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3 years ago

Which describes the composition of carbohydrates?

1 answer:

3 0

Answer: Carbohydrates (carbo- = “carbon”; hydrate = “water”) contain the elements carbon, hydrogen, and oxygen, and only those elements with a few exceptions. The ratio of carbon to hydrogen to oxygen in carbohydrate molecules is 1:2:1.

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Which statement is true for nuclear reactions but not for chemical reactions?

IrinaVladis [17]

Answer:maybe C

Explanation:

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2 years ago

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What is the molarity of 4 qof NaCl (MM=58.45) in 3,800 mL of solution?

tiny-mole [99]

Answer:

.018 M

Explanation:

grams/MM=ans./volume(L) = M

4/58.45=ans./3.8=.018 M

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3 years ago

Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these were decomposed the sulfur dioxide produce

Zinaida [17]

Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

Explanation : Given,

Mass of oxygen in sulfur dioxide = 3.49 g

Mass of sulfur in sulfur dioxide = 3.50 g

Mass of oxygen in sulfur trioxide = 9.00 g

Mass of sulfur in sulfur trioxide = 6.00 g

Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{3.49}{3.50}=0.997g$

and,

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{9.00}{6.00}=1.5g$

Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

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3 years ago

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc

pantera1 [17]

Answer:

$5.35 *10^{-4}$ M

Explanation:

Equation for the reaction is as follows:

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

By Applying the ICE Table; we have

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

Initial x 0.0025 M 0.0010 M

Change 0 0 0

Equilibrium x 0.0025 M 0.0010 M

$K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}$

Given that $K_c = 1.4*10^2$ ; Then:

$1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}$

$1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}$

$3.5 =( \frac{(0.001)}{(x)})^2$

$\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}$

$1.87=\frac{(0.001)}{(x)}$

$(x)= \frac{(0.001)}{1.87 }$

$x = 5.35 *10^{-4}$ M

∴ The equilibrium concentration of CO = $x = 5.35 *10^{-4}$ M

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3 years ago

Use the equation below to determine how many grams of CO are needed to react with an excess of Fe2O3 to produce 104.9 g Fe? You

guapka [62]

<h2>Answer: C= 12.01 g/mol</h2>

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3 years ago

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