3 years ago

The sugar molecule known as sucrose is a polar molecule. Which statement BEST summarizes how sucrose will dissolve in water?

Physics

1 answer:

kirill [66]3 years ago

8 0

I believe the answer is C

Hope this helps :)

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High blood levels of LDL cholesterol reduces your risk of developing heart

SVEN [57.7K]

Answer:

false

Explanation:

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3 years ago

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A car moves with the speed of 120m/s for 4 minutes ,calculate the distance covered by the car

Vsevolod [243]

Answer:

960 m

Explanation:

Given that,

Speed = 120 m/s
Time taken = 4 minutes

We have to find the distance covered.

Firstly, let's convert time in seconds.

→ 1 minute = 60 seconds

→ 4 minutes = (4 × 60) seconds

→ 4 minutes = 240 seconds

Now, we know that,

→ Distance = Speed × Time

→ Distance = (4 × 240) m

→ Distance = 960 m

Therefore, distance covered is 960 m.

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3 years ago

Which of the following is NOT true about energy conversion devices?

konstantin123 [22]

Answer:

Explanation:

Option A, B and C are true about energy conversion devices, only option D is NOT true about energy conversion devices.

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3 years ago

The half-lives for the radioactive decay of various elements are listed below.

myrzilka [38]

Answer:

D ko alam pasensya ka na ha

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3 years ago

A house is losing heat at a rate of 1600 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

Setler [38]

Answer:

1600 kJ/h per K, 888.88 kJ/h per °F and 888.88kJ/h per R

Explanation:

We make use of relations between temperature scales with respect to degrees celsius:

$1 K= 1^{\circ}C+273\\1^{\circ}F= (1^{\circ}C*1.8)+32\\1 R= (1^{\circ}C*1.8)+491.67$

This means that a change in one degree celsius is equivalent to a change of one kelvin, while for a degree farenheit and rankine this is equivalent to a change of 1.8 on both scales.

So:

$\frac{Q}{\Delta T(K)}=\frac{Q}{\Delta T(^\circ C)}=1600 \frac{kJ}{h} per K\\\frac{Q}{\Delta T(^\circ F)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per ^\circ F\\\frac{Q}{\Delta T(R)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per R$

5 0

3 years ago