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3 years ago

5. For the following equation:

Chemistry

2 answers:

Agata [3.3K]3 years ago

8 0

Answer:

The correct answer is D, copper and silver nitrate

Explanation:

Reactants are on the left, products on the right.

On the right there is Cu which is copper and AgNO3 which is silver nitrate

zmey [24]3 years ago

4 0

Answer:

the correct answer is option A copper and silver

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What will occur if the velocity of the gas molecules colliding with the sides of the container increases?

yanalaym [24]

The pressure of the gas will increases

6 0

3 years ago

Read 2 more answers

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 264 mL .

gavmur [86]

The formula for density is:

D = m/v

We can use the formula to figure out the mass because we already know two of the three values (we are given the density and volume), so we only have to solve for m. If we plug our given values into the formula, we get:

2.70 = m / 264

Now, all we need to do is solve for m. The goal is to get m on one side of the equation, and all we have to do is multiply each side of the equation by 264:

264 × 2.70 = (m÷264) × 264

264 × 2.70 = m

m = 712.8

The mass of the piece of aluminum is 712.8 grams.

4 0

3 years ago

Which grouping shows a decrease in Intermolecular

Anna11 [10]

Answer:

C) solid, liquid, gas

Explanation:

Which grouping shows a decrease in Intermolecular

Forces of Attraction?

A) gas, liquid, solid B) liquid, solid, gas

C) solid, liquid, gas D) solid, gas, liquid

the further the particles are from each other, the less the intermolecular attraction they are farthest in a gas, then a liquid, and closest in a solid

7 0

3 years ago

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$

From the reaction, we conclude that

As, 1 mole of $KI$ react to give 1 mole of $AgI$

So, 0.0178 moles of $KI$ react to give 0.0178 moles of $AgI$

Thus,

Moles of AgI = Moles of $I^-$ anion = Moles of $Ag^+$ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

$\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}$

$\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M$

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0

3 years ago

If an element has two isotopes, what is the atomic mass if one of the isotopes has a mass of 15.000 amu and makes up 5.000% of t

olasank [31]

The atomic mass of element is the weighted average atomic mass of the element with respect to the abundance of the isotopes of that element
atomic mass is the sum of the products of the mass of isotopes by their percentage abundance
atomic mass = 15.000 amu x 5.000 % + 16.000 amu x 95.000 %
= 0.7500 + 15.200
atomic mass of element is therefore 15.950

8 0

3 years ago