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Luda [366]
3 years ago
11

Please help me! I am offering 100 points and brainliest!

Chemistry
1 answer:
puteri [66]3 years ago
5 0

Explanation:

Since density is calculated with mass divided by volume, changing either of these variables will affect the density, not necessarily both of them. If only the volume is increased or only the mass is decreased, the density will decrease. If only the volume is decreased or only the mass in increased, the density will increase.

Example:

Original density is 50g/ml. Mass is 50g and volume is 1ml.

Increase mass only:

mass = 100g

Density = 100g/ml   <= Density is increased

Decrease mass only:

mass = 25g

Density = 25g/ml   <= Density is decreased

Increase volume only:

volume = 2ml

Density = 50g/2ml = 25g/ml   <= Density is decreased

Decrease volume:

volume = 0.5ml

Density = 50g/0.5ml = 100g/ml   <= Density is increased

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Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
It took 25 seconds for a horse to complete a 400 meter race. What was the speed of the horse?
Pani-rosa [81]

Answer:

v = 16 m/s

Explanation:

Given that,

Distance, d = 400 m

Time taken, t = 25 s

We need to find the speed of the horse. Speed is equal to distance covered divided by time taken. So,

v=\dfrac{d}{t}\\\\v=\dfrac{400\ m}{25\ s}\\\\v=16\ m/s

So, the speed of the horse is 16 m/s.

4 0
3 years ago
ANSWER PLS GIVING BRAINLIEST TO WHO ANSWERS CORRECTLY
ale4655 [162]

Answer:

Supergiant

Explanation:

3 0
3 years ago
Read 2 more answers
a 13.5 g sample of zinc(zn) heated from 24.2 degrees celsius to 83.6 degrees celsius that absorbs 312j of heat? use specific hea
Zielflug [23.3K]

Explanation:

83.6-24.2= 59.4 which is the change in heat

8 0
3 years ago
The reduction of iodine gas at a constant current of 1.50 amperes for 12.0 hours would result in how many grams of I–? (1 farada
hodyreva [135]

Answer:

85.2 grams.

Explanation:

Look up the relative atomic mass data on a modern periodic table.

  • Iodine: 126.904 g/mol.

The Faraday's Constant F gives the size of the charge (in Coulombs) on one mole of electrons. Also, keep in mind that \rm 1 \; Ampere = 1\; Coulomb/ second.

Charge supplied to the iodine:

\begin{aligned}Q &= I \cdot t = \rm 1.50 \; C \cdot s^{-1} \times (12.0 \times 3600) \; s\\ &= \rm 64800\; C\end{aligned}

How many moles of electrons does that much charge correspond to?

\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} = \rm \frac{64800\; C}{96500\; C\cdot mol^{-1}}= 0.67150\; mol\end{aligned}.

Consider the reduction half-equation for iodine:

\rm I_2 + 2 \; e^{-} \to 2 \; I^{-}.

Divide both sides by 2 to make sure that the coefficient in front of \rm e^{-} is equal to 1.

\displaystyle \rm \frac{1}{2}\; I_2 + e^{-} \to I^{-}.

The ratio between the coefficient in front of \rm e^{-} and that of \rm I^{-} is equal to 1. In other words, while gaseous iodine is in excess, each mole of electrons will produce one mole of iodine ions.

\rm 0.67150\; mol of electrons will produce \rm 0.67150\; mol of iodine ions.

The mass of one mole of iodine ions is approximately the same as that of one mole of iodine atoms.

M(\mathrm{I^{-}}) \approx M(\mathrm{I})= \rm 126.904\; g\cdot mol^{-1}.

m(\mathrm{I}^{-}) = n \cdot M \approx \rm 85.2\; g.

3 0
3 years ago
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