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Tanzania [10]
3 years ago
10

7. La constante de equilibrio Kc, se halla :

Chemistry
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

Molaresحخىطنصثىخرنص رهصجتخطرةصخىر رصىشطثرىنصكرىصخكترة شصر

You might be interested in
How can water dissovle many substances
11111nata11111 [884]

Answer: Polar molecules

Explanation: Water is able to dissolve many substances because of its polar molecules, The positive and negative ends of a water molecule attracts the molecules of other polar substances..

I'm pretty sure, idk college stuff

3 0
3 years ago
Read 2 more answers
PLeas Answer with A B C or D please be QUICK :>
Lena [83]

<em>Answer: </em>D

<em>Explanation:</em>

chemical formula of methane: CH4

electron configuration of C: 2,4

electron configuration of H: 1

there are 4 hydrogen atoms that donated 1 electron each to C

therefore it's D.

3 0
2 years ago
How many formula units are in 4.52 moles of H3SO3?<br> Type your answer
lakkis [162]

Answer:

The answer is 98.07848. We assume you are converting between grams H2SO4 and mole. You can view more details on each measurement unit: This compound is also known as Sulfuric Acid. The SI base unit for amount of substance is the mole. 1 grams H2SO4 is equal to 0.010195916576195 mole.

<u>Quick conversion chart of moles H2SO3 to grams</u>

1 moles H2SO3 to grams = 82.07908 grams

2 moles H2SO3 to grams = 164.15816 grams

3 moles H2SO3 to grams = 246.23724 grams

4 moles H2SO3 to grams = 328.31632 grams

5 moles H2SO3 to grams = 410.3954 grams

6 moles H2SO3 to grams = 492.47448 grams

7 moles H2SO3 to grams = 574.55356 grams

8 moles H2SO3 to grams = 656.63264 grams

9 moles H2SO3 to grams = 738.71172 grams

10 moles H2SO3 to grams = 820.7908 grams

7 0
3 years ago
What are the characteristics of Phosphate at room temperature?
crimeas [40]

Answer: Phosphate is colorless , soft, waxy solid that glows in the dark when at room temperature.

Explanation:

4 0
3 years ago
Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s
Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0
3 years ago
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