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Leokris [45]
2 years ago
10

Help me with this question please

Chemistry
1 answer:
Natasha_Volkova [10]2 years ago
7 0

Answer:

I dont know sorry i will try my best htough

Explanation:

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Sue dissolved a certain amount of salt in 400 grams of water to obtain 405 grams of salt solution. What was the mass of the salt
svp [43]
The answer would be 5 grams of salt
6 0
3 years ago
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Scientists need special tools to work in the laboratory. These tools are known as___________
inessss [21]
Instruments

The specific type of instruments depends on the type of laboratory that you're working in: some labs for example use electron microscopes, others use mass spectrophotometers, others use multiplex biochemical analyzers, etc. But very broadly, the specialized tools we use in the laboratory are usually referred to as "instruments"
3 0
3 years ago
When benzene (c6h6) reacts with bromine (br2) bromobenzene(c6h5br) is obtained: c6h6 + br2 → c6h5br + hbr what is the theoretica
saveliy_v [14]

Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,

30 g / 78 moles of benzene = 0.384 moles


For bromine it will be the same process,

65 g / 159.8 moles = 0.406 moles


By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1


We need to find the mass of bromobenzene,

which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol


So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g


Hence the theoretical yield will be 60.34 g


Answer B) : To calculate the actual yield we have to divide it with theoretical yield.


(56.7g / 60.343 g ) X100% = 93.96 %


Here, we can say that we got 93.96 % of actual yield.


As we know it is impossible to get 100% yield in any reaction.

3 0
3 years ago
In a science lab, a student heats up a chemical from 10 °C to 25 °C which requires thermal energy of 30000 J. If mass of the obj
olganol [36]

Answer:

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

Explanation:

Step 1: Data given

Initial temperature = 10.0 °C

Final temperature = 25.0 °C

Energy required = 30000 J

Mass of the object = 40.0 grams

Step 2: Calculate the specific heat capacity of the object

Q = m* c * ΔT

⇒With Q = the heat required = 30000 J

⇒with m = the mass of the object = 40.0 grams

⇒with c = the specific heat capacity of the object = TO BE DETERMINED

⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C

30000 J = 40.0 g * c * 15.0 °C

c = 30000 J / (40.0 g * 15.0 °C)

c = 50 J/g°C

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

3 0
3 years ago
1.34 milligrams is the same as _____ kg and _____g.
Karolina [17]
Answer C is for kg and but it's .00134 for grams
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3 years ago
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