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2 years ago

Convert 675 nanograms to milligrams.

Chemistry

1 answer:

STALIN [3.7K]2 years ago

7 0

675 nanograms is equivalent to 6.75 × 10-⁴ milligrams.

<h3>How to convert nanograms to milligrams?</h3>

Nanograms is a unit of mass equal to 0.000000001 grams and has a symbol of ng.

On the other hand, milligrams is another unit of mass equal to 0.001grams and has a symbol of mg.

Nanograms and milligrams are both units of mass and can be inter-convertible as follows:

1 nanogram = 1 × 10-⁶ milligram

According to this question, 675 nanograms is equivalent to 675 × 10-⁶ milligrams.

Therefore, 675 nanograms is equivalent to 6.75 × 10-⁴ milligrams.

Learn more about milligrams at: brainly.com/question/20320382

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A common characteristic of killers who choose poison as their weapon. This way they can avoid having to confront their victim me

Gekata [30.6K]

Answer:

The killers who choose poison for killing the victim are the one who does not want to confront the victim. When a knife or a bullet is used to kill a person he may struggle and can cause harm to killer also.

Explanation:

Poison is the most easiest way to kill a person without any struggle. The poison can be given to a person in a juice or through an injection. The poison entered in the body of victim will cause his heart to cease gradually and he will not have energy to struggle with the killer to save his life.

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3 years ago

A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.

Bingel [31]

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

$n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH$

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

$m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO$

And consequently the moles:

$n_O=0.12molO$

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

$C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1$

Thus, the empirical formula, taken the nearest whole number is:

$C_5H_{10}O_2$

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

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