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Ghella [55]
3 years ago
14

23. Which element is most likely to be malleable? a. Bromine b. Radon c. Nickel d. Carbon

Chemistry
1 answer:
Aleksandr [31]3 years ago
7 0
It’s c.nickel because it’s a transition metal
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What chemical bond will S and H form
Over [174]

Answer:

hydrogen sulfide (H2S)

Explanation:

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3 years ago
a certain piece of copper wire is determined to have a mass of 2.0 0g per meter how many centimeters of the wire would be needed
harina [27]

The given question tells you that a certain piece of wire has a mass  = 2.0 g per meter

That means that if you consider a piece of wire that is 1 m in length, its mass will be equal to  2.00 g .

According to question,

Now, you know that

1 m = 100 cm

Mass of  2.00 g  of copper will correspond to a wire that is a 100 cm long.

This implies that  0.28 gof copper will correspond to a wire that is

0.28 g * 100 cm /  2.0 g = 14 cm long

Hence, 14 cm of the wire would be needed to provide 0.28 g of copper.

Properties of Copper :

High conductivity and ductility.

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To know more about Copper here:

brainly.com/question/19761029?referrer=searchResults

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3 0
2 years ago
Convert 4.50 moles of fe to atoms of fe
Rina8888 [55]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We do as follows:

4.50 mol Fe ( 6.022x10^23 atoms / 1 mol Fe ) = 2.71x10^24 atoms Fe present </span>
7 0
3 years ago
A distance of one centimeter is the same as<br> 100 m<br> 10 m<br> .01 m<br> 1000 m
katovenus [111]
0.01m
...................
3 0
3 years ago
Read 2 more answers
()
const2013 [10]

Answer:

\large \boxed{\text{-1276 kJ/mol}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                             CH₃CH₂OH        +  3O₂ ⟶ 2CO₂ + 3H₂O

Bonds:         5C-H 1C-C 1C-O 1O-H    3O=O     4C=O   6O-H

D/kJ·mol⁻¹:    413    347  358  467       495        799      467

\Delta H = \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\\sum{D_{\text{reactants}}} = 5 \times 413 + 1 \times 347 + 1 \times 358 + 1 \times 467 + 3 \times 495 = 3237 + 1485\\=\text{4722 kJ}\\\sum{D_{\text{products}}} = 4 \times 799 + 6 \times 467 =3196 + 2802 = \text{5998 kJ}\\\Delta H = 4722 - 5998= \textbf{-1276 kJ} \\ \text{The overall energy change is $\large \boxed{\textbf{-1276 kJ/mol}}$}.

6 0
3 years ago
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