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3 years ago

Isotopes have

Chemistry

1 answer:

Feliz [49]3 years ago

7 0

Answer:

question 1 answer is b

Explanation:

istopes equal to same atomic number but different mass number

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How does an objects speed affect its kinetic energy?

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Because it’s kinetic energy INCREASES the speed

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3 years ago

The density (g/ml) of 6 pennies with a mass of 18.53 grams and a volume of 2.0 ml

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Answer: 9.27 g/mL

Explanation:

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2 years ago

What is the most corrosive element in the periodic table?

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3 years ago

Read 2 more answers

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

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3 years ago

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c

gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= 135,583 J = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

Now that we have the heat of combustion, we need to calculate the molar heat.

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the conversion factor as 135.58 kJ/ 4.40 g.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

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3 years ago