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evablogger [386]
3 years ago
9

Group 2 in the penodic table contains the elements beryllium (Be), magnesium (Mg) , calcium (Ca) , strontium (Sf) )barium (Ba)an

d radium (Ra) What characteristic do these elements share ?
Chemistry
1 answer:
mario62 [17]3 years ago
6 0

Answer:

They are all alkali earth metals.

Explanation:

Their valence shell each has 2 electrons. Also, they are all shiny, silvery-white, somewhat reactive metals at standard temperature and pressure. They form alkaline solutions, hydroxides, when reacting with water and their oxides are found in the earth’s crust.

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5. The density of a substance is 3.4 X 10^3 g/mL. Mixing it with another substance increases the density by 1.5 X 10^1 times. Wh
JulijaS [17]

Answer:

the total density of the mixture = 3.415 x 10^3 g/mL

Explanation:

given:

density of a substance is 3.4 x 10^3 g/mL

Mixing it with another substance increases the density by 1.5 x 10^1 times.

find:

What is the density of the mixture?​

-------------------------------------------------

let density (a) = 3.4 x 10^3 g/mL

density(b) = 1.5 x 10^1 g/mL

since there is no specific density provided for the mixture, we then add both to increase the density,

total density = density(a) + density(b)

total density = 3.4 x 10^3 g/mL + 1.5 x 10^1 g/mL

total density = 3.415 x 10^3 g/mL

therefore,

the total density of the mixture = 3.415 x 10^3 g/mL

5 0
3 years ago
Read 2 more answers
Pls help, the questions in the picture !
dedylja [7]

Answer:

B is sedimentary and A is metermorphic

4 0
3 years ago
Water waves are surface waves. The energy of the waves moves
horrorfan [7]

Answer:

Explanation:

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8 0
3 years ago
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State types of environment pollution?​
jekas [21]

Answer:

Air Pollution.

Water Pollution.

Soil/Land Pollution.

Noise Pollution.

Radioactive Pollution.

Explanation:

4 0
4 years ago
For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g)
dangina [55]

Answer:

14.4 g of CO_{2} can be produced.

Explanation:

Balanced equation: 2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O

                                        Molar mass (g/mol)

                  C_{4}H_{10}                    58.12

                    O_{2}                         32

                  CO_{2}                       44.01

So, 9.60 g of C_{4}H_{10} = \frac{9.60}{58.12}mol=0.165mol

      17.0 g of O_{2} = \frac{17.0}{32}mol=0.531mol

According to balanced equation-

2 moles of C_{4}H_{10} produce 8 moles of CO_{2}

So, 0.165 moles of C_{4}H_{10} produce (\frac{8}{2}\times 0.165)mol of CO_{2}  or 0.660 moles of CO_{2}

13 moles of O_{2} produce 8 moles of CO_{2}

So, 0.531 moles of O_{2} produce (\frac{8}{13}\times 0.531)moles of CO_{2} or 0.327 moles of CO_{2}

As least number of moles of CO_{2} are produced from O_{2} therefore O_{2} is the limiting reagent.

So, maximum amount of CO_{2} that can be formed = 0.327 moles

                                                                              = (0.327\times 44.01)g

                                                                              = 14.4 g

3 0
3 years ago
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