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Westkost [7]
3 years ago
15

What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?

Physics
1 answer:
pentagon [3]3 years ago
7 0

Answer:108.71 mL

Explanation:

Given

Volume of sample V=150 mL

concentration of sucrose solution 35 % w/w i.e. In  100 gm of sample 35 gm is sucrose

specific gravity =1.115

Density of solution \rho _s=1.115\times density\ of\ water

Thus

\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL

mass of sample M=1.115\times 150=167.25\ gm

mass of sucrose m_s=0.35\times 167.25=58.53\ gm

mass of Water m_w=108.71 gm

Volume of water =108.71\times 1=108.781 mL

You might be interested in
Why doesn't every planet have a moon?​
GalinKa [24]

Answer:

Up first are Mercury and Venus. Neither of them has a moon. Because Mercury is so close to the Sun and its gravity, it wouldn't be able to hold on to its own moon. Any moon would most likely crash into Mercury or maybe go into orbit around the Sun and eventually get pulled into it.

5 0
2 years ago
A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s

Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90

a = atan(Vy0/Vx0) = 44.9 degrees

5 0
3 years ago
How much heat is released when a 10.0-g sample of iron cools from 75.0°c to 25.5 °c? the specific heat capacity of iron is 0.449
GalinKa [24]
M = 10.0 g, the mass of the iron sample
ΔT = 75 - 25.2 = 49.5°C, the decrease in temperature
c = 0.449 J/(g-°C), the specific heat of iron

The heat released is
Q = m*c*ΔT
    = (10.0 g)*(0.449 J/(g-°C))*(49.5 C)
    = 222.255 J

Answer: 222.3 J (nearest tenth)

8 0
3 years ago
Which trophic level has the least available energy in kilojoules in this food web?
insens350 [35]

The highest trophic level has the least available energy in kilojoules.

Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.

Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.

At the highest trophic level, the the least available energy in kilojoules in this food web is found.

Learn more: brainly.com/question/2233704

7 0
2 years ago
MamaMia's Pizza purchases its pizza delivery boxes from a printing supplier. MamaMia's delivers on-average 200 pizzas each month
Ira Lisetskai [31]

Answer:

a) 138 units

b) 17 units

c) 17 units

d) Total Cost = $353.35

Explanation:

Given:

Average pizzas delivered = 200

Charge of inventory holding = 30% of cost

Lead time = 7 days

Now,

a) Economic Order Quantity =  \sqrt\frac{2\times\textup{Annual Demand}\times\textup{Cost per Order}}{\textup{Carrying cost}}

also,

Annual Demand = 200 × 12 = 2400

Cost per Order = Cost of Box + Processing Costs

= 30 cents + $10

= $10.30

and, Carrying Cost = \frac{\textup{Total Inventory Cost}}{\textup{total annual demand}}

=\frac{\textup{Total Cost per order}\times\textup{Annual demand}\times\frac{25}{100}}{\textup{Annual demand}}

= \frac{\$10.30\times2400}\times\frac{25}{100}}{2400}

= $2.575

Therefore,

Economic Order Quantity =  \sqrt\frac{2\times\textup{2400}\times\textup{10.30}}{\textup{2.575}}

= 138.56 ≈ 138 units

b) Reorder Point

= (average daily unit sales × the lead time in days) + safety stock

= (\frac{200}{30}\times7

= 46.67 ≈ 47 units

c) Number of orders per year = \frac{\textup{Annual Demand}}{\textup{Economic order quantity}}

= \frac{\textup{2400}}{\textup{138}}

= 17.39 ≈ 17 units

d) Total Annual Cost (Total Inventory Cost)

= Ordering Cost + Holding Cost

Now,

The ordering Cost = Cost per Order × Total Number of orders per year

= $10.30 × 17

= $175.1

and,

Holding Cost = Average Inventory Held × Carrying Cost per unit

Average Inventory Held = \frac{\textup{0+138}}{\textup{2}} =  69

Carrying Cost per unit = $2.575

Holding Cost = 69 × $2.575 =  

$177.675

Therefore,

Total Cost = Ordering Cost + carrying cost

= $175.1  + $177.675 = $353.35

5 0
3 years ago
Read 2 more answers
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