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3 years ago

Which property best indicates that a compound contains an ionic bond?

1 answer:

6 0

I think the best answer would be it electrical conductivity when in aqueous form. A compound having an ionic compound dissociates into ions when in aqueous solution. These ions move freely in the solution and can allow the flow of electricity into the solution.

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Mrs. Jacobson is pushing a fridge to the right with a force of 100 N. The force pushing it

alex41 [277]

Answer:

Explanation:

Let's illustrate this; see the attachment.

We see that Mrs. Jacobson is pushing to the right with a force of 100 N and there is another opposite force pushing with a force of 15 N. Since these are in opposite directions, we can say that the force opposite to Mrs. Jacobson is pushing the fridge -15 N to the right (instead of 15 N to the left).

The net force would then be:

100 N + (-15 N) = 85 N to the right

The answer is A.

8 0

3 years ago

Read 2 more answers

What best describes the collision between ideal gas molecules

Alika [10]

An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly eleastic and in which there are no intermolecular attractive forces. One can visualize it as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other.

Happy to help

6 0

3 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p

almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen 40300 g

Carbon 11160 g

Hydrogen 6200 g

Nitrogen 1860 g

Calcium 992 g

Phosphorus 744 g

Other elements 744 g

It is possible to check it adding all the components, the total is 62000 g.

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

Percentage ( %)

Oxygen 65.0

Carbon 18.0

Hydrogen 10.0

Nitrogen 3.0

Calcium 1.6

Phosphorus 1.2

Other elements 1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

Percentage ( %) Decimal number

Oxygen 65.0 65/100 = 0.65

Carbon 18.0 18/100 = 0.18

Hydrogen 10.0 10/100 = 0.10

Nitrogen 3.0 3/100 = 0.03

Calcium 1.6 1.6/100 = 0.016

Phosphorus 1.2 1.2/100 = 0.012

Other elements 1.2 1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

Percentage ( %) Decimal number Mass in grams (g)

Oxygen 65.0 65/100 = 0.65 0.65*62000=40300

Carbon 18.0 18/100 = 0.18 0.18*62000= 11160

Hydrogen 10.0 10/100 = 0.10 0.10*62000= 6200

Nitrogen 3.0 3/100 = 0.03 0.03*62000= 1860

Calcium 1.6 1.6/100 = 0.016 0.016*62000= 992

Phosphorus 1.2 1.2/100 = 0.012 0.012*62000=744

Other elements 1.2 1.2/100 = 0.012 0.012*62000=744

5 0

3 years ago

Un anillo, de masa 90 gramos, contiene 59,1% de oro. ¿Cuál es el valor del anillo si cada mol de oro vale S/.1800?. Considere el

LekaFEV [45]

Answer:

S/.486 es el valor del anillo

Explanation:

Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.

Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:

90g × 59.1% = 53.19g Oro en el anillo

Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:

53.19g × (1mol / 197g) = 0.27 moles de oro contiene el anillo.

Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:

0.27mol × (S/.1800 / 1mol oro) =

<h3>S/.486 es el valor del anillo</h3>

8 0

3 years ago