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DaniilM [7]
2 years ago
9

What is the balanced form of the chemical equation shown below? N2(g) + H2(g) → NH3(g) -

Chemistry
1 answer:
Rom4ik [11]2 years ago
6 0
To balance equations you have to have same number of atoms on both sides of the equation just multiply with a suitable digit

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How is the mass of a compound related to its chemical formula
Monica [59]
They're totally unrelated. No matter how low or high the molecular weight, you can always use more or less of it.
7 0
2 years ago
Convert 0.85 x 10-12 km to nm.
vitfil [10]

Answer:

8,5 E10 nm

Explanation:

⇒ 0.85 E-2 Km * ( 1000 m / Km ) * ( 1 E9 nm / m ) = 8.5 E10 nm

⇒ 0.85 E-2 Km = 8.5 E10 nm

8 0
3 years ago
Read 2 more answers
A student notices that after two chemicals are mixed together the temperature of the mixture is higher than the temperature of t
swat32

Answer: exothermic

EXPLANATION: any process in which heat energy is released is called an exothermic process. For example burning of wood produces heat, so combustion of wood is an exothermic process.

When chemicals were not mixed they were at room temperature and when we mix them exothermic reaction took place and heat was released which raised the temperature of mixture.

7 0
3 years ago
Hydroelectric power plants capture the ______________ of falling water to generate electricity. The falling water turns a wheel
LiRa [457]

Answer:

1: energy

2: converts

Kinetic energy is either 3 or 4, those questions are really similar.

3 0
2 years ago
The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was
Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}

[Ca^{2+}]=0.007118 \ M

[Mg^2+] = \dfrac{0.003474 \ M\times (26.23  - 20.49 )mL}{10.0 \ mL}

=0.001994 \ M

Mass of Ca²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0
3 years ago
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