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3 years ago

Help me, please

Chemistry

1 answer:

Finger [1]3 years ago

6 0

Put the marked magnet up to each side of the unmarked one. if they attract they are opposite -north attracts south, south attracts north- if they repel -forcefully push away- from each other they are the same -north and north, or south and south. once you have done that you can label the poles correctly.

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What is the maximum amount of kcl that can dissolve in 200g of water?

Likurg_2 [28]

The solubility of potassium chloride in at room temperature is approximately 34 g per 100 g of water. Therefore, the maximum amount that could be dissolved would be 34/100 ( 200) = 68 g of KCl. When more than this amount is added, excess potassium would not dissolve forming crystals in the solution.

8 0

3 years ago

Fission and Fusion of Atomic Nuclei:

bogdanovich [222]

Answer:

C B A D

Explanation:

6 0

3 years ago

Based upon the information provided in the class, which of the following bond types is the strongest? Question options:

Zigmanuir [339]

Answer: A hydrogen bonding is interaction between lone pair and hydrogen atom. An Ion-Dipole interaction is the interaction between an ion formed and a dipole. Dipole forms because of the electronegativity difference between two atom participating in the bond formation, and an ion is formed when an atom gains or lose electron. This ion-dipole interaction is strongest interaction.

Therefore, The right choice is (B)

7 0

3 years ago

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

3 0

3 years ago

The addition of an inert gas has no effect on the equilibrium position of a gaseous reaction because ______. Multiple choice que

boyakko [2]

Inert gas does not affect the equilibrium position:

It is because the partial pressures of the reaction components remain the same.

What is Inert Gas?

Under a given set of conditions, an inert gas is a gas that does not undergo chemical reactions.
The noble gases (helium, neon, argon, krypton, xenon, and radon) were previously known as "inert gases" due to their perceived lack of involvement in any biochemical processes.
Because inert gases are non-reactive, they do not affect equilibrium partial pressures and thus do not affect volume.
An inert gas does not react with the reactants or products; it does not change the concentration of the products and reactants. Furthermore, because the volume is constant, the concentrations are unaffected. As a result, this does not affect equilibrium.

The equilibrium position won't change if an inert gas is added. A volume change won't change the equilibrium position if the total moles of gas in the products and reactants are the same. When the volume is reduced, the process changes to create fewer moles of gas.

Learn more about the inert gas here,

brainly.com/question/15909389

#SPJ4

5 0

2 years ago