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kobusy [5.1K]
3 years ago
12

Give an example of an element that would be expected to share similar chemical properties with beryllium, calcium, and strontium

.​
Chemistry
1 answer:
Aleks04 [339]3 years ago
4 0
Strontium possibly but I’m not 100% sure
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Two liquids, a and b are immiscible. liquid a has a density of 0.89 g/ml. liquid b has a density of 0.72 g/ml. what would you ex
Burka [1]
In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.

The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.
5 0
3 years ago
Help me please
kati45 [8]

Answ????

Explanation:

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5 0
3 years ago
Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce
kkurt [141]

Explanation:

The given data is as follows.

             P = 3 atm

                = 3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}  

                 = 3.03975 \times 10^{5} Pa

    V_{1} = 9 L = 9 \times 10^{-3} m^{3}    (as 1 L = 0.001 m^{3}),  

        V_{2} = 15 L = 15 \times 10^{-3} m^{3}

            Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

                  W = P \times \Delta V

or,                W = P \times (V_{2} - V_{1})

Therefore, putting the given values into the above formula as follows.

                  W = P \times (V_{2} - V_{1})

                      = 3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})

                      = 1823.85 Nm

or,                   = 1823.85 J

As internal energy of the gas \Delta E is as follows.

                     \Delta E = Q - W

                                  = 800 J - 1823.85 J

                                  = -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

8 0
3 years ago
Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou
Ulleksa [173]

Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

\frac{0.0875~mol~Y_2}{2}=0.04375

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

6 0
3 years ago
calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation
balu736 [363]

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= ?
  • V= 5.005 L
  • n= 1.255 mol
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K

Solving:

P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}

P= 5.62 atm

<u><em>The pressure is 5.62 atm.</em></u>

8 0
3 years ago
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