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3 years ago

Whoever can answer this I will give all of my points to and brainliest award.

Chemistry

2 answers:

ZanzabumX [31]3 years ago

7 0

i dont see the picture.

Serga [27]3 years ago

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The picture doesn’t pop up

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14 someone help ASAP

levacccp [35]

D. the activation energy is the displacement of energy needed for the reaction, so the distance underneath the maximum point

5 0

2 years ago

HELP! <br><br>How many moles of a solute would 3 liters of a 2-molar solution contain?

katrin [286]

Answer: If it was 3 mol of solute in 2 L of solution it would be 1.5 mol/L.

However when the solute dissolves in the water creating the solution, the volume increases. So 3 mol of solute in 2 L of water creates more than 2 L of solution.

The correct method for making a 3 mol/L solution would be to place some water into a two liter volume container. Dissolve all 3 mol of the solute into the water. Then add water to the 2 L mark. Now there is 3 mol of solute and 2 L of solution.

Explanation: I hope this helps XDDDD

8 0

2 years ago

A sample of helium has a volume of 3.49 L at 741 mmHg. What would be the volume if the pressure

Naddika [18.5K]

P1: 741 mmHg

V1: 3.49 L P1 x V1 / P2= (741 mmHg) (3.49 L) / 760 mmHg = 3.40 L

P2: 760 mmHg

V2: ? L

5 0

2 years ago

A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a

Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

$H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)$

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = $m C .$ Δ $T$

m is mass of the solution = 151.8 mL * $\frac{1 g}{1 mL} = 151.8 g$

C is the specific heat of solution = 4.18 $\frac{J}{g. ^{0}C}$

ΔT is the temperature change = $31.50^{0}C - 21.45^{0}C = 10.05^{0}C$

q = $151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J$

Moles of NaOH = $101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol$ NaOH

Moles of $H_{2}SO_{4}$ = $50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}$

Enthalpy of the reaction = $\frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol$

5 0

3 years ago

1. How many molecules of S2 gas are in 756.2 L?

AfilCa [17]

Answer: There are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

Explanation:

It is known that 1 mole of any gas equals 22.4 L at STP. Hence, number of moles present in 756.2 L are calculated as follows.

$Mole = \frac{Volume}{22.4 L}\\= \frac{756.2 L}{22.4 L}\\= 33.76 mol$

According to mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ molecules.

Therefore, molecules of S present in 33.76 moles are calculated as follows.

$1 mol = 6.022 \times 10^{23}\\33.76 mol = 33.76 \times 6.022 \times 10^{23}\\= 2.032 \times 10^{25}$

Thus, we can conclude that there are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

5 0

3 years ago