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2 years ago

14

How many grams of lithium sulfide (Li2S) are in 275 g of a 16 % aqueous solution of Li2S?

1 answer:

krok68 [10]2 years ago

3 0

Answer:

600 mAh·g−1

Explanation:

i hope this is good let me know if its wrong

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Raising the temperature of 10.0 g of water from 10.0 °C to 20.0 °C requires 100.0 cal of energy, while raising the temperature o

djyliett [7]

Answer:

Explanation:

The change in enthalpy of a substance when heated is given by

ΔH = m x Csp x ΔT

so the enthalpy change is dependent on the specific heat , i.e its capactity to absorb heat, and is not influenced by factors such as being closer to the melting point, volume , potential energy or being a solid or liquid.

6 0

2 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

2 years ago

Read 2 more answers

N2 (g) + 2O2(g) = 2NO2 (g) ΔH = 66.4 kJ 2NO (g) + O2 (g) = 2NO2 (g) ΔH = -114.2 kJ the enthalpy of the reaction of the nitrogen

RSB [31]

Answer:

ΔH = 180.6 kJ

Explanation:

Given that:

N2 (g) + 2O2(g) = 2NO2 (g) ΔH = 66.4 kJ

<u>2NO (g) + O2 (g) = 2NO2 (g) ΔH = -114.2 kJ </u>

N2 (g) + O2 (g) = 2NO (g) ΔH = ????

The subtraction of both equations would yield the unknown ΔH , therefore:

ΔH = 66.4 - ( - 114.2 kJ)

ΔH = 180.6 kJ

3 0

2 years ago

Read 2 more answers

How much pure water must mixed with 8 pints of 50 developer to produce a mixture that is 24%?

Galina-37 [17]

8:24 = 0.333(3) pints is one percent
0.333(3)* 100=33.333(3) pints will be 24% mixture with the water
33.3333-8=25.333 pints of water is required for producing 24% mixture

25.3333 pints of pure water and 8 pints of juce.

6 0

3 years ago

The effectiveness of nitrogen fertilizers depends on both their ability to deliver nitrogen to plants and the amount of nitrogen

densk [106]

Answer:

Ammonia > Urea > Ammonium nitrate > Ammonium sulphate

Explanation:

Percentage by mass of nitrogen in NH3:

Molar mass of NH3= 17 g/mol

Hence % by mass = 14/17 × 100 = 82.35%

% by mass of NH4NO3

Molar mass of NH4NO3 = 80.043 g/mol

Hence; 28/80.043 × 100 = 34.98%

% by mass of (NH4)2SO4;

Molar mass of (NH4)2SO4= 132.14 g/mol

Hence; 28/132.14 × 100 = 21.19%

% by mass of CH4N2O

Molar mass of urea = 60.0553 g/mol

Hence 28/60.0553 × 100 = 46.62%

8 0

3 years ago