if 2.22 moles of ammonia (NH3) decomposes according to the reaction shown how many moles of hydrogen (H2) are formed?

A student is making a glucose solution. The student is using 12.55 g of glucose (C6H1206) and a 500 mL volumetric flask. Use thi

Ber [7]

Answer:

M = 0.138 M

Explanation:

Given data:

Mass of glucose = 12.55 g

Volume of solution = 500 mL

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 12.55 g/ 180.156 g/mol

Number of moles = 0.069 mol

Volume in L:

500 mL × 1 L /1000 mL

0.5 L

Molarity:

M = 0.069 mol / 0.5 L

M = 0.138 M

8 0

3 years ago

To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o

Romashka-Z-Leto [24]

Answer:

a) 1.740 g of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

$\pi r^{2}h$ ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

$r=\frac{d}{2}$

$=\frac{3.36x10^{2}}{2}$

= 168 m

Volume =

$=\frac{22\times 168^{2}\times 21.80}{7}$

$=1.93\times 10^{6} m^{3}$

2.Convert ppm to g/m3 and Solve for mass of F-

$1ppm = 1g/m^{3}$

$0.9ppm = 0.9g/m^{3}$

Because both ppm and g/m3 are same quantity .

$g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}$

$0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}$

$mass\ of\ F- =1.740g$

mass of F- required = 1.740 g

3. Apply mole concept to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

$2Na+F_{2}\rightarrow 2NaF$

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

$\frac {84}{38}\times 1.74$

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0

3 years ago

At 218°C, Keq=1.2 x 10^-4 for the equilibrium NH4SH(s) ⇋ NH3(g) + H2S(g). Calculate the equilibrium concentrations of NH3 and H2

maksim [4K]

At 218 °C, solid NH₄SH decomposes to form 0.011 M NH₃ and H₂S, as given by its equilibrium constant.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant (Keq) is the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, all raised to their stoichiometric coefficients.

Only gases and aqueous species are included.

Step 1. Make an ICE chart.

NH₄SH(s) ⇋ NH₃(g) + H₂S(g)

I 0 0

C +x +x

E x x

Step 2. Write the equilibrium constant.

Keq = 1.2 × 10⁻⁴ = [NH₃] [H₂S] = x²

x = 0.011 M

At 218 °C, solid NH₄SH decomposes to form 0.011 M NH₃ and H₂S, as given by its equilibrium constant.

Learn more about equilibrium here: brainly.com/question/5081082

#SPJ1

3 0

2 years ago

In a single carbon bond, two carbon atoms share pair of electrons.

torisob [31]

Answer:

C-C

A carbon has 4 valence electrons meaning it can share 4 electrons and is in short of 4 electrons to fill it outer energy shell making it 8. Which results to 2 double covalent bonds ,meaning there was a sharing of 2(4) = 8 electrons . Hope this helps . Your statement above is true .

8 0

3 years ago

Which sentence most accurately describes electrically charged objects?

ivolga24 [154]

Answer:

They are attracted to one other without coming into contact.

Explanation:

In conclusion, an electrically neutral object is an object that has a balance of protons and electrons. In contrast, a charged object has an imbalance of protons and electrons. The type of charge (positive or negative) is determined by whether the protons or the electrons are in excess

7 0

3 years ago

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