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kifflom [539]
2 years ago
12

What elements make up a molecule of a glucose?

Chemistry
1 answer:
slega [8]2 years ago
7 0

Answer:

All carbohydrates, including sugar, therefore contain the same three elements: carbon, hydrogen and oxygen. Different arrangements of these elements form single units to make different types of carbohydrates. Glucose, for instance, is a single-unit carb with six carbon atoms, 12 hydrogen atoms and six oxygen atoms.

Explanation:

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How many equivalents are present in 10 g of Ca2+? <br> 1 <br> 0.5 <br> 1.5 <br> 2
Lynna [10]

0.5

Explanation:

Given parameters:

Mass of Ca²⁺ = 10g

unknown:

Equivalent weight = ?

Solution:

Equivalent weight that is the amount of electrons which a substance gains or loses per mole.

Ca²⁺ has +3 charge

It lost 2e⁻;

therefore;

  In 1 mole of  Ca²⁺, we have 2 equivalent weight

1 mol  Ca²⁺ = 2eq. wts. 

1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt. 

Therefore;

10.0 g  Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.

learn more:

Molar mass brainly.com/question/2861244

#learnwithBrainly

8 0
3 years ago
What observation did Rutherford make from his gold-foil experiment which
JulsSmile [24]

Answer:

A piece of gold foil was hit with alpha particles, which have a positive charge. Most alpha particles went right through. This showed that the gold atoms were mostly empty space. Some particles had their paths bent at large angles. A few even bounced backward. The only way this would happen was if the atom had a small, heavy region of positive charge inside it.

8 0
2 years ago
PLEASE HELP ASAP I NEED IT HELP PEASEEEEEEEEEEE it’s number 4 (this is science)
hoa [83]
Here you go:
Continent Polar - Cold and Dry
Maritime Polar - Cold and Humid
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Hope this helps:)
5 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac
slega [8]
<h2>Answer:</h2>1.33*10^{-2}grams

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}

Determine the mass of water produced

\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

5 0
1 year ago
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