Please help with yes or no

How many grams are 1.20 x 1025 molecules of calcium iodide?

jeka57 [31]

Answer:

1230

Explanation:

1.20×1025=1230 is your answer

5 0

2 years ago

a. Calculate the pH of a solution that has [H3O+]= 4.3 x 10-5 M. b. Is the solution acidic or basic? How do you know?

Luda [366]

Answer:

* $pH=4.37$

* The solution is acidic since the pH is below 7.

Explanation:

Hello,

In this case, we can mathematically define the pH by:

$pH=-log([H_3O^+])$

Thus, for the given hydronium concentration we simply compute the pH:

$pH=-log(4.3x10^{-5})=4.37$

Thereby, we conclude the solution is acidic due to the fact that the pH is below 7 which is the neutral point and above it the solutions are basic.

Regards.

3 0

2 years ago

Covalent bonds involve the sharing of?

4vir4ik [10]

Answer:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

I hope this helps

7 0

2 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

What advice would you give a person who handles hydrogen and oxygen in their workplace?

Murljashka [212]

Regard the principle of utilization of two gas.

Make a consistent control of hardware containing gas.

Make a consistent control of weight diminishing valves giving gas.

No smoking zone.

3 0

3 years ago