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3 years ago

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WILL GIVE BRAINLEST PLZZZ HELP DUE IN 20 MINS ANSWER IF U CAN HELP BTW THERES A Q JUST LIKE THIS ONE ITS IN MY PROFILE IT GIVES

50 POINTS PLZZ HELP

1 answer:

givi [52]3 years ago

4 0

Answer:

this is the same one i just answered lol :P

Explanation:

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aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat

aleksandr82 [10.1K]

Answer:

Explanation:

HCl + NaOH = NaCl + H₂O.

1 mole 1 mole 1 mole 1 mole

6.93 g of hydrochloric acid = 6.93 / 36.5 = .189 mole of HCl

2.4 g of NaOH = 2.4 / 40 = .06 mole of NaOH

NaOH is in short supply so it is the limiting reagent .

1 mole of NaOH reacts with 1 mole of HCl to give 1 mole of Water

.06 mole of NaOH will react with .06 mole of HCl to give .06 mole of water

Water formed = .06 mole

= .06 x 18 = 1.08 g

= 1.1 g

4 0

3 years ago

Consider the four free body diagrams. Free-body diagrams are diagrams used to show the relative magnitude and direction of all f

9966 [12]

If you look closely at each of the four diagrams you would be able to conclude that

<span>D)
Yes. In B and D. In both cases, there is a net force.
In B, there is a net force to the left; in D there is a net force upward.

In A and C, the forces are in equilibrium both in the horizontal and vertical direction.</span>

5 0

3 years ago

Read 2 more answers

Calculate the pH of a 2.3 10-3M[OH'] solution.

disa [49]

Answer:

11.33

Explanation:

-log(2.3x10^-3) = 2.67

14-2.67

- Hope this helped! Let me know if you need a further explanation.

5 0

4 years ago

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

7 0

3 years ago

2CO + O2 --> 2002

olga_2 [115]

8 moles I think I’m not sure

4 0

3 years ago