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3 years ago

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What reaction type is MgCl2 ---> Mg + Cl2 ?

2 answers:

Serga [27]3 years ago

8 0

A sorry if I’m wrong

Firlakuza [10]3 years ago

7 0

Answer:

A.Synthesis

Explanation:

What reaction type is MgCl2 ---> Mg + Cl2 ?

a. Synthesis

b. Single Replacement

c. Double Replacement

d. Decomposition

Answer

A. Synthesis.

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Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 97. g of octane is mi

zhannawk [14.2K]

Answer: 61 grams

Explanation:

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

$\text{Number of moles of octane}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{97g}{114g/mol}=0.85moles$

$\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{150g}{32g/mol}=4.69moles$

The chemical equation for the combustion of octane in oxygen follows the equation:

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

By stoichiometry of the reaction;

25 moles of oxygen react with 2 moles of octane

4.69 moles of oxygen react with= $\frac{2}{25}\times 4.69=0.37$ moles of octane

Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.

25 moles of oxygen produce 18 moles of water

4.69 moles of oxygen produce= $\frac{18}{25}\times 4.69=3.38$ moles of water.

Mass of water produced= $moles\times {\text{Molar mass}}=3.38\times 18g/mol=61g$

The maximum mass of water that could be produced by the chemical reaction is 61 grams.

6 0

3 years ago

A solution that contains the maximum amount of solute that is soluble at a given temperature is said to be —

solniwko [45]

Answer: Saturated

A solution that contains the maximum amount of solute that is soluble at a given temperature is said to be saturated.

Explanation:

A Saturated solution is one that contains as much (i.e maximum) solute as it can dissolve at that temperature in the presence of undissolved solute particles.

For instance: if a given volume of water can only dissolve a certain amount of salt in it at room temperature, then, more salt added will not dissolve.

Thus, making the solution saturated.

5 0

3 years ago

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

4 years ago

Select the correct answer.

Stels [109]

I would definitely say that the answer is

C. drought

because if they are getting 43.76 inches of rain on average, and then one year they decrease 10.31 inches of rain to 33.45 inches of rain, its a drought for sure.

7 0

3 years ago

Read 2 more answers

Can someone please help 7-10 with some work shown a bit , pleaseeeeeee

Oksana_A [137]

Answer:

7.] 597.11 Calories

8] 50208 joules

9] 2.05016 kilojoules

10] 0.0142256 kilojoules

Explanation:

just google how to do it. you can find some good examples.

hope this helps a bit!

brainliest?

also, im in 7th, im in high school algebra already. (im in advanced stuff)

ty!

5 0

3 years ago