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3 years ago

The smallest unit that can exist as an element and still have the properties of that element is a(n):

Chemistry

2 answers:

Yuri [45]3 years ago

8 0

The smallest unit that can exist as an element and still have proprieties of that element is an Atom

liq [111]3 years ago

6 0

Explanation:

Every element is made up of atoms which cannot be further divided into simpler substances.

For example, a piece of zinc metal will only contains atoms of zinc and no there element.

Since, these atoms are the same as the element itself. Hence, the element still retain its properties because there no other different element is present in it.

Thus, we can conclude that the smallest unit that can exist as an element and still have the properties of that element is an atom.

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Choose the false statement regarding resistance

Dmitry_Shevchenko [17]

I would say all if no statements are available

4 0

3 years ago

Someone please help me!!!!

Artyom0805 [142]

Double replacement :

2Na₃PO₄+3CaCl₂⇒6NaCl + Ca₃(PO₄)₂

<h3>Further explanation</h3>

1. A single replacement reaction is a chemical reaction in which one element replaces the other elements of a compound to produce new elements and compounds

2. Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

3. Combination/syntesis : 2 or more reactants combine to form a new compound

4. Decomposition : the reactant is decomposed into 2/more products

If we look at the reaction options available, all of them can be included in the double replacement reaction, but we only choose the reaction from Sodium phosphate and Calcium chloride which leads to options: C because it is balanced (from the number of atoms in the same reactant and product) and is in accordance with the chemical formula of each compound (both products and reactants)

3 0

3 years ago

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0

3 years ago

Fill in the missing data point. Show all calculations leading to an answer.

Aleksandr-060686 [28]

Answer:

1090 mmHg

Explanation:

We know that with gases we must use a Kelvin temperatures, so let’s try a plot of pressure against the Kelvin temperature.

We can create a table as follows

t/°C T/K p/mmHg

10 283 726

20 293 750

40 313 800

70 343 880

100 373 960

150 423 ???

I plotted the data and got the graph in the figure below.

It appears that pressure is a linear function of the Kelvin temperature.

y = mx + b

where x is the slope and b is the y-intercept.

===============

Calculate the slope

I will use the points (275, 700) and (380, 975).

Slope = Δy/Δx = (y₂ - y₁)/(x₂ -x₁) = (975 -700)/(380 – 275) = 275/105 = 2.619

So,

y = 2.619x + b

===============

Calculate the intercept

When x = 275, y = 700.

700 = 2.619 × 275 + b

700 = 720 + b Subtract 720 from each side and transpose.

b = -20

So, the equation of the graph is

y = 2.619x -20

===============

Calculate the pressure at 423 K (150°C)

y = 2.619 × 423 - 20

y = 1110 - 20

y = 1090

At 150 °C, the pressure 1090 mmHg.

The point is approximately at the position of the black dot in the graph.

7 0

3 years ago

Balance the following equation: C7H602 + 02 -- CO2 + H20

azamat

Answer:

Explanation:

2C7H6O2 + 15O2 → 14CO2 + 6H2O.

5 0

3 years ago