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jonny [76]
3 years ago
12

Which topic are you most likely to find in a physics textbook?

Chemistry
2 answers:
Nadusha1986 [10]3 years ago
6 0
Hello there.

<span>Which topic are you most likely to find in a physics textbook?

Answer: </span><span>b.electricity and magnetism
</span>
Rom4ik [11]3 years ago
3 0
<span> electricity and magnetism
</span>
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Ok so I have 3 beakers of clear liquid each holding a different substance, Water, Magnesium Sulphate and Sodium Carbonate how ca
nataly862011 [7]
For water you could add oil..ex: cooking oil separates form water because water is heavier than oil.

For Magnesium Sulfate you could add Sodium Carbonate..ex: Sodium Carb reacts to Mg Sulfate adding a darker hue to the liquid and adding a lot of bubbles.

For Sodium Carbonate you could add Sulfuric Acid..ex: Sulfuric Acid would add a reaction to the Sodium Carb that would resembling water boiling 

H0P3 It H3LPS :)
7 0
3 years ago
WORTH 20 POINTS Confirmation bias is where a scientist may change the data to support a certain bias.
Orlov [11]

Answer:

True

Explanation:

Confirmation Bias is the tendency to look for information that supports, rather than rejects, one’s preconceptions, typically by interpreting evidence to confirm existing beliefs while rejecting or ignoring any conflicting data

8 0
3 years ago
In the reaction Fe2O3 + 3CO a 2Fe + 3CO2, 10 moles of solid iron and 15 moles of carbon dioxide are produced from 5 moles of iro
Schach [20]

Answer:

Ratio is 3:2

3CO = 2Fe or 1.5 CO = 1 Fe

Explanation:

Fe2O3 + 3CO = 2Fe + 3CO2

Fe2O3 = Iron (|||) oxide

CO = Carbon monoxide

Fe = Solid Iron

CO2 = Carbon dioxide

Excellent is already balanced.

10 Moles Fe and 15 Moles of CO2

5 Moles Fe2O3 + 15 Moles 3CO = 10 Moles Fe + 15 Moles 3CO2

What is the ratio of carbon monoxide to solid iron

Ratio is 3:2 or 1.5 CO = 1 Fe

5 0
3 years ago
A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
3 years ago
An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc
pantera1 [17]

Answer:

5.35 *10^{-4}M

Explanation:

Equation for the reaction is as follows:

2CO_{(g)}      +      O_{2(g)}        ⇄       2CO_{2(g)}

By Applying the ICE Table; we have

                             2CO_{(g)}      +      O_{2(g)}        ⇄       2CO_{2(g)}

Initial                      x                  0.0025 M              0.0010 M

Change                  0                       0                            0

Equilibrium             x                  0.0025 M              0.0010 M

K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}

Given that K_c = 1.4*10^2 ; Then:

1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}

1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}

3.5 =( \frac{(0.001)}{(x)})^2

\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}

1.87=\frac{(0.001)}{(x)}

(x)= \frac{(0.001)}{1.87 }

x = 5.35 *10^{-4}M

∴ The equilibrium concentration of CO = x = 5.35 *10^{-4}M

7 0
4 years ago
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