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1 year ago

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

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An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of calcium nitrate.

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<u>Answer:</u> The net ionic equation contains $Ca^{2+}(aq.)$ ions

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium sulfate and calcium nitrate is given as:

$(NH_4)_2SO_4(aq.)+Ca(NO_3)_2(aq.)\rightarrow CaSO_4(s)2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ca^{2+}(aq.)+2NO_3^{-}(aq.)\rightarrow CaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonia and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ca^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow CaSO_4(s)$

Hence, the net ionic equation contains $Ca^{2+}(aq.)$ ions

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2 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

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$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

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