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Lilit [14]
2 years ago
8

Practice question 1 How many moles are in 5.64e25 atoms of iron (Fe)?​

Chemistry
1 answer:
liberstina [14]2 years ago
6 0

1 mole is equal to 6.0220 x 1023 atoms of any element.

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5. The bond in a molecule of Cl2 could be described as
jenyasd209 [6]

the bond in a molecule of Cl2 is a covalent bond

6 0
3 years ago
A baker is decorating cupcakes using 3 cherries for every 1 cupcake. If the baker has 12 cherries and 5 cupcakes, what is the th
11111nata11111 [884]
C. 4 decorated cupcakes.

Because 12 divided by 3 = 4, so you can have 4 cupcakes each with 3 cherries
7 0
3 years ago
Read 2 more answers
David is making lemonade he asked you could you some lemon juice and water the sugar starts to dissolve what can we do to speed
suter [353]

We could (a) stir faster and (b) warm the mixture.

<em>Stirring faster</em> moves freshly-dissolved sugar away from the solid and allows new water molecules to contact with the surface,

<em>Warming the mixture</em> gives the water molecules more kinetic energy, so their collisions with the surface of the sugar will be more effective in removing the sugar molecules.

7 0
4 years ago
What is the limiting reactant in a chemical reaction?
IgorC [24]

Answer:

D.) the reactants that runs out first

Explanation:

Limiting reactant is the reactant which is present in the smallest amount and thus limit the yield of product.

In any chemical reaction limiting reactant is identified by steps:

First we will calculate the number of moles of given amount of reactants.

Then we will find the number of moles of product by comparing with moles of reactant through balanced chemical equation.

Then we will identified the reactant which produced smaller amount of product.

It can be better understand by following problem.

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Which is limiting reactant= ?

Chemical equation:

CaCO₃ + 2HCl  → CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                   CaCO₃         :               CaCl₂

                       1               :               1

                     0.25           :            0.25

                   HCl              :                CaCl₂

                     2                :                 1

                   0.36            :               1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of calcium chloride:

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

5 0
3 years ago
Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
allsm [11]

Answer:

\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}

Explanation:

1. Density from mass and volume

\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}

2. Volume from density and mass

V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}

3. Mass from density and volume

\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                =<u> 12.8 mL</u>

Volume of object               = 11.8 mL

\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

 

6 0
3 years ago
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