All categories

3 years ago

Which of the following contains properties of magnesium, Mg?

Chemistry

1 answer:

tekilochka [14]3 years ago

6 0

Answer:

Option B. Malleable, Conductor, High melting point, Lustrous

Explanation:

Mg has a higher melting point because of the strong electrostatic force of attraction between the magnesium ions (Mg^2+). The rest properties listed are all general properties of metals

You might be interested in

There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e

irakobra [83]

80% copper (Cu)

Cu: 80 : 63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 : 63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0

2 years ago

A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr

sergey [27]

Answer:

184.62 ml

Explanation:

Let $p_1, v_1,$ and $T_1$ be the initial and $p_2, v_2,$ and $T_2$ be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

$p_1=p_2$ ...(i)

$v_1$ = 200 ml

$T_1= 26 ^{\circ}C = 273+26 =299$ K

$T_2= 3 ^{\circ}C = 273+3 =276$ K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

$p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)$

For the final condition,

$p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)$

Equating equation (i), and (ii)

$\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}$

$\frac{v_1}{T_1}=\frac{v_2}{T_2}$ [from equation (i)]

$v_2=\frac{T_2}{T_1} \times v_1$

Putting all the given values, we have

$v_2=\frac{276}{299} \times 200 = 184.62 \; ml$

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

7 0

2 years ago

I’d appreciate the help! :)

seraphim [82]

Answer: 300g

Explanation:

first we write the given values on top

224L. x

3 NO2 (g) + H2O (l) = 2HNO3 (l) + NO (g)

22.4L 30g

then we form a formula

224L/22.4L= x/30g

224*30/22.4

6720/22.4= 300g

7 0

3 years ago

(2pts) During the Purification of Lactate Dehydrogenase (LDH) experiment, you will need 50ml of buffer A150. Buffer A150 is 30mM

torisob [31]

Answer:

The answer is " $20 \ mL$ "

Explanation:

Given:

Molarity= number of moles

because it is 1 Liter

$\to \frac{0.03\ moles}{1.5 moles}=0.02\ L= 20 \ mL \ of\ Tris\\\\$

therefore,

it takes 20 mL of Tris.

$\to \frac{0.150 \ moles}{5\ moles} =0.03\ L\\\\$

$= 30 \ mL \ of\ Nacl$

So, take $20 \ mL\ of\ NaCl.$

6 0

3 years ago

PLEASE HELP FILL OUT THIS CHART! ASAP THANK YOU

Svet_ta [14]

I will need a picture if the periodic table

6 0

3 years ago