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3 years ago

Li(OH) --> Li2O + H20

Chemistry

1 answer:

Zigmanuir [339]3 years ago

5 0

Answer:

2Li(OH)-->Li2O+H20

Explanation:

There are 2Li, 2H and 2O on each side

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Help me get this right no links

Travka [436]

Answer:

Explanation:

Because the Earth rotates through two tidal “bulges” every lunar day, coastal areas experience two high and two low tides every 24 hours and 50 minutes!!!

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3 years ago

What are the mechanical properties of solids

sergejj [24]

Stiffness
Elastic behaviour
Plastic deformation etc

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4 years ago

Photons of infrared radiation are responsible for much of the warmth we feel when holding our hands before a fire. These photons

KIM [24]

Answer:

$8.3\cdot 10^{22}$ photons

Explanation:

The energy of a photon is given by

$E_1 = \frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the photon wavelength

Here we have

$\lambda=1.5\cdot 10^{-6} m$

So, the energy of 1 of these infrared photons is

$E_1=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.5\cdot 10^{-6}}=1.32\cdot 10^{-19} J$

The amount of energy needed to increase the temperature of the cup of water is:

$E=mC\Delta T$

where

m = 175 g is the mass

$C=4.186 J/gC$ is the specific heat capacity

$\Delta T=40-25=15^{\circ}C$ is the increase in temperature

So,

$E=(175)(4.186)(15)=10,988 J$

Therefore, the number of photons needed is:

$n=\frac{E}{E_1}=\frac{10,988}{1.32\cdot 10^{-19}}=8.3\cdot 10^{22}$

5 0

3 years ago

True or false water readily dissolves most polar

Darina [25.2K]

The answer is false, hope this helps.

5 0

3 years ago

2. How many grams of water can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories?

sertanlavr [38]

Answer:

672 g

Explanation:

We can calculate the mass of water that can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories using the following expression.

$Q = c \times m \times \Delta T$

where,

c: specific heat of the water

m: mass

ΔT: change in the temperature

$m = \frac{Q}{c \times \Delta T } = \frac{8,064cal}{(1cal/g. \° C) \times (37.0 \° C - 25.0 \° C) } = 672 g$

The mass of water that can be warmed under these conditions is 672 grams.

5 0

3 years ago