All categories

2 years ago

Which of the compounds, li3p, ph3, c2h6, ibr3, are ionic compounds?

Chemistry

2 answers:

boyakko [2]2 years ago

8 0

Li3P=ionic 
PH3=Covalent 
C2H6=Covalent 
IBr=Covalent

REY [17]2 years ago

7 0

Answer:

Juts Li3P

Explanation:

An ionic compound is the one that is formed by a metal and a nonmetal, in that way when you join lithium (a metal) with phosphorous (a non metal), you are forming a ionic compound. The other three answers are just the join of two non metals, in that case, the three of them are considered like covalent compounds.

You might be interested in

How many moles of NaCl are present in a solution with a molarity of 8.59 M and a volume of 125 mL

Sergio [31]

Answer:

1.07ML

Explanation:

7 0

2 years ago

Read 2 more answers

The Sun is a main sequence star. During the last stage of its life cycle, it will become a white dwarf. It will shrink in size,

dangina [55]

The one that is tiny and white.

3 0

2 years ago

Read 2 more answers

Gold has always been a highly prized metal, and it has been widely used from the beginning of history as a store of value. It do

Greeley [361]

Answer: The balanced equation is

Au(s) + 3HNO3(aq) + 4HCl(aq) ---> HAuCl4(aq) + 3NO2(g) +3H2O(l)

The function of HCl in a solution of Aqua regia, that is used to dissolve gold is to dissolve other metals like quartz or iron stone that surround the gold.

Explanation: To balance the equation, we check the ratio of each element in the reacting side and the product side ( left and right hand side). Let their ratio be equal be adding moles to the compound of the element or the element it's self in either side of the equation.

HCl which is called hydrochloric acid, is an acid that does not react with gold, but it react with every other substance, like your skin, metals etc. it is used to clean a gold, by dipping the gold inside it, all the metals on the surface of the gold will dissolve.

When dissolving a gold in aqua regia solution, HCL is added to prepare this solution because it will help to dissolve all other substance on the surface of the gold.

8 0

1 year ago

Give the number of significant figures in this number: 40.00

Tatiana [17]

A significant figure is every symbol that made the number itself.

In this case, the number 40.00 has four figures but only two of them are significant 40, this is because you haven't got any more decimals than the first zero.

If you have a case with zeros in front, you take to the first non zero digits.

For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.

4 0

2 years ago

Read 2 more answers

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago