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Burka [1]
3 years ago
15

How many grams are found from 2 moles of glucose C6H1206?

Chemistry
1 answer:
Kitty [74]3 years ago
3 0

Answer:

360g

Explanation:

Hope it helps you!

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how do i solve these equations using proust's law? we haven't gone over balancing chem equations in class and our teacher doesn'
kow [346]

Answer:

Explanation:

Ok so Proust's Law, better known as the Law of Definite Proportions, states that the components of a compound always exist in a fixed ratio. This means that it does not matter what the source of the components are nor the coefficients in front of them. The ratio your teacher is referring to is most likely mass percent/ percent composition. This ratio is the amount of the component over the amount of the entire compound times 100%.

*I am not very familiar with this law so please do with my answers what you will :)*

7.)

First, you want to find the percent composition (aka mass composition) of Na in NaCl.

 45.89 g Na
---------------------  x 100%  =  39.3%
116.89 g NaCl

So, there must be 39.3% sodium in NaCl. You can find how much chlorine is in NaCl by subtracting that percent by 100 (to find the percent composition of chlorine) then multiplying it by the mass.

100% - 39.3% = 60.7%

60.7% / 100 = 0.0607

116.89 g NaCl x 0.0607 = 70.91 g Cl₂

You could also just subtract the mass of sodium from the mass of sodium chloride to find the mass of chlorine.

116.89 g NaCl - 45.89 g Na = 71 g Cl₂

8.)

10.57 g Mg + 6.96 g O₂ = 17.53 g MgO

  6.96 g O₂
---------------------  x 100%  = 39.7% O₂
17.53 g MgO

9.)

6.46 g Pb = 1 g O₂

68.54 g Pb  =  28.76 g O₂

68.54 g Pb / 28.76 g O₂ = 2.83 g ≠ 6.46 g

No, the two samples are not the same because the proportion of lead to oxygen is not the same for both samples. In the first sample, there is 6.46 g lead for every oxygen. In the second sample, there is 2.38 g lead for every oxygen.

6 0
2 years ago
Identify the factors that govern the speed and direction of a reaction. Check all that apply.
Fiesta28 [93]

Answer: option B and option C

Explanation:

4 0
3 years ago
What is the mole ratio of aluminum to chlorine in the compound aluminum
creativ13 [48]
1:3
The ratio of Al3+ ions to Cl− ions in the chemical formula is 1:3.
I’m sorry if this didn’t help I’m new to Brainly
6 0
3 years ago
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0
3 years ago
How many pi bonds are in the oxalate ion (C2O42-)?
suter [353]

Answer:there are 13

Explanation:

4 0
3 years ago
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