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Inessa [10]
3 years ago
5

How many atoms of Hydrogen in the products of the following equation?

Chemistry
1 answer:
love history [14]3 years ago
7 0

Answer:

2 atoms of H in the products side

Explanation:

3NO₂ + H₂O → NO + 2HNO₃

In product side: 3 N (1 from NO and 6 from nitric acid), 7 O (6 from nitric acid + 1 from NO), 2 H

In reactant side: 3 N, 7 O (6 from the dioxide + 1 from water) and 2 H

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A molecule of butane and a molecule of 2-butene both have the same total number of
Vinvika [58]
The correct answer is option 1. Butane and 2-butene have the same total number of carbon atoms. They both have four carbon atoms. They differ in there structure since the latter has double bonds on it. As a result of the different structure, they also have different properties.
8 0
3 years ago
Saturn is about 1 429 000 km from the sun. how many meters is saturn from the sun?
stira [4]

Answer:

1.493 trillion

Explanation:

8 0
4 years ago
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How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = 87.5^{o}C = (87.5 + 273) K = 360.5 K

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

5 0
3 years ago
Group 2A metal oxides react with water to produce hydroxides. When 35.55 g of CaO are placed into 125 mL of water (d= 1.000 g/mL
Vika [28.1K]
Answer is: mass of calcium hydroxide is 46.98 grams.
Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.
m(CaO) = 35.55 g.
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 35.55 g ÷ 56 g/mol.
n(CaO) = 0.634 mol; limiting reactant.
m(H₂O) = 125 mL · 1.000 g/mL.
m(H₂O) = 125 g.
n(H₂O) = 125 g ÷ 18 g/mol.
n(H₂O) = 6.94 mol.
From chemical reaction: n(CaO) : n(Ca(OH)₂) = 1 : 1.
n(Ca(OH)₂) = 0.634 mol.
m(Ca(OH)₂) = 0.634 mol · 74.1 g/mol = 46.98 g.
5 0
3 years ago
Anything that has mass and takes up space
Gnom [1K]

Answer:

true

Explanation:

6 0
3 years ago
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