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Artist 52 [7]
3 years ago
8

What is a measurement of the amount of solute that is dissolved in a given quantity of solvent?

Chemistry
2 answers:
weqwewe [10]3 years ago
6 0
B.concentration is the answer 

almond37 [142]3 years ago
3 0

Answer:

B

Explanation: The answer is B: concentration.

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If 700 g of water at 90 °C loses 27 kJ of heat, what is its final temperature?​
Phoenix [80]

Answer:

If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also increases; and the same happens when either of the two decreases .). The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature, ΔT= Tfinal - Tinitial

In this case:

  • Q= 27 kJ= 27,000 J (being 1 kJ=1,000 J)
  • c=4.186 \frac{J}{g* C}
  • m=700 g
  • ΔT= Tfinal - Tinitial= Tfinal - 90 °C

Replacing:

27,000 J=4.186 \frac{J}{g* C}*400 g* (Tfinal - 90C)\\

Solving:

27,000 J=1,674.4 \frac{J}{C}* (Tfinal - 90C)

\frac{27,000 J}{1,674.4 \frac{J}{C}} =(Tfinal - 90C)

16.125 °C= Tfinal - 90 °C

Tfinal= 16.125 °C + 90 °C

Tfinal= 106.125 °C

<u><em>If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C</em></u>

3 0
3 years ago
Gt4rgbrbrtbr\t?<br>frehjkgbrtnvboror
3241004551 [841]

Answer:I can't see any question

8 0
3 years ago
Write the appropriate symbol for each of the following isotopes: (a) Z 74, A 186; (b) Z 80, A 201; (c) Z 34, A 76; (d) Z 94, A 2
beks73 [17]

Answer:  a)  _{74}^{186}\textrm{W}

b) _{80}^{201}\textrm{Hg}

c)  _{34}^{76}\textrm{Se}

d) _{94}^{239}\textrm{Pu}

Explanation:

General representation of an element is given as:_Z^A\textrm{X}

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

In an atom, when neutrons or protons are lost or gains, it directly affects the mass number of an atom.

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

It is characteristic of a particular element.

Atomic number = Number of electrons = Number of proton

a) Z 74, A 186: _{74}^{186}\textrm{W}

b) Z 80, A 201: _{80}^{201}\textrm{Hg}

c) Z 34, A 76: _{34}^{76}\textrm{Se}

d) Z 94, A 239.: _{94}^{239}\textrm{Pu}

6 0
3 years ago
Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11
otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

\frac{110}{2}\times 2.6966 mol=148.31 mol of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

8 0
2 years ago
The compound potassium nitrate is a strong electrolyte. write the reaction when solid potassium nitrate is put into water:
Alenkinab [10]
<span>There is no chemical reaction between potassium nitrate and water. Potassium nitrate dissolves in water, which is a physical change.</span>
7 0
3 years ago
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