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3 years ago

A person's _ changes as he travels from the earth to space, but his _ remains the same.

Chemistry

2 answers:

irina1246 [14]3 years ago

6 0

Answer: weight , mass

Explanation:

Mass is defined as the amount of matter contained in a body. It is measured in units of kg, g and mg.

Weight of an object is the force acting on an object by virtue of its mass.

weight= $m\times g$

m = mass

g = acceleration due to gravity

As the value of g keeps changing from earth to space, the weight of the person would also keep changing. The mass of the person will remain same .

EleoNora [17]3 years ago

4 0

A persons weight changes as he travels from earth to space but his mass remains the same

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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

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

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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A person's _____ changes as he travels from the earth to space, but his _____ remains the same.

A person's _ changes as he travels from the earth to space, but his _ remains the same.