capture smaller ones in orbit. This could be one way the planet acquired the moon

7 0
3 years ago
Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you
Ludmilka [50]

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

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The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

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2 years ago
Find the mass of 3.89 mol of NaF​
Otrada [13]
The answer is 163.333993748 grams
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3 years ago
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