All categories

3 years ago

I need it done assap

Chemistry

1 answer:

Tcecarenko [31]3 years ago

7 0

Answer:

plasma, gas, liquid, solid

Explanation:

The particles in plasma move very quickly as to why its so free, gas is next because it has a little less freeness but it can still move around easily, liquid next because it is flowing and isn’t to free, then solid because it is compact and doesn’t flow at all

You might be interested in

If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4

abruzzese [7]

Answer: The limiting reagent in the given chemical reaction is carbon metal.

Explanation:

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For sodium sulfate:

Given mass of sodium sulfate = 16.29 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfate}=\frac{16.29g}{142g/mol}=0.115mol$

For carbon:

Given mass of carbon = 3.697 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon}=\frac{3.697g}{12g/mol}=0.31mol$

For the given chemical reaction:

$Na_2SO_4(aq.)+4C(s)\rightarrow Na_2S(s)+4CO(g)$

By Stoichiometry of the reaction:

4 moles of carbon reacts with 1 mole of sodium sulfate

So, 0.31 moles of carbon will react with = $\frac{1}{4}\times 0.31=0.0775mol$ of sodium sulfate

As, given amount of sodium sulfate is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon metal is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent in the given chemical reaction is carbon metal.

4 0

3 years ago

Choose all the answers that apply.

Anni [7]

Answer:

I believe it's the lowest portion of the atmosphere

3 0

3 years ago

In step 2, of the experiment, the procedure uses 3.0M NaOH. However, the student notices that the only solution of NaOH is conce

Luda [366]

Answer:

We need 78.9 mL of the 19.0 M NaOH solution

Explanation:

Step 1: Data given

Molarity of the original NaOH solution = 19.0 M

Molarity of the NaOH solution we want to prepare = 3.0 M

Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

Step 2: Calculate volume of the 19.0 M NaOH solution needed

C1*V1 = C2*V2

⇒with C1 = the concentration of the original NaOH solution = 19.0 M

⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED

⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M

⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

19.0 M * V2 = 3.0 M * 0.500 L

V2 = (3.0 M * 0.500L) / 19.0 M

V2 = 0.0789 L

We need 0.0789 L

This is 0.0789 * 10^3 mL = 78.9 mL

We need 78.9 mL of the 19.0 M NaOH solution

8 0

3 years ago

Write a net ionic equation for the reaction that occurs when excess hydrochloric acid (aq) and potassium sulfite (aq) are combin

Ket [755]

Answer: The net ionic equation for the given reaction is $2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow H_2O(l)+SO_2(g)$

Explanation:

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

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

The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

$2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)$