The south end of Earth's axis is tilted toward the Sun in?

Determine the concentration of sulfuric acid that needed 47 mL of 0.39M potassium hydroxide solution to neutralize a 25 mL sampl

77julia77 [94]

Answer:

0.37M

Explanation:

Since sulfuric acid, H₂SO₄, is a diprotic acid and potassum hydroxide, KOH, contains one OH⁻ in the formula, the number of moles of potassium hydroxide must be twice the number of moles of sulfuric acid.

1. Determine the number of moles of KOH in 47mL of 0.39M potassium hydroxide solution

number of moles = molarity × volume in liters

number of moles = 0.39M × 47mL × 1liter/1,000 mL = 0.1833mol

2. Determine the number of moles of sulfuric acid needed

number of moles of H₂SO₄ = number of moles of KOH/2 = 0.1833/2 = 0.009165mol

3. Determine the concentration that contains 0.009165 mol in 25mL of the acid.

Molarity = number of moles / volume in liters

M = 0.009165mol/(25mL) × (1,000mL/liter) = 0.3666M

Round to two significant figures: 0.37M

7 0

4 years ago

Atomic theory is

Viktor [21]

Atomic theory is the scientific inquiry of the nature of matter, which is subject to change as new information is discovered. The answer is A.

8 0

3 years ago

You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne

fenix001 [56]

Answer:

$P_2=1.90atm$

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

$\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}$

Whereas both n and R are cancelled out as they don't change, we obtain:

$\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}$

Thus, by solving for the final pressure, we obtain:

$\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}$

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

$P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm$

Best regards!

7 0

3 years ago

A 1300 mL sample of gas with a molar mass of 71.0 g/mol at STP has what density?

Blizzard [7]

Answer:

0.055g/mL

Explanation:

Data obtained from the question include:

Molar Mass of the gass sample = 71g/mol

Volume of the gas sample = 1300 mL

Density =?

The density of a substance is simply mass per unit volume. It is represented mathematically as:

Density = Mass /volume.

With the above equation, we can easily obtain the density of sample of gas as illustrated below:

Density = 71g / 1300 mL

Density = 0.055g/mL

Therefore, the density of the gas sample is 0.055g/mL

6 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 10.2 kilojoules of energy would be required.

3 0

3 years ago