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2 years ago

An oral solution has a dosage of 500 mg in 5 mL. If a 0.7 g dosage is required, how many mL must be given?

Physics

1 answer:

lina2011 [118]2 years ago

8 0

Answer:

7 mL

Explanation:

Given that:

mass of an oral solution = 500 mg

solution in mL = 5mL

Thus, the ratio = 5 mL/500 mg

= 0.01 mL/mg

since 1000 mg = 1g

∴

= 0.01 mL/mg × 1000 mg/1g

= 10 mL/g

Now, suppose the amount of 0.7 g dosage is required:

∴

the needed amount of mL will be:

= 10 mL/g × 0.7 g

= 7 mL

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The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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3 years ago

An artillery shell of mass 30 kg has a velocity of 250 m/s vertically upward. The shell explodes into two pieces; immediately af

olga_2 [115]

Answer:

9654.34 m

Explanation:

from conservation of momentum

$\begin{aligned}30 \times 250 &=-10 \times 120+20 \times V \\20 V &=30 \times 250+10 * 120 \\V &=\frac{30 \times 250+10 \times 120}{20}=435 \mathrm{~m} / \mathrm{s}\end{aligned}$

And from Conservation of Energy

$\frac{1}{2} m v^{2}=m g h\\h=\frac{v^{2}}{2 g}\\h=\frac{(435(m/s))^{2}}{2 \times 9.8(m/s^{2} )}\\h=9654.34 (m)$

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2 years ago

The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spri

Alja [10]

Answer:

D. "The net force is zero, so the acceleration is zero"

Explanation:

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3 years ago

What happens when gasoline is used to power a vehicle?

mojhsa [17]

<span>Energy is neither lost nor gained as it transforms from chemical, to heat, to mechanical energy.</span>

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3 years ago

Infrared radiation has frequencies from 3.0×1011 to 3.0×1014 Hz, whereas the frequency region for microwave radiation is 3.0×108

igor_vitrenko [27]

Answer:

1. The speed of infrared radiation is the same as microwave radiation.

2. The wavelength of infrared radiation is lower than microwave radiation.

Explanation:

1. The speed of infrared radiation is the same as microwave radiation. This is because both infrared radiation and microwave radiation are electromagnetic waves and all electromagnetic waves move at the same speed, the speed of light.

2. Frequency and wavelength has an inverse relationship. This means that the higher the frequency, the lower the wavelength.

Since the frequency range of infrared waves is higher than that of microwaves, the wavelength range of infrared waves is lower than that of microwaves.

We can prove this by using the maximum frequency value of each wave to calculate their corresponding wavelengths.

Speed of light, c, is given as:

c = λf

Where λ is wavelength and f is frequency.

Wavelength is therefore:

λ = c/f

For infrared wave, the maximum frequency is 3 * 10^14 Hz, hence the corresponding wavelength is:

λ = (3 * 10^8) / (3 * 10^14)

λ = 10^(-6) m

For microwave, the maximum frequency is 3 * 10^11 Hz, hence the corresponding wavelength is:

λ = (3 * 10^8) / (3 * 10^11)

λ = 10^(-3) m

Hence, the wavelength of infrared waves is lower than the wavelength of microwaves.

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2 years ago