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Alina [70]
3 years ago
10

How does weight influence the sliding force of friction

Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0

Because they heavier it is the slower it will go

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What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?​
trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution  = 2.5L

Mass of Calcium phosphate  = 600g

Unknown:

Concentration  = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

        Formula of calcium phosphate  = Ca₃PO₄

         molar mass = 3(40) + 31 + 4(16) = 215g/mol  

Number of moles of  Ca₃PO₄  = \frac{600}{215}   = 2.79moles

   Now;

  Concentration  = \frac{Number of moles }{volume }  

 Concentration  = \frac{2.79}{2.5}   = 1.12M

7 0
3 years ago
Non chemical industry examples​
Luba_88 [7]

Answer:

Light , heat , and sound are examples.

4 0
2 years ago
Read 2 more answers
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

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

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

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

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
Which of these structures are found in both plant and animal cells
erastova [34]

Answer: B

Explanation: The mitochondria are vacuoles and ribosomes are in both plant and animal cell

8 0
3 years ago
A chemist pours 1 mol of zinc granules into one beaker and 1 mol of zinc chloride powder into another beaker. What do the two sa
irina1246 [14]
The number of Zn particles (atoms of Zn ) in the first sample, and Zn ions are going to be the same, 
Also mass of Zn are going to be the same, because of electrons are too small, and would not have influence on mass.
6 0
2 years ago
Read 2 more answers
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