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Masja [62]
3 years ago
14

1 mole = 1000 millimoles (mmol); 1millimole = 1000 micromoles (µmol). if a solution contains 38231 µmol, what is that amount in

mmol?
Chemistry
1 answer:
Mrrafil [7]3 years ago
3 0
1 mmol --------------------- 1000 <span>µmol
( mmol ) -------------------- </span> 38231 µmol

mmol =  38231*1 / 1000

mmol = 38231/ 1000

=>  38.231 mmol
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2 years ago
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Calculate the specific heat in J/(g·ºC) of an unknown substance if a 2.50-g sample releases 12.0 cal as its temperature changes
Degger [83]

Answer:

\fbox{c =  - 4.01 \: joule/g°C}

<em><u>Step by step explanation</u></em><em><u>:</u></em>

<em>Given:</em>

Mass of given sample (m) = 2.50 g

Initial temperature (T1) = 25°C

Final temperature (T2) = 20°C

Heat Energy Q = 12 cal

<em>T</em><em>o </em><em>find:</em>

<em>Specific \:  Heat \:  c = \:  ?</em>

<em>Solution</em><em>:</em>

We know that,

<em>Specific</em><em> </em><em>heat</em> <em>of </em><em>any </em><em>substance </em><em>is </em><em>directly</em><em> </em><em>proportional</em><em> </em><em>to </em><em>the </em><em>mass </em><em>and </em><em>change </em><em>in </em><em>temperature.</em>

Represented by equation,

Q = mc \triangle T

Where,

<em>Q = Heat Energy</em>

<em>m = mass of given sampl</em><em>e</em>

<em>c = specific heat</em>

<em>∆T = change in </em><em>temperature</em>

Substituting corresponding values,

<em>Q = mc \triangle T \\  12 = 2.5\times c \times (20-25) \\ c  =  \frac{12}{2.5 \times ( - 5)}  \\  c =  - 0.96 \: cal/g°C \\</em>

We also know that,

1  \: cal = 4.184  \: joules

multiplying above answer by 4.184,

c =  - 0.96 \times 4.184 \\  \fbox{c =  - 4.01 \: joule/g°C}

<em>Thanks for joining brainly community!</em>

6 0
2 years ago
If the specific heat of water is 4,186 J/kg∙°C, how much heat is required to increase the temperature of 1.2 kg of water from 23
7nadin3 [17]

Answer:

8.0 × 10⁴ J

Explanation:

Step 1: Given data

  • Specific heat of water (c): 4,186 J/kg.°C
  • Mass of water (m): 1.2 kg
  • Initial temperature: 23 °C
  • Final temperature: 39 °C

Step 2: Calculate the change in the temperature

ΔT = 39 °C - 23 °C = 16 °C

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 4,186 J/kg.°C × 1.2 kg × 16 °C

Q = 8.0 × 10⁴ J

6 0
2 years ago
The number of moles of N2O produced when 26.5 g N2 reacts with excess oxygen?
insens350 [35]

Answer:

a.  0.964 moles of N₂O will be produced

b. mass of N₂ required = 7.31 g of N₂

c. Number of particles of N₂= 5.87 * 10²² molecules of N₂

Explanation:

a. Balanced equation of the reaction is given below: 2N₂ + O₂ ---> 2N₂O

Molar mass of N₂ = 28 g/mol

molar mass of O₂ = 32 g/mol

molar mas of N₂O = 44g/mol

From the equation of reaction, 2 moles of nitrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of N₂O

number of moles of N₂ present in 26.5 g = 26.5 g/28 g/mol = 0.946 moles

Since oxygen gas is in excess in the reaction, the limiting reactant is N₂

Mole ratio of N₂ to N₂O is 1 : 1

Therefore, 0.964 moles of N₂ will react with excess oxygen gas to produce 0.964 moles of N₂O

b. number of moles of N₂O in 11.5 g of N₂O = 11.5/44 = 0.261 moles

0.261 moles of N₂O will be produced by 0.261 moles of N₂

mass of N₂ present in 0.261 moles = 0.261 * 28

mass of N₂ required = 7.31 g of N₂

c. number of moles of O₂ present in 1.56 g of O₂ = 1.56/32 = 0.04875moles

Moles of N₂ required to react with 0.4875 moles of O₂ = 2 * 0.04875 = 0.0975 moles

Number of particles of N₂ present in 0.0975 moles = 6.02 * 10²³ * 0.0975

Number of particles = 5.87 * 10²² molecules of N₂

3 0
3 years ago
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