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swat32
3 years ago
10

A curved piece of glass or other transparent material that is used to refract light is called a (an) _____.

Chemistry
2 answers:
Furkat [3]3 years ago
7 0

Answer:

C) reflector

Explanation:

Think about a small mirror if you put it in the sun it reflects light so  it is reflector

lesya692 [45]3 years ago
4 0

Answer:

Lens

A curved piece of glass or other transparent material that is used to refract light.

Explanation:

u got ig?

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What mass of Al contains the same number of atoms as 33.5g of Au
wel
Stoich questions...

1) convert the grams of Au to moles
(33.5g)/(196.7g/mol)=___x___

2)find how many atoms are in that many moles by multiplying it by avogadro's number.

x(moles of Au)*(6.02*10^23)=?

3)now that you have found the number of atoms of gold, convert it to grams of aluminum by multiplying it by Avogadro's number and multiplying it by the atomic mass of aluminum.
( ? )(6.02*10^23)(26.98g/mol)= (mass of Al)

5 0
3 years ago
An unidentified compound contains only phosphorous and fluorine, meaning we could write its formula as PxFy. Heating 0.2324 g of
cricket20 [7]

Answer:

The molecular formula of the compound is P2F4

Explanation:

Step 1: Data given

Mass of the compound  = 0.2324 grams

Volume of container = 378 mL

Pressure at 77 °C = 97.3 torr

The gaseous PxFy was then reacted with an aqueous solution of calcium chloride, and all of the fluorine was converted to 0.2631 g of CaF2.

Molar mass CaF2 = 78.07 g/mol

Step 2: Calculate moles CaF2

Moles CaF2 = 0.2631 g/ 78.07 g/mol=0.003370

Step 3: Calculate moles F

For every 1 mol CaF2 we have 1 mol Ca2+ and 2 moles F-

moles F- = 2 * 0.003370=0.006740

Step 4: Calculate mass of F

mass F =  0.006740 mol * 19.00 g/mol=0.1281 g

Step 5: Calculate mass of P

mass P = mass of compound - mass of F = 0.2324 - 0.1281 =0.1043 g

Step 6: Calculate moles P

moles P = 0.1043 g/ 30.97 g/mol=0.003368 moles

Step 7: Calculate mol ratio

We divide by the smallest amount of moles

P: 0.003368/0.003368 =1

F: 0.006740/ 0.003368 =2

The empirical formula is PF2

The empirical formula has a molar mass of 68.97 g/mol

Step 8: Calculate moles of compound

p*V = n*R*T

n = (p*V)/(R*T)

⇒ with p = the pressure of the gas = 97.3 torr = 97.3/760 = 0.128 atm

⇒ with V = the volume of the gas = 0.378 L

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 77.0 °C = 350 Kelvin

Number of moles =(0.128 *0.378)/(0.08206*350)

Number of moles =

moles gas = pV/RT = 0.128 atm x 0.378 dm^3/ 0.08206x350 K=0.00168  

Step 9: Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.2324 grams / 0.00168 moles

Molar mass = 138 g/mol

Step 10: Calculate the molecular formula

138.0 / 68.97 = 2

2*(PF2) = P2F4

The molecular formula of the compound is P2F4

4 0
2 years ago
What is the best way to package and preserve a questioned document?
Zanzabum
In a plastic sheet protector
5 0
3 years ago
Read 2 more answers
Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O
Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.5kJ)=-787kJ

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.8kJ)=-571.6kJ

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)

\Delta H=52.4kJ

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0
3 years ago
Which of the following statements is FALSE?
vovikov84 [41]

Answer: ITS FALSE

ExplaC6H12 + 9 O2 → 6 H2O + 6 CO2 is a single replacement reaction

nation:

3 0
2 years ago
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