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4 years ago

9

An NHTSA study found that the average inpatient costs for crash victims who were not using safety belts were __________ percent

higher than for those who were belted.
A. 15
B. 35
C. 55
D. 75

1 answer:

jeka944 years ago

5 0

The 1996 NHTSA study, Crash Outcome Data Evaluation System (CODES), related traffic flow and medicinal accounts in seven states to measure entire costs of wound from motor vehicle crashes. The research originate that the mean inpatient costs for crash fatalities who were not using safety belts were 55 percent <span>higher than for those who were belted.</span>

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Difference between NaOH and HCI

morpeh [17]

Compare HCl, NaOH, and NaCl: HCl is a stronger acid than water. NaCl is a weaker base than NaOH. Strong acids react with strong bases to form weaker acids and bases. ... Compare NaOH, NH3, and H2O, and NH4Cl: NaOH is a stronger base than NH3.

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3 years ago

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What number of moles of O2 will be produced by the decomposition of 4.4 moles of water?

adelina 88 [10]

2.2 moles of O2 will be produced by the decomposition of 4.4 moles of water.

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4 years ago

Which of the following is the correct model of C6H14?

asambeis [7]

Answer:

This question is incomplete

Explanation:

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H H H H H H

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I I I I I I

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which can also be written as

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3 years ago

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The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

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$\Delta G^o=-2.303\times R\times T\times \log K_c$

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$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

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3 years ago

an ideal gas is at a pressure 1.00 x 10^5 N/m^2 and occupies a volume 11.00 m^3. If the gass is compressed to a volume of 1.00 m

bogdanovich [222]

Answer:

$P_2=1.1x10^6Pa$

Explanation:

Hello.

In this case, we can solve this problem by applying the Boyle's law which allows us to understand the pressure-volume behavior as a directly proportional relationship:

$P_1V_1=P_2V_2$

In such away, knowing the both the initial pressure and volume and the final volume, we can compute the final pressure as shown below:

$P_2=\frac{P_1V_1}{V_2}$

Consider that the given initial pressure is also equal to Pa:

$P_2=\frac{1.00x10^5Pa*11.00m^3}{1.00m^3}\\ \\P_2=1.1x10^6Pa$

Which stands for a pressure increase when volume decreases.

Regards.

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3 years ago