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Mnenie [13.5K]
3 years ago
15

Number the steps for balancing equations:

Chemistry
2 answers:
Goshia [24]3 years ago
6 0

Answer:

3,4,2,1

Explanation:

Natali5045456 [20]3 years ago
3 0

Answer:

identify the atoms on each side

count the atoms on its side

use coefficients to increase the atoms on each side

check to make sure you have the same number of each type of atom on each side

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An alpha particle has the same composition as a A) hydrogen nucleus B) deuterium nucleus C) beryllium nucleus D) helium nucleus​
Elza [17]

Answer:

Helium Nucleus

Explanation:

4 0
2 years ago
Which part of the mantle is still a solid but flows like a thick, heavy liquid?
marusya05 [52]

Hello!

Your answer is A, asthenosphere

<u>The asthenosphere is a part of the mantle</u>. It helps move the plates in the Earth.

It is <u>below the lithosphere,</u> between <u>80 and 200 km</u> below the surface.

Therfore, the asthenosphere is <u>the part of the mantle that is still a solid but flows like a thick, heavy liquid.</u>

<u />

Hope this helps!

Have a great day!

3 0
3 years ago
Read 2 more answers
How many total atoms are in C3H-NO3?
Alik [6]

Answer:

Percent composition by element

Element Symbol # of Atoms

Hydrogen H 5

Carbon C 3

Nitrogen N 3

Oxygen O 9

5 0
3 years ago
Transpiration is the process by which ____.<br>​
saul85 [17]

Answer:

Plants add water to the atmosphere

Explanation:

3 0
3 years ago
Read 2 more answers
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
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