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A student mixes a 10.0 ml sample of 1.0 m naoh(aq) with a 10.0 ml sample of 1.0 m hcl(aq) in a polystyrene container. the temper

AleksAgata [21]

Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
Q water = m * C * (Tf - Ti)
= (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol

5 0

3 years ago

A student has a large number of coins of different diameters, all made by the same metal. She wishes to find the density of the

Elenna [48]

Answer:

A. In a graduated cylinder, put some quantity of water and measure the initial volume. Then put a coin and measure the volume. To find the volume of the coin, simply subtract the initial volume (water only) from the ending volume (water + coin). To measure the mass, take a dry coin and place it on an electronic scale. Density = mass / volume, so divide the mass by the volume to calculate the density of the coin.

B. When measuring the volume, make sure to look at the graduated cylinder at eye level and read from the bottom of the meniscus.

6 0

2 years ago

The gaseous product of a reaction is collected in a 25.0L container at 27.0 C. The pressure in the container is 3.0atm and the g

NeX [460]

Answer: The molar mass of the gas is 31.6 g/mol

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 3.0 atm

V = Volume of gas = 25.0 L

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $27.0^0C=(27.0+273)K=300K$

$n=\frac{PV}{RT}$

$n=\frac{3.0atm\times 25.0L}{0.0821 L atm/K mol\times 300K}=3.04moles$

Moles = $\frac{\text {given mass}}{\text {Molar mass}}$

$3.04=\frac{96.0g}{\text {Molar mass}}$

${\text {Molar mass}}=31.6g/mol$

The molar mass of the gas is 31.6 g/mol

4 0

3 years ago

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree

lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = $T_1=33.9^oC=33.9+273K=306.9 K$

Final temperature of the water = $T_2$

Specific heat capacity of water under these conditions = c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

$Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)$

$-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC$

The new temperature of the water bath 32.0°C.

5 0

3 years ago

Help me please.. dont understand

dedylja [7]

When gases get cooler the molecules slow down or lose energy and condense so the volume in the tire or soccer ball would decrease and may feel flatter

4 0

3 years ago

Please helpppp!!!!!!!!¡!!!!!¡!!!!!!!!¡​

Please helpppp!!!!!!!!¡!!!!!¡!!!!!!!!¡