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3 years ago

If there is a chemical change it will remain the same properties because a new substance was formed

Chemistry

1 answer:

rewona [7]3 years ago

5 0

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Why must sample spots be above the solvent in paper chromatography

marishachu [46]

Because if they are submerged in the solvent, they would dissolve! This would prevent them from seperating and not allow you to actually record anything

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2 years ago

A chemist requires 0.802 mol Na2CO3 for a reaction. How many grams does this correspond to?

Komok [63]

Answer:

Ok:

Explanation:

So grams = mols*MolarMass. Here, MolarMass (MM) = 105.99g which can be found using the periodic table. mols is given to be 0.802. We can then plug in to get that it corresponds to 85.0g.

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2 years ago

What three compounds are produced by the combustion of hydrocarbon fuels?

MakcuM [25]

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2 years ago

The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?

vazorg [7]

C + O2= CO2
$n = \frac{m}{mw}$
$n = \frac{5.4}{12} \\ n = 0.45 \: mol \: of \: carbon$
$n = \frac{13.6}{12 + 16 \times 2} \\ n = \frac{13.6}{44} \\ n = 0.31 \: mol \: of \: carbon \: dioxide$
CO2 is limit
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3.72 g = x
x=68.9 %

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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:

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

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3 years ago