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Flura [38]

Answer:

6.50 g of Hydrogen

Explanation:

We know that in every 20.0g of sucrose, there are 1.30g of hydrogen.

We now have 100.0g of sucrose. 100.0g is 5x larger than the 20.0g sample, which is a 5 : 1 ratio. Applying this ratio to the amount of hydrogen, we would have 5*1.3g of hydrogen in the 100.0g of sucrose.

5*1.3 = 6.5, so our answer is that there are 6.50g of hydrogen in 100.0g of sucrose.

Hope this helps!

5 0

1 year ago

If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele

atroni [7]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

Energy difference btween the two electronic states can be expressed as :

${ \qquad \sf \dashrightarrow \: \Delta E = h\nu}$

[ h = planks constant, ${\: \nu }$ = frequency ]

$\qquad \sf \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14} \times \nu$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 } \times 10 {}^{14}$

$\qquad \sf \dashrightarrow \: \nu \approx 5.395 \times10 {}^{14} \:\:hertz$

5 0

1 year ago

(Someone please help!)

ElenaW [278]

Answer:

COLORS

Explanation:

i looked it up lol

5 0

2 years ago

Read 2 more answers

if I have 22.1 grams of CO2 gas at -188 degrees Celsius and 165 ml what is the pressure of the gas in kPa

Harman [31]

Answer:

P = 2.145kPa

Explanation:

Mass = 22.1g

Molar mass of CO2 = 44g/mol

Vol = 165mL = 0.165L

T = -188°C = (-188 + 273.15)K = 85.15K

R = 8.314J/mol.K

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles if the gas

R = ideal gas constant

T = temperature of the gas

n = number of moles

n = mass / molar mass

n = 22.1 / 44 = 0.50moles

PV = nRT

P = nRT/ V

P = (0.5 × 8.314 × 85.15) / 0.165

P = 2145.26Pa = 2.145kPa

Pressure of the gas is 2.145kPa

6 0

2 years ago

Calculate the mole fraction of kbr (molar mass 119.00 g/mol) in a solution made by dissolving 0.30 g kbr in 0.400 l of H2O (d =

julia-pushkina [17]

The mole fraction of KBr in the solution is 0.0001

<h3>How to determine the mole of water</h3>

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
Density of water = 1 g/mL
Mass of water =?

Density = mass / volume

1 = Mass of water / 400

Croiss multiply

Mass of water = 1 × 400

Mass of water = 400 g

Finally, we shall determine the mole of the water

Mass of water = 400 g
Molar mass of water = 18.02 g/mol
Mole of water = ?

Mole = mass / molar mass

Mole of water = 400 / 18.02

Mole of water = 22.2 moles

<h3>How to de terminethe mole of KBr</h3>

Mass of KBr = 0.3 g
Molar mass of KBr = 119 g/mol
Mole of KBr = ?

Mole = mass / molar mass

Mole of KBr = 0.3 / 119

Mole of KBr = 0.0025 mole

<h3>How to determine the mole fraction of KBr</h3>

Mole of KBr = 0.0025 mole
Mole of water = 22.2 moles
Total mole = 0.0025 + 22.2 = 22.2025 moles
Mole fraction of KBr =?

Mole fraction = mole / total mole

Mole fraction of KBr = 0.0025 / 22.2025

Mole fraction of KBr = 0.0001

Learn more about mole fraction:

brainly.com/question/2769009

#SPJ1

6 0

1 year ago