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3 years ago

Predict the formula of magnesium argonide

Chemistry

2 answers:

lozanna [386]3 years ago

8 0

Answer:

H2MgO that is the answer. have a nice day.

dezoksy [38]3 years ago

6 0

Answer:The formula of magnesium argonide is H2MgO

Explanation:

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CH3CH2OH + 302 → 2002 + 3H20

Softa [21]

Answer:

Oxygen gas

Explanation:

Given expression:

CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O

Number of moles of oxygen gas = 1 mole

Number of moles of ethanol = 1 mole

Unknown:

The limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. To find this specie, we always us the number of moles.

From the equation of the reaction:

1 mole of ethanol would require 3 moles of oxygen gas

But we have been given just 1 mole of oxygen gas instead of 3moles.

Therefore, oxygen gas is the limiting reactant.

3 0

3 years ago

SYNTHESIS OFCARBONATECTIONLABORATORY SIMULATIONLab Data- X99.00.10CollectedVolume sodium carbonate (mL)Molarity sodium carbonate

evablogger [386]

Answer:

$\begin{gathered} \text{Limiting Reagent = Sodium Carbonate} \\ \text{Percent Yield = 98\%} \end{gathered}$

Explanation:

The chemical reaction talks about the synthesis of calcium carbonate

It is from the reaction between sodium carbonate and calcium chloride

Let us write the equation of reaction as follows:

$Na_2CO_{3(aq)}+CaCl_{2(aq)}\text{ }\rightarrow2NaCl_{(s)\text{ }}+CaCO_{3(aq)}$

Firstly, we want to get the expected mass of calcium carbonate

This speaks about getting the theoretical yield based on the equation of reaction

From the data collected, 90 ml of 0.20 M (mol/L) of sodium carbonate gave calcium carbonate

We need to get the actual number of moles of sodium carbonate that reacted

We can get this by multiplying the volume by the molarity (kindly note that we have to convert the volume to Liters by dividing by 1000)

Thus, we have it as:

$\frac{90}{1000}\times\text{ 0.1 = 0.009 moles}$

Hence, we see that 0.009 moles of sodium carbonate reacted theoretically

Since 1 mole of sodium carbonate gave 1 mole calcium carbonate, it is expected that 0.009 mole of sodium carbonate will give 0.009mole of calcium carbonate

What we have to do now is to get the theoretical grams of calcium carbonate produced

That would be the product of the number of moles of calcium carbonate and its molar mass

The molar mass of calcium carbonate is 100 g/mol

The theoretical yield (expected mass) is thus:

$100\text{ g/ mol }\times\text{ 0.009mol = 0.9 g}$

Finally, we proceed to get the percentage yield which is calculated using the formula below:

$\text{Percent Yield = }\frac{Actual\text{ yield}}{\text{Theoretical yield}}\times\text{ 100 \%}$

The actual yield is the observed mass which is given as 0.88 g

The percent yield is thus:

$\frac{0.88}{0.9}\times\text{ 100 = }98\text{ \%}$

7 0

1 year ago

Sam wants to demonstrate how water changes from a solid to a gas. He places ice in a pot on a stove. What variable is causing wa

olganol [36]

Hello,

The answer is option A "<span>the heat from the stove".

Reason:

The answer is option A because the variable the heat from the stove is causing the ice to change into water. Its not option B because thats not in the procedures. Its not option C because it is not a variable. Its also not D because thats also not a variable.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>

8 0

3 years ago

Read 2 more answers

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago

Is this balanced ?2 Fe(NO3)3 + 3 MgSO4 → 3 Mg(NO3)2 + 2 Fez(SO4)3

Alex

it shoudl be : 2Fe(NO3)3 + 3MgSO4 → 3Mg(NO3)2 + Fe2(SO4)3

(the difference is Fe2(SO4)3 has no coefficient)

8 0

3 years ago

Predict the formula of magnesium argonide​

Predict the formula of magnesium argonide