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KIM [24]
3 years ago
8

How many attoms are equal 6.19 moles of hydrogen?​

Chemistry
1 answer:
Wittaler [7]3 years ago
4 0

Answer:

3.727x10^24 atoms.

Explanation:

When calculating how many molecules are in a compound or element, we refer to Avogardo's number.

Your equation will look like this: 1 mole/ 6.022x10^23 molecules (Avogardo's number).

Multiply 6.19 by Avogardo's number to get your answer.

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GarryVolchara [31]
Balanced chemical reaction:
2Na₃PO₄(aq) + 3CaCl₂(aq) → 6NaCl(aq) + Ca₃(PO₄)₂(s).
Ionic reaction:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ca²⁺(aq) + 6Cl⁻(aq) → 6Na⁺(aq) + 6Cl⁻(aq) + Ca₃(PO₄)₂(s).
Net ionic reaction: 2PO₄³⁻(aq) + 3Ca²⁺(aq) → Ca₃(PO₄)₂(s).
<span>(aq) means that substances are dissociated on cations and anions in water.
</span>(s) means solid.
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3 years ago
How much does one teaspoon of baking soda affect Cookies?
Gekata [30.6K]

Answer:

Explanation:

At three-quarters of a teaspoon of baking soda, the cookie's texture becomes fragile like that of a soda cracker, with a deeply roasted flavor that distracts from the ginger.

6 0
3 years ago
At what point does salt melt
fredd [130]
Salt melts at 116°c.
7 0
3 years ago
Read 2 more answers
How many moles of N2 in 57.1 g of N2?
SpyIntel [72]

We are given –

  • Mass of \bf N_2 is 57.1 g and we are asked to find number of moles present in 57.1 g of \bf N_2

\qquad\pink{\bf\longrightarrow  { Molar \:mass \:of \: N_2:-}  }

\qquad\bf  \twoheadrightarrow 14\times 2

\qquad\bf \twoheadrightarrow   28

\qquad____________________

Now,Let's calculate the number of moles present in 57.1 g of \bf N_2

\qquad\purple{\bf\longrightarrow  { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}

\qquad\bf   \twoheadrightarrow \dfrac{57.1}{28}

\qquad\bf  \twoheadrightarrow 2.04\: moles

__________________________________

7 0
2 years ago
The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0
2 years ago
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