What type of isomers are ch3ch2och3 and ch3ch2ch2oh?

HELP PLS !!!! :))Formula unit of Sr and CI<br><br> Formula unit of AI and S

xenn [34]

Answer : The formula unit of Sr & Cl is $SrCl_2$ and the formula unit of Al & S is $Al_2S_3$

Explanation :

Formula unit : it is defined as the lowest number ratio of ions of an elements in an ionic compound or covalent compound.

For the formula unit of Sr & Cl, two chloride ions $(Cl^-)$ are needed to neutralize the one strontium ion $(Sr^{2+})$ .

For the formula unit of Al & S, three sulfide ions $(S^{2-})$ are needed to neutralize the two aluminium ion $(Al^{3+})$ .

The formula unit of $SrCl_2$ and $Al_2S_3$ are shown below.

4 0

3 years ago

Which statement describes the location of metalloids on the periodic table? The metalloids are located below metals and above no

Talja [164]

The metalloids are Boron, Silicon, Geranium, etc and are found to the right of the metals and the left of the nonmetals. Since that is not an option, the best choice would be: The metalloids are located below nonmetals and above metals within a group.

4 0

3 years ago

Read 2 more answers

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2

Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0

3 years ago

If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?

statuscvo [17]

The process in which the concentration of the solution is lessened by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

$C_{1}V_{1}=C_{2}V_{2}$ (1)