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Digiron [165]
3 years ago
6

BRAINLIST IF U ARE CORRECT AND ANSWER ALL OF THEM!!!!!!!!!!Match the name of the way water is used to its description.

Chemistry
2 answers:
Semenov [28]3 years ago
3 0

Answer:

I think it is the first one all those places do use water

sdas [7]3 years ago
3 0
All these places use water however the best described one is the first one
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My teacher is being weird about it?? every answer i put is wrong someone please help
Kruka [31]
Answer is 146. G/mol
7 0
3 years ago
A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of w
Oksanka [162]

Complete Question

A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.

Answer:

M=58g

Explanation:

From the question we are told that:

Heat Capacity H=0.897

Mass of water M=200g

Initial Temperature of Aluminium T_a=85.6

Initial Temperature of Water T_{w1}=16.0

Final Temperature of Water  T_{w2}=16.0

Generally

Heat loss=Heat Gain

Therefore

M*0.897*(85.6-20.1) =200*4.184*(20.1-16)

M=58g

5 0
2 years ago
Bronze is an alloy, which is what form of matter?
grandymaker [24]
It is a Compound. have a nice day
3 0
3 years ago
Read 2 more answers
I NEED HELP QUICK! BRAINLIEST TO WHO IS RIGHT!
Travka [436]

Answer:

3rd option

Explanation:

3 0
2 years ago
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot
kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

p_A=\chi_A\times P_T

p_{O_2} = partial pressure of O_2 = ?

\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}

P_{T} = total pressure of mixture  = 525 mmHg

{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles

{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles

Total moles = 1.94 + 1.35 = 3.29 moles

\chi_{O_2}=\frac{1.94}{3.29}=0.59

p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0
3 years ago
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