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3 years ago

2O POINTS AND BRAINLIEST!!! FAST ANSWER PLEASE!!!! I WILL FAIL THIS IF YOU DON'T HELP ME!!!!!!

2 answers:

8 0

<span>1. 1 molecule of C6H12O6(dextrose sugar), 2 molecles of c2h6o (ethyl alcohol), 2 molecules of Co2

2. 48 hydrogen atoms </span>

Anna35 [415]3 years ago

8 0

The easy way to do this is to remember that the number of molecules is the big number at the very beginning which makes your answer:

__1__ molecule(s) of dextrose sugar yields __2__ molecule(s) of ethyl alcohol plus __2__ molecule(s) of carbon dioxide through the action of yeast. The small amount of alcohol evaporates during baking. __12__ hydrogen atoms are involved in this reaction.

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A sample of potassium has an average atomic mass of 39.0983amu. There are three isotopic forms of potassium element in the sampl

igomit [66]

Answer:

Percent Composition of 41K = 6.7302%

Explanation:

The explination is in the image.

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3 years ago

Which ocean zone lies inside V-shaped ocean trenches? O A. Bathypelagic O B. Abyssopelagic Ο Ο Ο Ο O C. Mesopelagic O D. Hadalpe

alisha [4.7K]

Answer:

O D. Hadalpelagic is the answer.

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3 years ago

) write the balanced equation for the reaction that occurs when methanol, ch3oh (l), is burned in air. what is the coefficient o

MaRussiya [10]

2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)

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3 0

3 years ago

Read 2 more answers

Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

7 0

3 years ago

5.If 30mL of 0.5M NaOH(aq) was used to neutralize 10mL of HCl(aq), what is the concentration of the acid? Show work.

Zolol [24]

Answer:

0.15M

Explanation:

NaOH + HCl -->NaCl + H2O

moles of NaOH reacted with HCl

(0.5 × 30)/1000 = 1.5 × 10^ -3

so the moles of HCl reacted with NaOH

1.5 × 10^ -3

concentration of HCl = ( 1.5 × 10^ -3)/10 ×1000

7 0

3 years ago