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3 years ago

Plsss helpp will mark brainliest!!!

Chemistry

2 answers:

Tju [1.3M]3 years ago

6 0

The answer is B. HOPE ITS RIGHT!

Ymorist [56]3 years ago

3 0

Answer:

high melting point

Explanation:

high melting point

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Put the elements in order from easiest to hardest to remove an electron. Ba,Be,Ca,Sr.

laiz [17]

Answer:

Ba,Sr,Ca,Be

Explanation:

as atomic size increases ionisation potential decreases

8 0

3 years ago

Given the following equation: 2 KClO 3 --> 2 KCl + 3 O 2

SCORPION-xisa [38]

Explanation:

12.0 mol KClO3 × (3 mol O2/2 mol KClO3)

= 18.0 mol O2

6 0

3 years ago

The volume of gas occupied 240.0ml when the pressure is 1.20 atm what volume at constant pressure.will the gas occupies when the

Marizza181 [45]

Answer:

334.89 ml

Explanation:

V2 = V1P1/P2

V2 = 240×1.2/0.86

V2 = 334.89 ml

7 0

3 years ago

How many mols of chlorine are in 120g of chlorine gas

xz_007 [3.2K]

Answer:

Number of moles of chlorine = 3.38 mol

Explanation:

Given data:

Mass of chlorine = 120 g

Moles of chlorine = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of chlorine = 35.5 g/mol

Now we will put the values in formula.

Number of moles = 120 g/ 35.5 g/mol

Number of moles = 3.38 mol

6 0

3 years ago

A gas has a of 590 mi. At a temperature What volume will the gas occupy at 30.0öc.1 Which gas law is this

scoray [572]

The given question is incomplete. The complete question is:

A gas has a volume of 590 ml at a temperature of $0^0C$ .What volume will the gas occupy at $30.0^0C$ . Which gas law is this?

Answer: 655 ml , Charle's Law

Explanation:

To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=590ml\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=30.0^0C=(30+273)K=303K$

Putting values in above equation, we get:

$\frac{590}{273}=\frac{V_2}{303}\\\\V_2=655ml$

Thus the volume at $30.0^0C$ is 655 ml

6 0

3 years ago