What was significant about the rocks Darwin found in the mountains

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

How does changing the concentration of the reactants change the parts of a rate law? It changes the rate, R. It changes the rate

Sidana [21]

Answer

A. It changes the rate, R

Explanation

When we change the concentration of the reactants in a chemical reaction, it affects the rate of reaction that happens in the process. Typically, the rate of reaction will decrease with time if the concentration of the reactants decreases because the reactants will be converted to products. Similarly, the rate of reaction will increase when the concentration of reactants are increased.

4 0

3 years ago

In acetyl CoA formation, the carbon-containing compound from glycolysis is oxidized to produce acetyl CoA. From the following co

kondor19780726 [428]

Answer:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Explanation:

Hello,

In this case, it is important to recall that acetyl-CoA is produced either by oxidative decarboxylation of pyruvate derived from glycolysis, which is carried out into the mitochondrial matrix, by cause of the oxidation of high-order fatty acids, or by oxidative degradation of very specific amino acids. Acetyl-CoA then enters in the citric acid cycle where it is oxidized in the light of energy production.

In this manner, during such processes, there are some net inputs and outputs, therefore, they are sorted as show below, considering there some of them not classified neither as input nor output:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Best regards.

8 0

3 years ago

The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x

Andrews [41]

Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

6 0

3 years ago

When the name of an anion that is part of an acid ends in -ite, the acid name includes the suffix _____.

Marina86 [1]

(B) suffixous

Hope It Helps

3 0

3 years ago

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