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Butoxors [25]
3 years ago
12

Which of the following measurements is equal to 23 dL?

Chemistry
1 answer:
77julia77 [94]3 years ago
8 0
2300 dL = 230 L 
<span>23 L = 23 L </span>
<span>230 cL = 2.3 L </span>
<span>230000 mL = 230 L </span>
<span>2300000 nL = 0.0023 L </span>
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When 2.5 grams react, the will produce how many grams of reactant in a complete chemical reaction?
const2013 [10]

Answer:

2.5g

Explanation:

When the reaction goes into completion, they will produce 2.5g. This is complement the law of conservation of mass.

According to the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".

  • The mass of reactants and products in a chemical reaction must be the same.
  • There is no change in mass in moving from reactant to product
  • So, if we start with 2.5g of reactants, we must end with 2.5g of products.
5 0
3 years ago
A plain-carbon steel contains 45 wt % proeutectoid ferrite. What is its average carbon content in weight percent
ASHA 777 [7]

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Explanation:

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3 years ago
?cual es el tiempo de<br>onda con 6<br>oscilaciones.?​
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Answer:

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Explanation:

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7 0
3 years ago
What type of rays result in shorter days? (science)
OLEGan [10]

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verical rays

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4 0
3 years ago
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of
Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH}  =0.1900gO

4. Compute the moles of oxygen by using its molar mass:

n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1

6. Search for the closest whole number (in this case multiply by 2):

C_3H_6O_2

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

M=12*3+1*6+16*2=74g/mol

Which is about three times in the molecular formula, for that reason, the actual formula is:

C_9H_{18}O_6

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0
3 years ago
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