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3 years ago

Beaux has purchased a synthetic material for a product he is launching. What is most likely a benefit of this material?

Chemistry

2 answers:

Brut [27]3 years ago

6 0

Answer:

it can be mass produced

Explanation:

suter [353]3 years ago

5 0

Answer:

it can be mass produced

Explanation:

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If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

Leadership productivity and self direction are examples off

jekas [21]

Being a leader is one of the most important things you can do. when your are a leader there are people looking up to. you know what you are supposed to be doing so if you do it one day it will pay off. someone is out there to be just like you.

4 0

3 years ago

Read 2 more answers

In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2

allsm [11]

Answer:

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

Explanation:

$SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)$

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Balance the charge by adding 2 electrons on product side:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-$ ....[1]

Reduction :

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)$

Balance O by adding water on required side:

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

Now, balance H by adding $H^+$ on the required side:

$SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

At last balance the charge by adding electrons on the side where positive charge is more:

$SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)$ ..[2]

Adding [1] and [2]:

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

4 0

3 years ago

Give two examples of diffusion process in plants and animals.

lisabon 2012 [21]

answer:

in plants

Transport manufactured food from the leaves to others parts of the plant

Facilitates gaseous exchange through the stomata in the leaves to other parts of the plant

in animal

Exchange of respiratory gases across respiratory services

Excretion of nitrogenous waste in some unicellular organisms

Explanation:

Hope it benefit

3 0

3 years ago

How many moles are in 9.50 x 1022 molecules of CO? Be sure to ALL work below and report your answer using the correct number of

eimsori [14]

Answer:

0.158 moles

Explanation:

We are given;

9.50 x 10^22 molecules of CO

We are required to determine the number of moles;

We need to know;

1 mole of a compound = 6.022 × 10^23 molecules

Therefore;

9.50 x 10^22 molecules of CO will be equivalent to;

= 9.50 x 10^22 molecules ÷ 6.022 × 10^23 molecules/mole

= 0.158 moles

Therefore, the number of moles are 0.158 moles

4 0

3 years ago