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MrRissso [65]
3 years ago
13

The winner of a 10 km road race took half an hour to complete the race. Calculate her average speed. Give your answer in metres

per second.
Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
3 0

Explanation:

10 km = 10000m

30 min = 1800 sec

10000 ÷ 1800 = 5.556 m/s

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9.8 x 10​^-6 regular notation
topjm [15]

Answer:

0.0000098 should be the answer

Explanation:

3 0
3 years ago
Can someone help me, please?
VikaD [51]
15. D is correct, exothermic reactions release heat

I am not sure about 16
7 0
3 years ago
What is the mass percentage for copper ?
RideAnS [48]

Answer:

Explanation:

To find the mass percent composition of an element, divide the mass contribution of the element by the total molecular mass. This number must then be multiplied by 100% to be expressed as a percent.

3 0
3 years ago
Which of the following molecules would exhibit only London forces?a)CH4, BH3, and CCl4b)H2O, NH3, and CCl4c)PH3, NH3, and CCl4
Alex787 [66]

Answer:

a)CH₄, BH₃, and CCl₄

Explanation:

<u>London dispersion forces:- </u>

The bond for example, in the molecule is F-F, which is non-polar in nature because the two fluorine atoms have same electronegativity values.

The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.

Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-

<u>Hydrogen bonding:- </u>

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Thus option B and C rules out.

<u>Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄</u>

4 0
4 years ago
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
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