How many moles of NaCl are contained in 50.0 mL of 2.50 M NaCl?

Which material is not a metamorphic rock

ASHA 777 [7]

Letter C would be the correct answer

3 0

3 years ago

Determine the oxidation number of sodium in Na202

Olegator [25]

Answer:

+1

Explanation:

Na₂O₂

NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.

Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:

Na₂O₂ = 0 (oxidation number of ground state compound is zero)

2Na + 2O = 0

O = –1

2Na + 2(–1) = 0

2Na – 2 = 0

Collect like terms

2Na = 0 + 2

2Na = 2

Divide both side by 2

Na = 2/2

Na = +1

Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1

5 0

3 years ago

A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

$Molarity=\frac{moles}{\text{Volume of solution(L)}}$

Moles of glucose = $\frac{18.5 g}{180 g/mol}=0.1028 mol$

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = $\frac{0.1028 mol}{0.1 L}=1.028 mol/L$

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = $M_1=1.208 mol$

Volume of the solution taken = $V_1=30.0 mL$

Molarity of the solution after dilution = $M_2$

Volume of the solution after dilution= $V_2=0.500L = 500 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}$

$M_2=0.07248 mol/L$

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

$0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}$

Moles of glucose = $0.07248 mol/L\times 0.1 L=0.007248 mol$

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0

4 years ago

Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) => CH 3OH(l) Using the following information: C(graphite) +

Alika [10]

Answer:

$\Delta H$ for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

$C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)$ $\Delta H_{1}= -393.5 kJ$

$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)$ $\Delta H_{2}= (2\times -285.8)kJ$

$CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)$ $\Delta H_{3}= 726.4 kJ$

---------------------------------------------------------------------------------------------------

$C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)$

$\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ$

4 0

4 years ago

Solid potassium chlorate (kclo3) decomposes into potassium chloride and oxygen gas when heated. how many moles of oxygen form wh

IrinaVladis [17]

Answer:

0.665 mol of O₂.

Explanation:

1. Molecular chemical equation:

2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

2. Mole ratios:

2 mol KClO₃ : 2 mol KCl : 3 mol O₂

3. Number of moles of KClO₃

Number of moles = mass in grams / molar mass

Molar mass of KClO₃ = 122.55 g/mol

Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol

3. Number of moles of O₂

As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many moles of O₂ will be produced from 0.44038 mol of KClO₃.

3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃

x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂

Round to 3 significant figures: 0.665 mol of O₂ ← answer

3 0

3 years ago

Read 2 more answers