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lions [1.4K]
3 years ago
6

How many elements are in NaC2H302?

Chemistry
1 answer:
DanielleElmas [232]3 years ago
7 0
There are four different elements
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Three cards with holes are arranged in a straight line. A light is shined through the first card’s hole and travels through all
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C) that light travels in a straight line.
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Dose your heart muscle get tired?<br> Yes or no.
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Answer:

<em>no</em><em> </em>

Explanation:

our heart muscles never get tired, because it has to pump blood in our body 72 times a minute, it is made of special cardiac muscles which helps it to perform it's function without getting tired ....

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

5 0
3 years ago
Mass is measured in= a. Liters b. centimeters c. newtons d. kilograms
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Mass is measured in kilograms.
5 0
2 years ago
A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o
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The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
                                                 = 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
                                               = 0.031395 g
                                               = 31.395 mg
                                               ≈ 31.4 mg
5 0
3 years ago
300 moles of sodium nitrite are needed for a reaction. the solution is 0.450 m. how many ml are needed?
miv72 [106K]

Answer:

\boxed{\text{667 000 mL}}

Explanation:

\text{Molar concentration} = \dfrac{\text{moles}}{\text{litres}}\\\\c = \dfrac{n}{v}

Data:

c = 0.450 mol·L⁻¹

n = 300 mol

Calculation:

\begin{array}{rcl}0.450 & = & \dfrac{300}{V}\\\\0.450V& = & 300\\\\V & = & \dfrac{300}{0.450}\\\\V & = & \text{667 L} =\textbf{667 000 mL}\end{array}\\\text{The volume of solution needed is }\boxed{\textbf{667 000 mL}}

4 0
3 years ago
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