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omeli [17]
3 years ago
10

The alchemists were the first to do what?

Chemistry
1 answer:
Bezzdna [24]3 years ago
8 0
Alchemists attempted to purify, mature, and perfect certain materials. ... Islamic and European alchemists developed a structure of basic laboratory techniques, theory, terminology, and experimental method, some of which are still in use today.
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Calculate the molarity of each of the following solutions. Use the periodic table if necessary.
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Find the average speed of a person who swims 105 m in 70 s
spin [16.1K]

Answer:

v = 1.5 m/s

Explanation:

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Distance covered by a person, d = 105 m

Time taken, t = 70 s

We need to find the average speed of the person. We know that the average speed is given by :

v=\dfrac{d}{t}\\\\v=\dfrac{105\ m}{70\ s}\\\\v=1.5\ m/s

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4 0
3 years ago
GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g
JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of H_2 gas and 1 mole of Br_2 liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ

Divide the equation by 2.

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

4 0
3 years ago
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