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NARA [144]
3 years ago
7

How much time will it take a car traveling at an average of 75 mph to go a distance of 225 miles​

Chemistry
1 answer:
mojhsa [17]3 years ago
7 0

Time taken = 3 hours

<h3>Further explanation</h3>

Given

speed : 75 mph

distance : 225 miles

Required

time taken

Solution

An equation of constant velocity motion  

\large {\boxed {\bold {d = v \times \: t}}}

d = distance = m  

v = speed = m / s  

t = time = seconds  

Input the value :

t = d : v

t = 225 miles : 75 miles/hour

t = 3 hours

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Biotic - animals anything tht is living abiotic anything tht is non-living

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3 years ago
What is the volume of a 5.30g piece of aluminum? The density for aluminum is 2.70mL
Makovka662 [10]

Answer:

1.96mL

Explanation:

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So:

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7 0
3 years ago
Which of the following statements best describes why a solution of 6.00 g of Ca(NO3)2?in 30.0 g of water has a greater boiling-p
sukhopar [10]
Boiling-point elevation is a colligative property.

That means, the the boiling-point elevation depends on the molar content (fraction) of solute.

The dependency is ΔTb = Kb*m

Where ΔTb is the elevation in the boiling point, kb is the boiling constant, and m is the molality.

A solution of 6.00 g of Ca(NO3) in 30.0 g of water has 4 times the molal concentration of a solution of 3.00 g of Ca(NO3)2 in 60.0 g of water.:

(6.00g/molar mass) / 0.030kg = 200 /molar mass
(3.00g/molar mass) / 0.060kg =   50/molar mass

=> 200 / 50 = 4.

Then, given the direct proportion of the elevation of the boiling point with the molal concentration, the solution of 6.00 g  of CaNO3 in 30 g of water will exhibit a greater boiling point elevation.

Or, what is the same, the solution with higher molality will have the higher boiling point.
5 0
3 years ago
Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kin
Dennis_Churaev [7]

Answer:

35.75 days

Explanation:

From the given information:

For first-order kinetics, the rate law can be expressed as:

\mathsf{In \dfrac{C}{C_o} = -kt}

Given that:

the rate degradation constant = 0.12 / day

current concentration C = 0.05 mg/L

initial concentration C₀ = 3.65 mg/L

\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}

㏑(0.01369863014) = -(0.12) t

-4.29 = -(0.12)

t = -4.29/-0.12

t = 35.75 days

3 0
3 years ago
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