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2 years ago

Guys can you help me please it’s due today

Chemistry

2 answers:

Sonja [21]2 years ago

8 0

Answer:

i hope it will help you ..

Stells [14]2 years ago

8 0

Good idea lol but yeah I’m walking back from work when we are back home

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Water boils at what degrees Celsius

Archy [21]

100 degrees Celsius. And apparently this answer needs at least 20 characters to explain it well. But yes, the answer is 100 degrees Celsius

4 0

3 years ago

Read 2 more answers

A 3.82-g sample of magnesium nitride is reacted with 7.73 g of water. mg3n2 (s) + 3 h2o (â) â 2 nh3 (g) + 3 mgo (s) the yield of

horrorfan [7]

The reaction is properly written as

Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s)

Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol

Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429

Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,

Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g

Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 =<em> 78.77%</em>

5 0

3 years ago

Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou

swat32

39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

$Ca(s) + Cl_2(g)$ → $CaCl_2(s)$

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

$m_{ca_C_l_2}$ = 16.7 g Ca x $\frac{1 mol \;of \;Ca}{40g Ca}$ x $\frac{1 mol \;of \;CaCl_2}{1 mol \;Ca}$ x $\frac{111g of \;CaCl_2}{1 mol \;CaCl_2}$

$m_{ca_C_l_2}$ = 39.96 g

Hence, 39.96 g product form when 16.7 g of calcium metal completely reacts.

Learn more about the stoichiometric process here:

brainly.com/question/15047541

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4 0

2 years ago

How many moles of CaCl₂ are required to produce 45.97 grams of NaCl?

marin [14]

Answer is 0.7866187543

6 0

2 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0

3 years ago