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dimulka [17.4K]
3 years ago
15

Which of the following is a testable hypotheses?

Chemistry
1 answer:
Solnce55 [7]3 years ago
3 0

Answer:

The awser is D.

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number 9 is : Organized Data is a customizable information product versioning and management system used to track and store document, images and other products or information

5 0
4 years ago
The atomic and mass numbers for four different atoms are given below. Which two are isotopes of the same element?
Dmitrij [34]
The correct option is B AND C.
An isotope is defined as two or more forms of the same element which contain equal number of protons but different number of neutrons in their nuclei, hence they differ in relative atomic mass but not in chemical properties.
Thus, isotopes has the same number of protons but differ in number of neutrons. Due to this fact, isotopes always have the same atomic number but different mass number. You will notice that the atomic number given for elements B and C are the same [101].
4 0
4 years ago
Calculate the atoms in 2CaCO3​
Fudgin [204]
5 atoms is in 2CaCO3​
Element: 1 atom
Compounds: 2+ atoms
3 0
2 years ago
What happens when you add acetic acid vinegar and sodium bicarbonate baking soda nahco​
Sophie [7]

Explanation:

oh shi sorry I wish I could help but I'm stupid

8 0
3 years ago
N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?
just olya [345]

Answer:

\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     28.01               17.03

            N₂ + 3H₂ ⟶ 2NH₃

m/g:                          240.0

(a) Moles of NH₃

\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

(b) Moles of N₂

\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

(c) Mass of N₂

\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

7 0
3 years ago
Read 2 more answers
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