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Strike441 [17]
2 years ago
5

Heat being transferred from a hot pan to our food while cooking is an example of... *

Chemistry
1 answer:
stellarik [79]2 years ago
6 0

A.conduction  i hope it helps

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What is a decay series?
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D

Explanation:

because in a decay series a daughter nuclei may be stable or decay itself. that starts a decay series

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3 years ago
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To form an oxygen molecule o2 two oxygen atoms share two pairs of electrons what kind of bond is shown below by the electron dot
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They have a Covelant bond
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12. Study the chemical reaction below, what are the product(s) in the reaction and what are the reactant(s) in the reaction (Sho
natita [175]

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The reactants are Aluminium and Copper chloride.

The products are Copper and Aluminium chloride.

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3 years ago
Element
Svetllana [295]

Answer:

Number of moles = 10.6 mol

Explanation:

Given data:

Molar mass of H = 1.008 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16.00 g/mol

Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)

Number of moles of citric acid = ?

Solution:

Formula:

Number of moles = mass/molar mass

Now we will calculate the molar mass of citric acid:

C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)

C₆H₈O₇ = 72.06 + 8.064+112

C₆H₈O₇ = 192.124g/mol

Number of moles = 2030 g/ 192.124g/mol

Number of moles = 10.6 mol

7 0
3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
2 years ago
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