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3 years ago

When was that one moment when you laughed the hardest

Chemistry

2 answers:

kolezko [41]3 years ago

7 0

Answer:

last month!!

it was on yt

Andrei [34K]3 years ago

4 0

Answer:

My weightlifting coaches are all gay for each other and its hilarious to see 3 large buff men play fighting and wrestling around the weight room.

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How many grams of hydrogen are in 46 g of CH40?<br>

Romashka [77]

Answer:

5.8 grams of hydrogen

Explanation:

Your welcome

6 0

2 years ago

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How many molecules of CO2 will be produced by the decomposition

34kurt

Answer:

2.99×10²⁵ molecules of CO₂ are produced

Explanation:

Decomposition reaction is:

Ca(HCO₃)₂ => CaO(s) + 2CO₂(g) + H₂O(g)

Ratio is 1:2. Let's make a rule of three:

1 mol of bicarbonate can produce 2 moles of CO₂

Therefore, 24.9 moles of bicarbonate may produce, 49.8 moles (24.9 .2 )/1

Let's determine the number of molecules

1 mol has 6.02×10²³ molecules

49.8 moles must have (49.8 . 6.02×10²³) / 1 = 2.99×10²⁵ molecules

4 0

3 years ago

In an exothermic reaction, the water will _______ in a calorimeter. A. release heat B. drop in temperature C. drop in volume D.

elena-14-01-66 [18.8K]

the correct answer is D, absorb heat

7 0

3 years ago

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1. If we used 0.0100 moles of K2CO3, how many moles of SrCO3 can be expected to form?

faltersainse [42]

Answer:

0.01 moles of SrCO₃

Explanation:

In this excersise we need to propose the reaction:

K₂CO₃ + Sr(NO₃)₂ → 2KNO₃ + SrCO₃

As we only have data about the potassium carbonate we assume the strontium nitrite as the excess reactant.

1 mol of K₂CO₃ react to 1 mol of Sr(NO₃)₂ in order to produce 2 moles of potassium nitrite and 1 mol of strontium carbonate.

Ratio is 1:1. In conclussion,

0.01 mol of K₂CO₃ must produce 0.01 moles of SrCO₃

3 0

3 years ago

A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af

larisa86 [58]

Explanation:

The given reaction is as follows.

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

n = $0.05 L \times 0.1 M$

= 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

n = $0.025 L \times 0.1 M$

= 0.0025 mol

According to ICE table,

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial: 0.005 mol 0.0025 mol 0 0

Change: -0.0025 mol -0.0025 mol +0.0025 mol

Equibm: 0.0025 mol 0 0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

[HA] = $\frac{0.0025 mol}{V}$

[NaA] = $\frac{0.0025 mol}{V}$

$[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

pH = $pK_{a} + log \frac{A^{-}}{HA}$

$pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

= $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

= 3.42

Thus, we can conclude that $pK_{a}$ of the weak acid is 3.42.

8 0

3 years ago