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2 years ago

How many grams of calcium hydroxide is produced from 9.4 g of calcium chloride in this balanced

Chemistry

1 answer:

Yuliya22 [10]2 years ago

7 0

Answer:

6.27 [gr].

Explanation:

the required mass can be calculated as:

m[Ca(OH)₂]=M[Ca(OH)₂]*m[CaCl₂]/M[CaCl₂], where M - molar mass [gr/mol], m - mass [gr];

m[Ca(OH)₂]=9.4*74/111≈6.2(6) [gr].

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Given 2.91 moles of a gas in a 500 milliliter-container, if the temperature is found to be 31 degrees Celsius, what is the press

kari74 [83]

Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K

P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm

3 0

3 years ago

Read 2 more answers

Please help it’s a test due in five minutes! I’ll give 10 points

barxatty [35]

Answer: negative acellaration or mass.

Explanation:

the first reason why is that i got that quistion right. and when objects are unbalanced it gives negative acellaration

3 0

3 years ago

A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper

finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1: Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL * 0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene = 35.146 grams / 78 g/mol = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4 = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

7 0

3 years ago

How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?

gulaghasi [49]

Answer:

$\boxed {\boxed {\sf 10.2 \ kJ}}$

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

$q=mc \Delta T$

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

m= 225 g
c= 0.897 J/g° C
ΔT= 50.5 °C

$q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

Multiply the first two numbers. The units of grams cancel.

$q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

$q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)$

$q= (201.825\ J / \textdegree C)(50.5 \textdegree C)$

Multiply again. This time, the units of degrees Celsius cancel.

$q= 201.825 \ J * 50.5$

$q= 10192.1625 \ J$

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

$\frac { 1 \ kJ}{ 1000 \ J}$

Multiply by the answer we found in Joules.

$10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}$

$10192.1625 * \frac{ 1 \ kJ}{ 1000 }$

$\frac {10192. 1625}{1000} \ kJ$

$10.1921625 \ kJ$

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

$10.2 \ kJ$

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0

3 years ago

Oxidation state of Nitrogen in N2O5

balandron [24]

Explanation:

Oxidation state of Nitrogen in N2O5 is +5

7 0

2 years ago