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lbvjy [14]
3 years ago
13

The basic particle from which all elements are made; the smallest particle of an element that has the properties of that element

.
Chemistry
2 answers:
NeX [460]3 years ago
7 0

Answer:

Atoms

Explanation:

Elements are made from atoms

Anastasy [175]3 years ago
6 0

Answer:

<h2><u>Atom.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u></h2>

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a 4.50 g coin of copper absorbed 54 calories of heat. what was the final temperature of the copper if the initial temperature wa
vlada-n [284]

Answer:

Final temperature =  T₂ = 155.43 °C

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of coin = 4.50 g

Heat absorbed = 54 cal

Initial temperature = 25 °C

Specific heat of copper = 0.092 cal/g °C

Final temperature = ?

Solution:

Q = m.c. ΔT

ΔT = T₂ -T₁

Q = m.c. T₂ -T₁

54 cal = 4.50 g × 0.092 cal/g °C ×  T₂ -25  °C

54 cal = 0.414 cal/ °C ×  T₂ -25  °C

54 cal /0.414 cal/ °C =  T₂ -25  °C

130.43 °C  =  T₂ -25 °C

130.43 °C + 25 °C = T₂

155.43 °C = T₂

4 0
3 years ago
When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be
Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH  = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL  +  50.mL  = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH_{neutralization}

ΔH_{neutralization} = -q_soln / mole of water produced

so we substitute

ΔH_{neutralization} = -( 4076.68 J ) / 0.0500 mol  

ΔH_{neutralization} = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0
3 years ago
What is reaction rate
zimovet [89]

Answer:The change in concentration of a reactant or product per unit time

Explanation:

6 0
3 years ago
A team of engineers is doing a cost-benefit analysis to determine whether
beks73 [17]

Answer:

1) The risk of high cost due to increased resource requirements

2) The risk of late entry into the (a changed) market

Explanation:

The analysis being performed by the engineers = A cost benefit analysis to determine if a new technology should be developed

A cost-benefit analysis is a process of appraising or measuring the advantages, benefits of a policy, action or decision, so as to find the (equilibrium) balance point between the costs of the decision or action

The risk to be considered are;

1) The risk of high cost due to increased resource requirements

The increased cost required for the development of the new technology now which due to the unlikely existence of a similar invention in the market that will give them an advantage of increased profits

2) The risk of late entry into the (a changed) market

The changes in the consumer preferences, market landscape, and the likely introduction into the market of a similar invention by the competition in the near future which will reduce the amount of profits that can be gained from the invention

3 0
3 years ago
Please refer to image please
GarryVolchara [31]

Answer:

76.9L

Explanation:

Based on the graph, whenever the temperature increases by 100K, the volume increases by 10L, so do 769/10

7 0
2 years ago
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