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seropon [69]
3 years ago
13

"In a test tube was add a piece of Zn and 2 mL of HCl (.1M). This reaction produced ZnCl2 and Hydrogen gas. This reactions is a:

Chemistry
1 answer:
omeli [17]3 years ago
4 0

Answer:

d.  A single replacement reaction.

Explanation:

The zinc replaces the hydrogen in the hydrochloric acid.

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What is the mass of 0.251 moles of water? Using dimensional analysis show work
AVprozaik [17]

1 \text{ mol of H}_2 \text{O} \equiv 18 \text{ g} \implies 0.251  \text{ moles of H}_2 \text{O} \equiv 4.518 \text{ g}

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3 years ago
Name the seventh inner transition metal in period 6
vladimir1956 [14]

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8 0
3 years ago
Lewis structure for nitrogen triiodide, NI(3)
Rasek [7]

Answer:

Lewis structure for nitrogen triiodide,NI_3 is given in the attachment.

Explanation:

Given:

The given compound is Nitrogen triiodide. In which 1 atom of Nitrogen combines with 3 atoms of Iodine. Both Nitrogen and Iodine are non-metals,So they form covalent bond by sharing of electrons.

The electron configuration of Nitrogen and Iodine is given below;

N(7) = 1s^2,2s^22p^3\\I(53) = 1s^2,2s^22p^6,3s^23p^63d^{10},4s^24p^64d^{10},5s^25p^5

There are 5 electrons in valance shell of Nitrogen atom and 7 electrons in valance shell of Iodine atom.

So, 3 atom of Iodine shares 1 electron with 1 electrons of Nitrogen.

The Lewis dot Structure is in the attachment.

3 0
3 years ago
Element Z has a mass of 430g. Scientists determined it to have a ½ life of 5 years. How many grams of element Z will remain afte
Afina-wow [57]

Answer:

hope this helped youu:)

7 0
3 years ago
if 0.3 moles of carbon reacts completely with 0.3 moles of molecular oxygen, what is the yield in Percent of 6.6 g of CO2 are fo
Anettt [7]
Answer is:<span>the yield is 50%.
</span>
Chemical reaction: C + O₂ → CO₂.
n(C) = 0.3 mol; amount of substance.
n(O₂) = 0.3 mol.
From chemical reaction: n(C) : n(CO₂) = 1 : 1.
n(CO₂) = 0.3 mol.
M(CO₂) = 44 g/mol; molar mass of caron(IV) oxide.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) =0.3 mol · 44 g/mol.
m(CO₂) = 13.2 g; mass of carbon(IV) oxide.
the yield = 6.6 g ÷ 13.2 g · 100%.
the yield = 50%.
8 0
3 years ago
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