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Bingel [31]
1 year ago
14

Convert 675 nanograms to milligrams.

Chemistry
1 answer:
STALIN [3.7K]1 year 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

#SPJ1

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The mass of an iron(II) sulfate crystal is 8.36 g.How many moles of FeSO4 are in the crystal?
Yuki888 [10]

The number of moles present in the FeSO4 are 0.055 mol.

<u>Explanation:</u>

  • The mass of a substance containing the same number  atoms in 12.0 g of 12C is known as mole. One mole of any substance is equal to  6.023 x 10^23. The moles of a substance can be determined by using the formula,

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Given,

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What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g
ArbitrLikvidat [17]
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Explanation:

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Moles of <span>KCl<span>O3</span><span>(s)</span></span> = <span><span>10.0⋅g</span><span>122.55⋅g⋅mo<span>l<span>−1</span></span></span></span> = <span>0.0816⋅mol</span>

And thus <span><span>32</span>×0.0816⋅mol</span> dioxygen are produced, i.e. <span>0.122⋅mol</span>.

At STP, an Ideal Gas occupies a volume of <span>22.4⋅L⋅mo<span>l<span>−1</span></span></span>.

And thus, volume of gas produced = <span>22.4⋅L⋅mo<span>l<span>−1</span></span>×0.0816⋅mol≅3L</span>

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8 0
3 years ago
Read 2 more answers
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emmasim [6.3K]
When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant. 

Let's do the ICE analysis

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-------------------------------------
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The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

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K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}

Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.
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