All categories

3 years ago

A certain object has a volume of 25.0 mL and a mass of 100 g. What is the density of the object?

1 answer:

3 0

Density = mass/volume

Since there was no condition of unit attached to the question...

Density = 100g/25.0ml

Density = 4g/mL.......

You might be interested in

What way do S waves move

BlackZzzverrR [31]

S waves shake the ground in a shearing, or crosswise, motion that is perpendicular to the direction of travel. These are the shake waves that move the ground up and down or from side to side.

4 0

1 year ago

]The image shows groundwater zones.

11111nata11111 [884]

Answer:

C.) 3 is the correct answer

7 0

3 years ago

Read 2 more answers

In the Reaction CuCl2 + 2 NaNO3 —> Cu(NO3)2 + 2 Naci

Lunna [17]

Explanation:

The molar mass of the compounds are incorrect

5 0

2 years ago

A mixture of 454 kg of applesauce at 10 degrees Celsius is heated in a heat exchanger by adding 121300 kJ. Calculate the outlet

stira [4]

Explanation:

The given data is as follows.

Mass of apple sauce mixture = 454 kg

Heat added (Q) = 121300 kJ

Heat capacity ( $C_{p}$ ) of apple sauce at $32.8^{o}C$ = 4.0177 $kJ/kg^{o}C$

So, Heat given by heat exchanger = heat taken by apple sauce

Q = $mC_{p} \Delta T$

or, Q = $mC_{p} (T_{f} - T_{i})$

Putting the given values into the above formula as follows.

Q = $mC_{p} (T_{f} - T_{i})$

121300 kJ = $454 kg \times 4.0177 kJ/kg^{o}C \times (T_{f} - 10)$

$T_{f}$ = $76.5^{o}C$

Thus, we can conclude that outlet temperature of the apple sauce is $76.5^{o}C$ .

3 0

3 years ago

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose.

Rainbow [258]

Answer:

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

ΔT = T° frezzing pure solvent - T° freezing solution

(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1

T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C

4 0

3 years ago