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3 years ago

What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 14.630 g to 3.000 g ?

Chemistry

1 answer:

nlexa [21]3 years ago

8 0

Answer:

Explanation:

The quantity of sugar in the bar of milky chocolate of 45 g is 25 g of sugar

u can use this information to calculate the amount of the sugar in any amount .

mass of sugar in 14.630 g →

chocolate mass sugar mass

45 g 25 g

14.630 x

X= mass of sugar = 8.13 g

now for 3 gram of chocolate :

45 g 25 g

3.000 x

X = 1.667 grams of sugar

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The reaction P4 +502P4010 + 712 kcal is

igor_vitrenko [27]

Answer:

The reaction is exothermic (option 4)

Explanation:

P4 + 5O2 → P4O10 + 712 kcal

In chemical reactions heat can be absorbed or released:

⇒in the first case, when heat is absorbed, this is called an endothermic reaction. The products have more energy than the reactants. The reaction requires or absorbs energy from it's surroundings. That means that in this reaction energy , in the form of heat, will be absorbed by the reactants.

⇒ when heat is released, this is called an exothermic reaction. The reactants have more energy than the products. The reaction gives or releases energy to it's surroundings. That means that in this reaction energy , in the form of heat, will be released by the reactants.

in the case of P4 + 5O2 → P4O10 + 712 kcal

We notice that on the right side, which is the product side, there is a positive amount of energy. This means that the energy is released by the the reactants, in this reaction. <u>The reaction is exothermic.</u>

4 0

2 years ago

Read 2 more answers

Which the following statement ia correct for the equation shown here

Shkiper50 [21]

B.

As you can see both NO and NH3 have 4 moles therefore it is 4:4 between the molecules or in other words a 1:1 ratio in simplest forms

7 0

2 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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Explanation:

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Vitek1552 [10]

Answer:

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