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3 years ago

Which volume of 0.600 M H2SO4(aq) exactly neutralizes 100. milliliters of 0.300 M Ba(OH)2(aq)?

2 answers:

3 0

Answer: 50 ml of 0.600 M of sulfuric acid will exactly neutralize the 100 ml of 0.300 M of barium hydroxide solution.

Explanation:

$H_2SO_4+Ba(OH)_2\rightarrow BaSO_4+2H_2O$

According to reaction 1 mole sulfuric acid neutralizes 1 mole oh barium hydroxide.

$M_1=0.600 M,H_2SO_4$

$V_1=?$

$M_2=0.300 M,Ba(OH)_2$

$V_2=100 ml$

$M_1V_1=M_2V_2$

$V_1=\frac{M_2\times V_2}{M_1}=50 ml$

50 ml of 0.600 M of sulfuric acid will exactly neutralize the 100 ml of 0.300 M of barium hydroxide solution.

cestrela7 [59]3 years ago

3 0

50.0 ml
this because i got it from castle learning lol
-9th grader

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Answer:

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3 years ago

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4 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?

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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.

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

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3 years ago

He developed the beginnings of the modern atomic theory in the 1800s.

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Answer:

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3 years ago

Consider the following generic chemical equation.

makvit [3.9K]

Answer:

$\fbox{ A) A \: is \: limiting \: reactant }$

$\fbox{ B) B \: is \: limiting \: reactant }$

$\fbox{ C) A \: is \: limiting \: reactant }$

$\fbox{ D) A \: is \: limiting \: reactant }$

Explanation:

Given equation:

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Limiting reactant for corresponding number of moles=?

Solution:

We know that the liming reactant is any atom, ion or molecule which is completely consumed during a reaction and other reactant is still left in reactant vessel.

For the given reaction A+3B→C

for every one mole of A three moles of B are required.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in A) there is one mole of A and 4 mole of B,

if 1 mole of A will react with 3 moles of B, 1 mole of B will be still there in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in B) there is 2 mole of A and 3 mole of B,

if 1 mole of A will react with 3 moles of B, 1 mole of A will be still there in reaction, the reactant was completely consumed is B which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in C) there is 0.5 mole of A and 1.6 mole of B

if one mole of A requires three moles of B to complete the reaction then,

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if 0.5 mole of A will react with 1.6 moles of B, 0.1 mole of B will be still there in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

in D) there is 24 mole of A and 72 mole of B

if one mole of A requires three moles of B to complete the reaction then,

24 moles of A will require 72 moles of B.

if 24 mole of A will react with 75 moles of B, 3 mole of B will be left over in reaction, the reactant was completely consumed is A which is limiting reactant.

﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌

$\fbox{ A) A \: is \: limiting \: reactant }$

$\fbox{ B) B \: is \: limiting \: reactant }$

$\fbox{ C) A \: is \: limiting \: reactant }$

$\fbox{ D) A \: is \: limiting \: reactant }$

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