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Minchanka [31]
3 years ago
14

1.34 milligrams is the same as _______kg and ______g

Chemistry
2 answers:
Alina [70]3 years ago
5 0

Answer : The 1.34 milligrams is the same as 1.34\times 10^{-6}kg and 1.34\times 10^{-3}g

Explanation :

The conversion used for mass from milligram to kilogram is:

1mg=10^{-6}kg

The conversion used for mass from milligram to gram is:

1mg=10^{-3}g

As we are given that the mass 1.34 mg. Now we have to determine the mass in kg and g.

Mass of 'kg' :

1mg=10^{-6}kg

1.34mg=\frac{1.34mg}{1mg}\times 10^{-6}kg=1.34\times 10^{-6}kg

Mass of 'g' :

1mg=10^{-3}g

1.34mg=\frac{1.34mg}{1mg}\times 10^{-3}g=1.34\times 10^{-3}g

Therefore, the 1.34 milligrams is the same as 1.34\times 10^{-6}kg and 1.34\times 10^{-3}g

Maurinko [17]3 years ago
3 0
There are 1000 mg in 1 g
and there are 1000 g in 1 kg

Start by converting 1.34 mg to grams by dividing 1.34 mg by 1000 g = 0.00134 g

Then convert 0.00134 g to kg by dividing 0.00134 g by 1000 kg = 1.34×10^-6 kg OR 0.00000134 kg
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Answer:

D because excited electrons fall back.....

6 0
3 years ago
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when hydrogen atom absorbs a photon, and an electron moves level 1 to level 2, what happens to the energy of the atom?
pychu [463]

Answer:

When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon).

Explanation: pls mark brainliest :))

7 0
2 years ago
Given the equation:
My name is Ann [436]

Answer:

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6 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
2 years ago
Which of the following is an acid-base neutralization reaction? (1 point)
GuDViN [60]
It would be NaOH + HCl → <span>NaCl + H2O
</span>
NaOH is sodium hydroxide, which is a strong base. HCl  is hydrochloric acid, which is a strong acid. 

You have a strong base and a strong acid on the left side, however, at the result side, you end up with NaCl + H2O. Sodium chloride is simply table salt and H2O is just water, thus it has been neutralized. 


6 0
3 years ago
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