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3 years ago

What mass of water in grams will fill a tank 126cm long, 47 cm wide, and 13 cm high

Chemistry

1 answer:

Shtirlitz [24]3 years ago

8 0

Answer:

Mass = 76986 g

Explanation:

Given data:

Dimensions of tank = 126 cm× 47 cm× 13 cm

Mass of water required to filled = ?

Solution:

First of all we will calculate the volume of tank which is equal to the volume of water required to fill it.

Volume = length ×height ×width

Volume = 126 cm × 13 cm× 47 cm

Volume = 76986 cm³

Mass of water:

Mass = density × volume

density of water is 1 g/cm³

Mass = 1 g/cm³× 76986 cm³

Mass = 76986 g

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Please answer asap! worth 35 points!

Gnesinka [82]

Answer:

The enthalpy of the solution is -35.9 kJ/mol

Explanation:

Step 1: Data given

Mass of lithiumchloride = 3.00 grams

Volume of water = 100 mL

Change in temperature = 6.09 °C

Step 2: Calculate mass of water

Mass of water = 1g/mL * 100 mL = 100 grams

Step 3: Calculate heat

q = m*c*ΔT

with m = the mass of water = 100 grams

with c = the heat capacity = 4.184 J/g°C

with ΔT = the chgange in temperature = 6.09 °C

q = 100 grams * 4.184 J/g°C * 6.09 °C

q =2548.1 J

Step 4: Calculate moles lithiumchloride

Moles LiCl = mass LiCl / Molar mass LiCl

Moles LiCl = 3 grams / 42.394 g/mol

Moles LiCl = 0.071 moles

Step 5: Calculate enthalpy of solution

ΔH = 2548.1 J /0.071 moles

ΔH = 35888.7 J/mol = 35.9 kJ/mol (negative because it's exothermic)

The enthalpy of the solution is -35.9 kJ/mol

3 0

3 years ago

g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint

diamong [38]

Answer:

The value is $L = 0.985 \pm 0.00801 \ g$

Explanation:

From the question we are told that

The molar mass of $CaCO_3$ is $MW = 100.085 \ g/mol$

The total mass is $m_g = 12.3 \ g$

The uncertainty of the total mass is $\Delta g = 0.1$

Generally the molar weight of calcium is $M_c = 40 g/mol$

The percentage of calcium in calcite is mathematically represented as

$C = \frac{40.07}{100.085} * 100$

$C = 40.03 \%$

Generally the mass of each sample is mathematically represented as

$m= \frac{m_g}{5}$

$m= \frac{12.3}{5}$

$m= 2.46 \ g$

Generally mass of calcium present in a single sample is mathematically represented as

$m_c = 2.46 * \frac{40.04}{100}$

$m_c = 0.985 \ g$

The uncertainty of mass of a single sample is mathematically represented as

$k = \frac{\Delta g }{5}$

$k = \frac{0.1 }{5}$

$k = 0.02\ g$

The uncertainty of mass of calcium in a single sample is mathematically represent

$G = \frac{0.02 * 40.04}{ 100}$

$G = 0.00801 \ g$

Generally the average mass of calcium in each sample is

$L = 0.985 \pm 0.00801$

6 0

3 years ago