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3 years ago

Facts about heat absorptionand simple questions about heat absorption with answer

1 answer:

8 0

<span>As heat energy reaches an object it can be absorbed in a similar manner to the way sponges absorb water. Heat enters an object, warming it. The longer the object is exposed to the heat source, the more heat it absorbs.</span>

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ASAP will give brainliest!!!!

Andru [333]

Answer:

Explanation:

fluorine's atomic number is 9

electronic configuration: 2,7

so it needs 1 electron to stabilise

that's why 1 covalent bond

hope it helps!!

7 0

3 years ago

A chunk of sulfur has a volume of 8.33 cm3. What is the mass of this sulfur? (Density of sulfur = 2.07 g/cm3.)

nata0808 [166]

Explanation:

equations to note:

density= mass/volume

mass= volume *density

volume= mass/density

you have a volume- 8.33cm3

you have a density- 2.07 g/cm3

Answer:

8.33cm3 * 2.07g/cm3= 17.24g

mass= 17.24g

5 0

2 years ago

Match the solutions to the descriptions of the freezing points.a. One mole of the ionic compound Na3PO4 dissolved in 1000 g H2O

rosijanka [135]

Explanation:

Depression in Freezing point

= Kf × i × m

where m is molality , i is Van't Hoff factor, m = molality

Since molality and Kf remain the same

depression in freezing point is proportional to i

i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2

i=1 for C2h6O

i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)

So the freezing point depression is highest for MgCl2 and lowest for C2H6O

so freezing point of the solution = freezing point of pure solvent- freezing point depression

since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point

5 0

2 years ago

zn (s) 2 hcl (aq) -----> zncl2 (aq) h2(g) if 520 ml of h2 is collected over water at 28oc and the atmospheric pressure is 1.0

nadya68 [22]

The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.

Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.

We recognise that

h * 2 = PT - P * h * 20 = 1atm - 0.037atm

= 0.963 atm

1 * h * 2 = Ph * 2V / R * T

= 0.963 atm x 0.520 L / 0.0821 L atm/

molK * 301

= 0.02 mol h2

= 0.02molZn

So 0.02 mol Zn x 65.39 g/mol

= 1.33 g Zn