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Varvara68 [4.7K]
4 years ago
8

You need to measure the brightness of a light bulb using the metric system. In which unit do you express your measurement?

Chemistry
1 answer:
kherson [118]4 years ago
7 0
Candela is the correct answer. In measuring the brightness or luminosity of the bulb and you want to use the metric system, candela is the unit to be used. You are correct that candela is the unit of light intensity. The term candela is the Latin for candle which signifies light.<span> </span>
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Why do large area's of water change how the wind blows?
sp2606 [1]

Answer:

C

water changes temperature slowly i believe C is your anwser

5 0
3 years ago
Read 2 more answers
Which two terms represent types of chemical formulas?
Kisachek [45]

Answer: 3 empirical and structural

Explanation:

Just guessed on same question got it right

6 0
3 years ago
how many liters of O2 are needed to produce 5.62 g of SO2 in the following reaction? 4FeS2+11O2 -&gt;2Fe2O3+8SO2
Liula [17]

8moles of SO_2 need 11mol O_2

  • 1mol need=11/8=1.4molO_2

Moles of sO_2

\\ \rm\hookrightarrow \dfrac{5.62}{64}=0.1

So

Moles of O_2

  • 1.4(0.1)=0.14mol

We know

  • 1mol at STP=22.4L
  • 1.4mol=0.14(22.4)=3.13L O_2
5 0
3 years ago
What atomic or hybrid orbitals make up the sigma bond between br and f in bromine trifluoride, brf3?
mina [271]
<span>Draw the Lewis structure. Bromine has 3 bonds and two lone pairs for a trigonal bipyramidal electron geometry and an sp^3d hybridization. Fluorine is peripheral it does not require hybridization, but we often consider it to be hybridized too - it has 1 bond and 3 lone pairs for sp^3 hybridization. So the sigma bonds come from an overlap of an sp^3d orbital on Br with an sp^3 orbital on F. If you don't consider the F to be hybridized the overlap would have to be to a p orbital on the F</span>
8 0
4 years ago
For the process 2SO2(g) + O2(g) --&gt; 2SO3(g),
CaHeK987 [17]

Answer:

–187.9 J/K

Explanation:

The equation that relates the three quantities is:

\Delta G = \Delta H - T \Delta S

where

\Delta G is the Gibbs free energy

\Delta H is the change in enthalpy of the reaction

T is the absolute temperature

\Delta S is the change in entropy

In this reaction we have:

ΔS = –187.9 J/K

ΔH = –198.4 kJ = -198,400 J

T = 297.0 K

So the Gibbs free energy is

\Delta G=-198,400-(297.0)(-187.9)=254.2 kJ

However, here we are asked to say what is the entropy of the reaction, which is therefore

ΔS = –187.9 J/K

8 0
4 years ago
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