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3 years ago

Historically the troubadour was a ____.

1 answer:

6 0

Composer and performer of Old Occitan lyric poetryduring the High Middle Ages (1100–1350). Since the word troubadour is etymologically masculine, a female troubadour is usually called a trobairitz.

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Can somebody help me with this also plz

just olya [345]

ANSWER:

4 a) Specific elements have more than one oxidation state, demonstrating variable valency.

For example, the following transition metals demonstrate varied valence states: $Fe^{2+}$ , $Fe^{3+}$ , $Cr^{2+}, Cr^{3+}$ , etc.

Normal metals such as $Pb^{2+} and Pb^{4+}$ also show variable valencies. Certain non-metals are also found to show more than one valence state $Pb^{3+} and Pb^{5+}.$

4 b) Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons.

For example, Carbon-14 is a naturally occurring radioactive isotope of carbon, having six protons and eight neutrons in the nucleus. However, C-14 does not last forever and there will come a time when it loses its extra neutrons and becomes Carbon-12.

5 a) $2Fe + 3Cl_2$ → $2FeCl_3$

5 b) $3Pb + 8HNO_3$ → $3Pb (NO_3)_2 + 4H_2O + 2NO_2$

5 c) $Zn + H_2SO_4$ → $ZnSO_4 + H_2$ (already balanced so don't need to change)

5 d) $2H_2 + O_2$ → $2H_2O$

5 e) $2Mg + 2HCl$ → $2MgCl + H_2$

EXPLANATION (IF NEEDED):

1. Write out how many atoms of each element is on the left (reactant side) and right (product side) of the arrow.

2. Start multiplying each side accordingly to try to get atoms of the elements on both sides equal.

EXAMPLE OF BALANCING:

8 0

2 years ago

What did GEM Anscombe stand for give a brief description?

KATRIN_1 [288]

Answer:

Explanation:

Anscombe is a philosopher who has approached contemporary themes, today enormously current, precisely in the ethical and social field, stimulating the development of virtue ethics as an alternative to utilitarianism, Kantian ethics and social contract theories. Anscombe argues that what we understand as a moral obligation, what we "must" do, is an obligation to God.

6 0

3 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago

The less mass in a given volume of air the

DiKsa [7]

Answer:

The less mass in a given volume of air the less dense the air is going to be.

7 0

3 years ago

Read 2 more answers

How do the atmosphere conditions near the beginning of Precambrian time contrast with the atmosphere conditions that are present

babunello [35]

The early precambrian atmosphere consisted primarily of nitrogen and carbon dioxide with almost no oxygen.

<span>Today, the atmosphere contains about 20% oxygen, less carbon dioxide and similar amounts of nitrogen. </span>

<span>Photosynthetic green-leaf plants and trees are largely responsible for the change, converting carbon dioxide to oxygen.</span>

5 0

3 years ago