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3 years ago

What type of reaction is this? 2Na(s) + Cl2(g) —> 2NaCl(s)

Chemistry

1 answer:

soldier1979 [14.2K]3 years ago

7 0

Answer: Synthesis

Explanation:

2 or more substances combine to form a new compound. In this case 2Na(s) combined with Cl2(g) to make 2NaCl(s)

A + X ---> AX

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How to answer all these questions?

Fudgin [204]

You have to do it to see the results

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3 years ago

Based on the crystal-field strengths cl- < f- < h2o < nh3 < h2nc2h4nh2, which octahedral ti (iii) complex below has

kati45 [8]

Based on the crystal field strength, Cl ligand would give the longest d-d transition when complexed with Ti(III). as this is the weak field ligand and would cause minimum splitting of d orbitals.

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3 years ago

Silicon carbide (SiC) is an important ceramic material made by reacting sand (silicon dioxide, SiO2) with powdered carbon at a h

makkiz [27]

Answer:

Percent yield of SiC is 77.0%.

Explanation:

Balanced reaction: $SiO_{2}+3C\rightarrow SiC+2CO$

Molar mass of SiC = 40.11 g/mol

Molar mass of $SiO_{2}$ = 60.08 g/mol

So, 100.0 kg of $SiO_{2}$ = $\frac{100.0\times 10^{3}}{60.08}$ moles of $SiO_{2}$ = 1664 moles of $SiO_{2}$

According to balanced equation, 1 mol of $SiO_{2}$ produces 1 mol of SiC

Therefore, 1664 moles of $SiO_{2}$ produce 1664 moles of SiC

Mass of 1664 moles of SiC = $(1664\times 40.11)g$ = 66743g = 66.74 kg (4 sig. fig.)

Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)] $\times 100$ %

= $\frac{51.4kg}{66.74kg}\times 100$ %

= 77.0%

4 0

2 years ago

Why does foliation occur? (Metamorphic rock)

dusya [7]

Foliation occurs when the pressure squeezes the flat or elongated minerals within a rock so that they become aligned.

5 0

2 years ago

Read 2 more answers

A rock contains 0.623 mg of 206Pb for every 1.000 mg of 238U present. Assuming that no lead was originally present, that all the

maxonik [38]

Answer:

t = 3,496x10⁹ years

Explanation:

The decay of ²³⁸U is:

²³⁸U → ²⁰⁶Pb + 8He + 6e⁻

Moles of ²⁰⁶Pb presents in 0,623mg are:

0,623x10⁻³g×(1mol / 206g) = 3,02x10⁻⁶ moles of ²⁰⁶Pb.

These moles are equals to moles of ²³⁸U before decay, that means, 3,02x10⁻⁶ moles²³⁸U

In grams:

3,02x10⁻⁶ moles²³⁸U× (238g / 1mol) = 7,20x10⁻⁴ g ²³⁸U = 0,720 mg²³⁸U

That means initial ²³⁸U was 1,000mg + 0,720mg = 1,720mg

Applying the formula:

ln (N₀/N) t₁₂ = t ln2

Where N₀ is initial amount of uranium (1,720mg), N is concentration of uranium (1,000mg), half-life time is a constant (t₁₂= 4,468x10⁹ years) and t is the time transcurred for the reaction. Replacing:

ln(1,720/1)*4,468x10⁹ years = t ln2

t = 3,496x10⁹ years

I hope it helps!

6 0

2 years ago