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Bingel [31]
2 years ago
14

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

#SPJ1

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Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature
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Answer:

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Explanation:

The equation of the reaction can be represented as:

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Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

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