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SIZIF [17.4K]

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

<h3>Further explanation</h3>

Given

Ionic charge

Required

The formula of binary ionic compounds

Solution

Ionic compounds consisting of cations (ions +) and anions (ions -)

Ionic compounds usually consist of metal cations and non-metal anions

Metal: cation, positively charged.

Nonmetal: negatively charged

The anion cation's charge is crossed

The ionic compounds :

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

4 0

2 years ago

Hannah does an experiment. She has two colorless chemicals that she mixes together in a beaker. Which of the following indicates

uranmaximum [27]

I am not to sure but is it all 4 of them?

8 0

2 years ago

Read 2 more answers

HI(aq)+NaOH(aq)→ <br> what the final balanced chemical equation with the phases included

julia-pushkina [17]

Answer:

$HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)$

Explanation:

Hello there!

In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:

$HI(aq)+NaOH(aq)\rightarrow NaI+H_2O$

Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:

$HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)$

Which is also balanced as the number of atoms of all the elements is the same at both sides.

Best regards!

7 0

2 years ago

Which would result in radioactive decay? Check all that apply

Gekata [30.6K]

6+2=115 and its good it took test

4 0

3 years ago

Read 2 more answers

At the start of a reaction, there are 0.0249 mol N2,

gladu [14]

Answer:

Explanation:

The reaction is given as:

$N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$

The reaction quotient is:

$Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}$

From the given information:

TO find each entity in the reaction quotient, we have:

$[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}$

$[N_2] = \dfrac{0.024 }{3.5}$

$[N_2] = 0.006857$

$[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}$

$[H_2] = 9.17 \times 10^{-3}$

∴

$Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135$

However; given that:

$K_c = 1.2$

By relating $Q_c \ \ and \ \ K_c$ , we will realize that $Q_c \ \ < \ \ K_c$

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

7 0

2 years ago

Please help! Answer please!