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3 years ago

14

Andrea is investigating how increasing the concentration of sugar affects the boiling point of the solution. What would be the b

est way to present the results to help Andrea see if there is a pattern?

1 answer:

NNADVOKAT [17]3 years ago

5 0

An increase in the concentration of sugar in a solution leads to the increase in boiling point of the solution.

<h3>What is boiling point?</h3>

Boiling point of a solution is that pressure exerted on a liquid by its surrounding is equal to the vapour pressure of the liquid.

When there is an increase in concentration of solute, more space will be occupied in the fluid leading to an increase in the boiling point of the fluid.

Therefore, An increase in the concentration of sugar in a solution leads to the increase in boiling point of the solution.

Learn more about boiling point here:

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Object A weighs 750 N on earth. Object B weighs 750 N on Jupiter.

Katarina [22]

The object A has the greater mass compared to object B.

<u>Explanation: </u>

The weight of any object on any planet is the measurement of gravity’s influence acting on the mass of the object. So for Earth, the acceleration will be acting on the object A’s mass (m) in Earth leading to the weight of the object A as 750 N.

While the acceleration of Jupiter will be acting on the object B’s mass kept in Jupiter to attain the weight of 750 N. So, the mass of both the objects at their respective planet will vary depending on the acceleration of each planet. We can check this as below:

$\text { object A's weight }=\text { m of object A } \times \text { Acceleration of Earth}$

So,

$750 \mathrm{N}=\text { m of object } A \times 9.8 \mathrm{m} / \mathrm{s}^{2}$

Thus,

$\text { m of object } A=\frac{750}{9.8}=76.5 \mathrm{kg}$

Similarly for object B,

$\text { Weight of object } B=m \text { of object } B \times \text { Acceleration due to gravity of Jupiter }$

$750 = m \text { of object } B \times 24.79 \mathrm{m} / \mathrm{s}^{2}$

Thus,

$\text { m of the object } B=\frac{750}{24.79}=30.25 \mathrm{kg}$

Thus, the mass of object A is greater than the mass of object B.

5 0

4 years ago

A football is kicked into the air with a horizontal velocity of 20 m/s and a vertical velocity of 30 m/s what is the resultant v

kati45 [8]

For this one, you can use Pythagorean theorem.

a^2 + b^2 = c^2 -------- c will be the resultant

(20 m/s)^2 + (30 m/s)^2 = c^2

solve for c.

= 36.1 m/s

Please brainliest me if this helped; trying to level up!

6 0

4 years ago

A hair dryer is a pump for air. Do a battery and a Genecon act like pumps for charge?

lyudmila [28]

Answer:

No

Explanation:

8 0

4 years ago

Read 2 more answers

Two cars started at the same position and were driven through a city using different streets. The red car traveled north 2 miles

inysia [295]

Answer:

The green car traveled a shorter distance, but the displacement of the cars was equal.

Explanation:

4 0

3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago