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2 years ago

The rise and fall of ocean Water are called

Chemistry

2 answers:

puteri [66]2 years ago

8 0

The tide is where the sea level rises and falls

Gnoma [55]2 years ago

3 0

The tide it is where the sea level rises and falls (mainly caused by the moon)

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The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of

Mademuasel [1]

Answer:

\large \boxed{\textbf{609 kJ}}

Explanation:

The formula for the heat absorbed is

q = mCΔT

Data:

m = 2.07 kg

T₁ = 23 °C

T₂ = 191 °C

C = 1.75 J·°C⁻¹g⁻¹

Calculations:

1. Convert kilograms to grams

2.07 kg = 2070 g

2. Calculate ΔT

ΔT = T₂ - T₁ = 191 - 23 = 168 °C

3. Calculate q

$q = \text{2070 g} \times 1.75 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 168 \,^{\circ}\text{C} = \text{609 000 J} = \textbf{609 kJ}\\\text{The heat energy required is }\large \boxed{\textbf{609 kJ}}$

4 0

2 years ago

How many moles of calcium oxide are formed when 3 moles of calcium react with oxygen?

Temka [501]

Answer:

3 moles of CaO

Explanation:

The chemical balanced equation is;

2Ca + O₂ → 2CaO

The ratio of calcium(Ca) to Calcium Oxide(Cao) in this equation is 2:2.

Now, when 3 moles of calcium react with oxygen, Cao produced will be;

3 moles of O₂ × 2 moles of CaO/2 moles of Cao = 3 moles of CaO

7 0

2 years ago

What is atom ? chemistry

Neporo4naja [7]

Answer:

atom is the smallest unit of matter that has the characteristic properties of a chemical element.

3 0

2 years ago

Help! Please :) Thank you

BartSMP [9]

First one??

I believe this is the correct answer

5 0

2 years ago

A solution contains 0.021 M Cl and 0.017 M I. A solution containing copper (I) ions is added to selectively precipitate one of t

lidiya [134]

Answer: Copper (I) iodide will precipitate first.

Explanation:

We are given:

$K_{sp}$ of CuCl = $1.0\times 10^{-6}$

$K_{sp}$ of CuI = $5.1\times 10^{-12}$

Concentration of $Cl^-\text{ ion}=0.021M$

Concentration of $I^-\text{ ion}=0.017M$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

For CuCl:

$K_{sp}=[Cu^+][Cl^-]$

Putting values in above equation, we get:

$1.0\times 10^{-6}=[Cu^+]\times 0.021$

$[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M$

Concentration of copper (I) ion = $4.76\times 10^{-5}M$

For CuI:

$K_{sp}=[Cu^+][I^-]$

Putting values in above equation, we get:

$5.1\times 10^{-12}=[Cu^+]\times 0.017$

$[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M$

Concentration of copper (I) ion = $3.00\times 10^{-10}M$

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.

7 0

2 years ago