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DENIUS [597]
3 years ago
10

how many grams of nahco3 (fm84.007) must be added ro 4.00g of k2co3 (fm 138.206) to give a ph of 10.80 in 500ml of water​

Chemistry
1 answer:
Anna11 [10]3 years ago
6 0

Answer:

0.804g of NaHCO₃ you must add

Explanation:

pKa of HCO₃⁻/CO₃²⁻ is 10.32.

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of acid (HCO₃⁻) and [A⁻] is concentration of conjugate acid (CO₃²⁻).</em>

Moles of CO₃²⁻ = K₂CO₃ are:

4.00g ₓ (1mol / 138.206g) = 0.0289 moles CO₃²⁻

Replacing:

10.80 = 10.32 + log [0.0289] / [HCO₃⁻]

[HCO₃⁻] = 0.009570 moles you need to add to obtain the desire pH

As molar mass of NaHCO₃ is 84.007g/mol, mass of NaHCO₃ is:

0.009570 moles ₓ (84.007g / mol) =

<h3>0.804g of NaHCO₃ you must add</h3>

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kirill [66]

Answer:

<h3>Ocean crust is the relatively thin part of the earth's crust which underlies the ocean basins. It is geologically young compared with the continental crust and consists of basaltic rock overlain by sediments.</h3>
8 0
3 years ago
C(s) + S8(s) → CS2(l)
Vanyuwa [196]
The balanced chemical reaction is written as:

<span>4C(s) + S8(s) → 4CS2(l)

We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.

</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8

The limiting reactant would be S8. We use this amount to calculate.

0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2
4 0
3 years ago
Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
Calculate the mass, in grams, of 1.2000 mol Mg₃N₂
fredd [130]
So what do you want me to do
Explanation
7 0
3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
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