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Keith_Richards [23]
2 years ago
8

What can you conclude about the relationship between height and energy?​

Chemistry
1 answer:
Bas_tet [7]2 years ago
7 0

Answer:

Since the gravitational potential energy of an object is directly proportional to its height above the zero position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling of the height will result in a tripling of the gravitational potential energy.

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The ___ is the short form that chemists use to identify an element
bonufazy [111]

Answer:

Chemical Symbols

Explanation:

Im not positive on this one so dont rely on me lol

8 0
3 years ago
Read 2 more answers
Vhat is the molarity of an Nal solution that contains 7.0 g of Nal in 23.0 mL of solution?
Alika [10]

Answer:

Molarity = 5.22 M

Explanation:

Given data:

Mass of sodium chloride = 7.0 g

Volume of solution = 23.0 mL ( 23.0/1000 = 0.023 L)

Molarity = ?

Solution;

Number of moles of NaCl = 7.0 g/ 58.4 g/mol

Number of moles of NaCl = 0.12 mol

Molarity = moles of solute / volume in litter

Molarity = 0.12 mol /  0.023 L

Molarity = 5.22 M

4 0
3 years ago
Write a balanced equation for the combustion of C7H16(l) (heptane) -- i.e. its reaction with O2(g) forming the products CO2(g) a
JulsSmile [24]

Answer:

<u>The standard enthalpy of reaction = -4854.7kJ</u>

<u>The difference: </u>ΔH-ΔE = Δ(PV) = Δn.R.T = <u>9910.288 J ≈ 9.91 kJ</u>    

Explanation:

<u>The balanced chemical equation for the combustion of heptane</u>:

C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)

Given: The standard enthalpy of formation (\Delta H _{f}^{\circ }) for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol

<u>To calculate the standard enthalpy of reaction (\Delta H _{r}^{\circ }) can be calculated by the Hess's law</u>:

\Delta H _{r}^{\circ } = \left [\sum \nu \cdot\Delta H _{f}^{\circ }(products)  \right ] - \left [\sum \nu\cdot\Delta H _{f}^{\circ }(reactants)  \right ]

Here, \nu is the stoichiometric coefficient

⇒ \Delta H _{r}^{\circ } =

\left [ 7\times \Delta H _{f}^{\circ }\left (CO_{2}\right )+ 8\times \Delta H _{f}^{\circ }\left (H_{2}O \right )\right ]

- \left [1\times \Delta H _{f}^{\circ }\left (C_{7}H_{16}\right ) +11\times \Delta H _{f}^{\circ }\left (O_{2} \right ) \right ]

=\left [ 7\times \left (-393.5 kJ/mol \right )+ 8\times \left (-286 kJ/mol \right )\right ]

-\left [1\times \left (-187.8 kJ/mol \right ) +11\times \left (0 kJ/mol \right ) \right ]

⇒ \Delta H _{r}^{\circ } = \left [ \left (-2754.5 \right )+ \left (-2288 \right )\right ]\left -[ \left (-187.8 \right ) +\left (0 \right )\right ]

⇒ \Delta H _{r}^{\circ } = \left [ -5042.5 ]\left -[ -187.8] = \left ( -4854.7kJ \right )

<u>To calculate the difference: </u>ΔH-ΔE=Δ(PV)

We use the ideal gas equation: P.V = n.R.T

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T

Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹

Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = <u>9910.288 J = 9.91 kJ</u>                              (∵ 1 kJ = 1000J )

                                                                             

8 0
3 years ago
Find the mass of a 60 ML volume of water if the density of water is 1 g/mL
valkas [14]

Answer:  60 grams

Explanation:  (60 ml)*(1g/ml) = 60g

3 0
2 years ago
3. Write DISSOCIATION or IONIZATION equations for the following chemicals:
Romashka [77]

Answer:

a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+

b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -

c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.

d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base

e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt

f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt

g) NaCl​ -> Na+ + Cl- ionization of the salt, Sodium Chloride

Explanation:

Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.

8 0
2 years ago
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