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3 years ago

How can you increase friction

2 answers:

5 0

Answer:Well There is many ways to increase friction between objects. One being rubbing the two objects together quicker and harder. If there is any sort of wetness or anything related to that make sure to dry the surface between the two objects you want to create friction between so it will be more effective.

Explanation:

Luda [366]3 years ago

3 0

Rubbing stuff together

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11. The strength of a force is measured in

dem82 [27]

Answer:

In units.

Explanation:

Such as newtons or pounds.

4 0

3 years ago

A certain chemical reaction releases 36.2 kJ/g of heat for each gram of reactant consumed. How can you calculate what mass of re

Lilit [14]

Answer:

0.038 g of reactant

Explanation:

Data given:

Heat release for each gram of reactant consumption = 36.2 kJ/g

mass of reactant that release 1360 J of heat = ?

Solution:

As 36.2 kJ of heat release per gram of reactant consumption so first we will convert KJ to J

As we know

1 KJ = 1000 J

36.2 kJ = 36.2 x 1000 = 36200 J

So it means that in chemical reaction 36200 J of heat release for each gram of reactant consumed so how much mass of reactant will be consumed if 1360 J heat will release

Apply unity formula

36200 J of heat release ≅ 1 gram of reactant

1360 J of heat release ≅ X gram of reactant

Do cross multiplication

X gram of reactant = 1 g x 1360 J / 36200 J

X gram of reactant = 0.038 g

So 0.038 g of reactant will produce 1360 J of heat.

5 0

3 years ago

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

For process 2:

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

For process 3:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

For process 4:

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

PLEASE PLEASE PLEASE ANSWER HAVE MERCY 35 POINTS

Arada [10]

You should give answers since some people will just try and find any answer, which happened to me. So like if you have answers for the question just put them like A B C D kinda like true or false for others to know what they got instead of giving them more work, and then they just give you random answers.

6 0

3 years ago

What is the bond between Ca and Cl

gtnhenbr [62]

ionic bond is formed between ca and cl forming molecule cacl2 ca has 2 velancy and cl has one velancy (ca has 2 electrons in its outer most shell while cl has 1 electron vecancy in its outermost shell). So ca would make bond with 2 cl atoms

4 0

3 years ago