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3 years ago

How many CaH2 formula units are present in 8.294 g of CaH2?

Chemistry

1 answer:

raketka [301]3 years ago

7 0

RMM of CaH2=20+(1x2)=22
Number of moles or formula units
=mass/RMM
=8.294/22
=0.377

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Isotope what dose it mean

Sergio039 [100]

Answer:

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Explanation:

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3 years ago

5.6 moles of gold contain how many atoms? A. 34 atoms B. 9.3 × 10-22 atoms C. 9.3 × 1023 atoms D. 3.4 × 1024 atoms

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3 years ago

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The pre-exponential constant and activation energy for the diffusion of iron in cobalt are 1.7 x 10-5 m2/s and 273,300 J/mol, re

Kitty [74]

Answer: The temperature of the system will be 1622 K

Explanation:

The equation relating the pre-exponential factor and activation energy follows:

$\log D=\log D_o-\frac{E_a}{2.303RT}$

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$E_a$ = activation energy of iron in cobalt = 273,300 J/mol

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T = temperature = ?

Putting values in above equation, we get:

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3 years ago

Your patient gets a prescription for 62.5 mcg (micrograms, mg) of digoxin in liquid form. The label reads 0.0250 mg/mL. How many

blsea [12.9K]

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4 years ago

PLSSSSS HELP I DONT GET THIS PROBLEMMMM

Aleks [24]

Answer:

C. 7370 joules.

Explanation:

There is a mistake in the statement. Correct form is described below:

Using the above data table and graph, calculate the total energy in Joules required to raise the temperature of 15 grams of ice at -5.00 °C to water at 35 °C.

The total energy needed to raise the temperature is the combination of latent and sensible heats, all measured in joules, and represented by the following model:

$Q = m\cdot [c_{i} \cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})]$ (1)

Where:

$m$ - Mass of the sample, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of the sample, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of water, in degrees Celsius.

$Q$ - Total energy, in joules.

If we know that $m = 15\,g$ , $c_{i} = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334.72\,\frac{J}{g}$ , $T_{1} = -5\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ and $T_{3} = 35\,^{\circ}C$ , then the final energy to raise the temperature of the sample is:

$Q = (15\,g)\cdot \left[\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (5\,^{\circ}C)+ 334.72\,\frac{J}{g} + \left(4.184\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (35\,^{\circ}C) \right]$

$Q = 7371.9\,J$

Hence, the correct answer is C.

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3 years ago