Hello,
I believe the answer is C. A and B say that it would help to change it but when we think about it data and opinions are not what people use to change a curriculum or program. Mainly they use surveys, which they didn't use here. Option D doesn't make any sense, data doesn't complement a program it supports some sort of theory which leads us to C. C is the only option where they use the data for effectiveness. Hope this helps!

4 0

3 years ago

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How many moles of SnF₂ will be produced along with 48 grams of H₂? *

KIM [24]

Answer: 24 moles of $SnF_2$ are produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

${\text{Moles of} H_2}=\frac{48g}{2g/mol}=24moles$

$Sn+2HF\rightarrow SnF_2+H_2$

According to stoichiometry :

1 mole of $H_2$ is accompanied with = 1 mole of $SnF_2$

Thus 24 moles of $H_2$ is accompanied with = $\frac{1}{1}\times 24=24moles$ of $SnF_2$

Thus 24 moles of $SnF_2$ are produced.

8 0

3 years ago

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i

Ipatiy [6.2K]

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

$Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)$

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the net ionic equation is written above.

8 0

3 years ago