Which organisms have greater variation and produce offspring through asexual reproduction?

A calorimeter contains 500 g of water at 25°C. You place a hand warmer containing 100 g of liquid sodium acetate (NaAC) inside t

antoniya [11.8K]

The heat absorbed by the water is
Q = 500 (4.18) (32.2 - 25)
Q = 15048 J

The enthalpy of fusion of the sodium acetate is:
ΔHf = Q / m
ΔHf = 15048 / 100
ΔHf = 150.48 J/g

3 0

3 years ago

Read 2 more answers

A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate

oee [108]

The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

Initial temperature of water = 25.0°C

Final temperature of water = 29.0°C

Mass of water = 400.0 g

Mass of metal = 30.0 g

Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

$Q_{metal} = Q_{water}$

Mathematically, heat capacity or quantity of heat is given by the formula;

$Q = mc\theta$

Where:

Q is the heat capacity or quantity of heat.

m is the mass of an object.

c represents the specific heat capacity.

∅ represents the change in temperature.

Substituting the values into the formula, we have:

$30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}$

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

8 0

2 years ago

A chemist requires 0.881 mol Na2CO3 for a reaction. How many grams does this correspond to?

irga5000 [103]

Molar mass Na2CO3 = 106 g/mole
Using dimensional analysis:
0.787 moles Na2CO3 x 106 g/mole Na2CO3 = g Na2CO3
Answer = 83.4 g Na2CO3

5 0

2 years ago

What is the density of a 6cm3 cork that has a mass of 3g?

Andreyy89

Answer:

The density is 0,5 g/cm3

Explanation:

The density (δ) is the ratio between the mass and the volume of a compound:

δ=m/v= 3 g/6 cm3= 0,5 g/cm3

5 0

3 years ago

A gas that was heated to 150 Celsius has a new volume of 1587.4 L. What was its volume when its temperature was 100 Celsius?

sdas [7]

Answer:

$\boxed {\boxed {\sf 1058.3 \ L}}$

Explanation:

We are asked to find the new volume of a gas after a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The gas was heated to 150 degrees Celsius and had a volume of 1587.4 liters.

$\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{T_2}$

The temperature was 100 degrees Celsius, but the volume is unknown.

$\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C}$

We are solving for the volume at 100 degrees Celsius, so we must isolate the variable V₂. It is being divided by 100°C and the inverse of division is multiplication. Multiply both sides of the equation by 100°C.

$100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C} * 100 \textdegree C$

$100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = V_2$

The units of degrees Celsius cancel.

$100 *\frac {1587.4 \ L }{150 } = V_2$

$100 *10.58266667 \ L = V_2$

$1058.266667 \ L = V_2$

The original measurement of volume has 5 significant figures, so our answer must have the same. For the number we calculated, that is the tenth place. The 6 in the hundredth place to the right tells us to round to 2 up to a 3.

$1058.3 \ L = V_2$

The volume of the gas at 100 degrees Celsius is approximately 1058.3 liters.

6 0

3 years ago