5. Water is known to have a composition which is 88.8% oxygen by mass. If you had a

A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man

kap26 [50]

<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

Mass of NaBr sample is 11.97 g

% composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

A sample of 11.97 g of NaBr contains 22.34% of Na by mass

Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

= 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0

2 years ago

When molecules are joined I an ordered structure

olga2289 [7]

Answer:

ice

Explanation:

5 0

3 years ago

1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.

Snowcat [4.5K]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of $CH_3OH$ = 26.5 g

Molar mass of $CH_3OH$ = 32.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{26.5\ g}{32.04\ g/mol}$

$Moles\ of\ CH_3OH= 0.8271\ moles$

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

$m=\frac {0.8271\ moles}{0.735\ kg}$

<u>Molality = 1.13 m</u>

4 0

3 years ago

30 points please no links!) A sea star has one of its arms break off. That arm grows into its own separate organism through ____

Juli2301 [7.4K]

Answer:

B) Regeneration

the action or process of regenerating or being regenerated, in particular the formation of new animal or plant tissue.

PLEASE MARK THIS AS BRAINLIEST!

4 0

3 years ago

Read 2 more answers

The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K

BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + $\int\limits^{T2}_{T1} {DCp(T)} \, dT$

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

$\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT$ = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

7 0

3 years ago