Select the two statements that are TRUE about compounds

jeyben [28]

Water intoxication occurs when:
water is consumed in larger amounts without much electrolyte take in.

4 0

3 years ago

How could you determine whether one of these two suspected compounds was identical to the unknown compound without using any for

sukhopar [10]

Answer: Use Mixture Melting Point

Explanation:

A procedure called mixture melting point would be used to determine whether or not the suspected compound is identical to the unknown.

The two suspected compounds would need to be used to create a new mixture and determine the mixtures melting point. Compare this melting points with that of the unknown compound in order to determine which one of these two suspected compounds is identical to the unknown compound.

4 0

3 years ago

sound travels 1,500 m/s through water at 25 degrees celciesunder these conditions how long would it take to travel 300m

lesya [120]

Answer:

$v = s \div t$

$t = s \div v$

$t = 300 \div 1500 = 0.2s$

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:

The standard enthalpy of reaction = -4854.7kJ

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

Explanation:

The balanced chemical equation for the combustion of heptane:

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

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

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

Here, $\nu$ is the stoichiometric coefficient

⇒ $\Delta H _{r}^{\circ } =$

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

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

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

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

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

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

To calculate the difference: 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) = 9910.288 J = 9.91 kJ (∵ 1 kJ = 1000J )

8 0

3 years ago

Calculate the specific heat capacity for a 15.3-g sample of gold that absorbs 87.2 J when its temperature increases from 35.0 °C

diamong [38]

Answer:

The specific heat of gold is 0.129 J/g°C

Explanation:

Step 1: Data given

Mass of gold = 15.3 grams

Heat absorbed = 87.2 J

Initial temperature = 35.0 °C

Final temperature = 79.2 °C

Step 2:

Q = m*c*ΔT

⇒ Q =the heat absorbed = 87.2 J

⇒ m = the mass of gold = 15.3 grams

⇒ c = the specific heat of gold = TO BE DETERMINED

⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C

87.2 J = 15.3g * c * 44.2°C

c = 87.2 / (15.3 * 44.2)

c = 0.129 J/g°C

The specific heat of gold is 0.129 J/g°C

4 0

3 years ago