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valina [46]
3 years ago
7

How does SI help scientists in different parts of the world?

Chemistry
1 answer:
soldier1979 [14.2K]2 years ago
8 0

Answer:

It help them to gain accurate measurements of substance though in different countries

Explanation:

You might be interested in
If 0.200 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, what is the mass in grams o
morpeh [17]

Answer:

31.2 g of Ag₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + 2HNO₃ (aq)

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄ to produce 1 mole of Ag₂SO₄ and 2 moles of HNO₃.

Next, we shall determine the limiting reactant.

This can obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react with = (0.2 x 1)/2 = 0.1 mole of H₂SO₄.

From the calculations made above, only 0.1 mole out of 0.155 mole of H₂SO₄ given is needed to react completely with 0.2 mole of AgNO₃. Therefore, AgNO₃ is the limiting reactant.

Next,, we shall determine the number of mole of Ag₂SO₄ produced from the reaction.

In this case we shall use the limiting reactant because it will give the maximum yield of Ag₂SO₄ as all of it is consumed in the reaction.

The limiting reactant is AgNO₃ and the number of mole of Ag₂SO₄ produced can be obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted to produce 1 mole of Ag₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react to produce = (0.2 x 1)/2 = 0.1 mole of Ag₂SO₄.

Therefore, 0.1 mole of Ag₂SO₄ is produced from the reaction.

Finally, we shall convert 0.1 mole of Ag₂SO₄ to grams.

This can be obtained as follow:

Molar mass of Ag₂SO₄ = (2x108) + 32 + (16x4) = 312 g/mol

Mole of Ag₂SO₄ = 0.1

Mass of Ag₂SO₄ =?

Mole = mass /Molar mass

0.1 = Mass of Ag₂SO₄ /312

Cross multiply

Mass of Ag₂SO₄ = 0.1 x 312

Mass of Ag₂SO₄ = 31.2 g

Therefore, 31.2 g of Ag₂SO₄ were obtained from the reaction.

5 0
3 years ago
How much energy in joules is released when 18.5 grams of copper cools from 285 °C<br> down to 45 °C
nadya68 [22]

Answer: 1709.4 Joules

Explanation:

The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = ?

M = 18.5 grams

Recall that the specific heat capacity of copper C = 0.385 J/g.C

Φ = 285°C - 45°C = 240°C

Then, Q = MCΦ

Q = 18.5grams x 0.385 J/g.C x 240°C

Q = 1709.4 Joules

Thus, 1709.4 Joules is released when copper is cooled.

5 0
3 years ago
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum?
SIZIF [17.4K]
The chemical reaction would be written as follows:

2Al + 3Cl2 = 2AlCl3

We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:

19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3
5 0
3 years ago
Read 2 more answers
Which gas in Earth's atmosphere protects Earth by reflecting untraviolet radiation?
egoroff_w [7]

Answer:

I think it's nitrogen I'm not really sure..

6 0
3 years ago
Is the electrochemical cell spontaneous or not spontaneous as written at 25 ∘C ? spontaneous not spontaneous Calculate the poten
givi [52]

Answer:

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

Explanation:

Let's consider the oxidation and reduction half-reactions and the global reaction.

Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻

Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)

Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red,cat - E°red,an

E° = 0.771 V - 0.154 V = 0.617 V

The Nernst equation allows us to calculate the cell potential (E) under the given conditions.

E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

4 0
2 years ago
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