CHEMISTRY COLLEGE

Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or 3 significant figures, e. g. 9.8% or 10.2%.

Answers

Answer 1

Answer:

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

What is percentage mass?

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

$\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} * 100 \%$

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

$\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} * 100 \%\n\n\text{Percent Mass} = 10.6 \%}$

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

Learn more about percentage mass:

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Answer 2

Answer:

Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of

hydration must be included.

1 Manganes 52.94 g = 63.55 g

1 Sulphur 32.07 g =

32.07 g 2 Hydrogen is = 2.02 g

4 Oygen =

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol 18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

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Answers

Answer:

A. 2-keto acid decarboxylase and alcohol dehydrogenase

Explanation:

2-keto acid decarboxylase and alcohol dehydrogenase are used to produce many higher alcohols. These enzymes also display a high degree of specificity on their substrate

How is energy transfer connected to your life

Answers

Answer:

Baking, microwave, heating system for your house, water boiler, fridge.

A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Answers

Answer: The concentration of KOH solution is 1.215 M

Explanation:

For the given chemical equation:

$2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)$

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\nM_1=1.50M\nV_1=16.2mL\nn_2=1\nM_2=?M\nV_2=40.0mL$

Putting values in above equation, we get:

$2* 1.50* 16.2=1* M_2* 40.0\n\nM_2=(2* 1.50* 16.2)/(1* 40.00)=1.215M$

Hence, the concentration of KOH solution is 1.215 M

HELP!!!!!!
I DON'T KNOW THE ORDER!!

Answers

Answer:

for 1 solid its freezing.

for 2 solid and liquid its melting

for 6 liquid to gas its evaporation and for 5 gas to liquid its condensation.

Explanation:

hope this helped :)

Answer:

solid->liquid= melting

liquid->solid= freezing

gas->liquid= consendation

liquid->gas= evaporation

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C. How are these numbers affected by the addition of 0.1 mol/dm3 of KCL? At what distance from the particle surface (r) has the potential decayed to 1% of its initial value?

Answers

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^(3)$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 * 10^(25) ions/m^(3)$

T = $30^(o)C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_(D)$ ) is as follows.

$\lambda_(D)$ = $(1)/(k) = \sqrt (\varepsilon \varepsilon_(o) K_(g)T)/(2 n^(o) z^(2) \varepsilon^(2))$

where, $n^(o)$ = concentration = $6.022 * 10^(25) ions/m^(3)$

Hence, putting the given values into the above equation as follows.

$\lambda_(D)$ = $\sqrt (\varepsilon \varepsilon_(o) K_(g)T)/(2 n^(o) z^(2) \varepsilon^(2))$

= $\sqrt (78 * 8.854 * 10^(-12) c^(2)/Jm * 1.38 * 10^(-23)J/K * 303 K)/(2 * 6.022 * 10^(25) ions/m^(3) * (1)^(2) * (1.6 * 10^(-19)C)^(2))$

= $9.669 * 10^(-10)$ m

or, = $9.7 A^(o)$

= 1 nm (approx)

Also, it is known that $\lambda_(D)$ = $\sqrt (1)/(n^(o))$

Hence, we can conclude that addition of 0.1 $mol/dm^(3)$ of KCl in 0.1 $mol/dm^(3)$ of NaBr " $\lambda_(D)$ " will decrease but not significantly.

calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass

Answers

Answer:

We are given that there is 95% ethanol by mass in rectified spirit

so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water

we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction

Moles of Ethanol:

Molar mass of ethanol = 46 grams / mol

Number of moles = Given mass / molar mass

Number of moles = 95 / 46

Moles of Ethanol = 2 moles (approx)

Moles of Water:

Molar mass of water = 18 grams per mol

Number of moles = Given mass / molar mass

Moles of water = 5 / 18

Moles of water = 0.28 moles (approx)

Mole Fractions:

Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution

Mole fraction of Ethanol:

Moles of ethanol / (moles of ethanol + moles of water)

2 / (2 + 0.28)

2 / (2.28) = 0.9 (approx)

Mole fraction of Water:

Moles of water / (Moles of ethanol + moles of water)

0.28 / (2 + 0.28)

0.28 / (2.28) = 0.1 (approx)

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