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Llana [10]
3 years ago
15

A plastic ball’s density is 0.802 g/mL. What volume would a 15.8 g plastic ball occupy?

Chemistry
1 answer:
krek1111 [17]3 years ago
7 0

Answer:

<h2>19.70 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{15.8}{0.802}  \\  = 19.700748...

We have the final answer as

<h3>19.70 mL</h3>

Hope this helps you

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Which are the three main types if lipids
konstantin123 [22]

The three main types of lipids are triglycerides, steroids and phospholipids

8 0
3 years ago
Read 2 more answers
If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml
Bumek [7]

Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T} =k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

\frac{P}{T} =k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Having two different states, an initial state and an final state, it is true:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

In this case:

  • P1= 0.9 atm
  • V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
  • T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

  • P2= 1 atm
  • V2= ?
  • T2= 0°C= 273 °K

Replacing:

\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}

Solving:

V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}

V2= 2.415 L =2,415 mL

<u><em>The new volume is 2415 mL</em></u>

6 0
3 years ago
Calculate the feed ratio of adipic acid and hexamethylene diamine that should be employed to obtain a polyamide of approximately
artcher [175]

Answer:

r= 0.9949 (For 15,000)

r=0.995 (For 19,000)

Explanation:

We know that

Molecular weight of hexamethylene diamine = 116.21 g/mol

Molecular weight of adipic acid = 146.14 g/mol

Molecular weight of water = 18.016 g/mol

As we know that when  adipic acid  and hexamethylene diamine react then nylon 6, 6 comes out as the final product and release 2 molecule of water.

So

M_{repeat}=146.14+166.21-2\times 18.106\ g/mol

M_{repeat}=226.32\ g/mol

So

Mo= 226.32/2 =113.16 g/mol

M_n=X_nM_o

Given that

Mn= 15,000 g/mol

So

15,000 = Xn x 113.16

Xn = 132.55

Now by using Carothers equation we know that

X_n=\dfrac{1+r}{1+r-2rp}

132.55=\dfrac{1+r}{1+r-2\times 0.99r}

By calculating we get

r= 0.9949

For 19,000

19,000 = Xn x 113.16

Xn = 167.99

By calculating in same process given above we get

r=0.995

3 0
4 years ago
Balance the redox reaction equation (occurring in acidic solution) and choose the correct coefficients for each reactant and pro
algol13
Balanced chemical reaction: 
PbO₂<span>(s) + Sn(s)+ 4H</span>⁺(aq) → Pb²⁺(aq) + Sn²⁺(aq) + 2H₂O<span>(l).
Oxidation half-reaction: Sn </span>→ Sn²⁺ + 2e⁻.<span>
Reduction half-reaction: PbO</span>₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O.
Net reaction: Sn + PbO₂ + 4H⁺ + 2e⁻ → Sn²⁺ + 2e⁻ + Pb²⁺ + 2H₂O.
Oxidation is increase of oxidation number, reduction is decrease of oxidation number.
6 0
4 years ago
Arrange the following measurements, in seconds, from the greatest bias to the least bias. 6.63 ± 0.01 s, 6.6 ± 0.1 s, 6.52 ± 0.0
lord [1]
I understand here "bias" to be the uncertainty of measurements. So the order will be the following:

6.4 ± 0.5 s
<span>6.6 ± 0.1 s,
</span><span>6.63 ± 0.01 s,
</span><span>6.52 ± 0.05 s,
</span>
(notice how the second number, the one behind the symbol ± gets smaller, as the bias gets smaller). 



6 0
4 years ago
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