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DENIUS [597]
3 years ago
11

Which statements are scientific statements

Chemistry
1 answer:
Harrizon [31]3 years ago
6 0
Wym there is no question or statements.
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How many molecules are in 0.25 moles of CH4
Brums [2.3K]
The base gram of the compound is 16.042

To find the new amount of grams, you need to multiply the amount of moles by the the base grams

16.042*.25=4.01

So one fourth of the compound is 4.01 grams of Carbon tetrahydride.
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3 years ago
Mr. Gordon's science students are studying scale models of the Sun-Earth-Moon system. Mr. Gordon is using this model to show his
Nata [24]

Answer:

It's very likely

Explanation: UwU

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3 years ago
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Pure water at 25°C has a pH of<br> (1) 1 (2) 5 • (3) 7<br> (4) 14
Marysya12 [62]

Answer:

7. .........................

8 0
3 years ago
Calculate the total amount of energy required in calories to convert 50.0 g of ice at 0.00 degrees Celsius to steam at 100. degr
r-ruslan [8.4K]

Answer:

HFusion*mass + Spec.Heat*mass*ΔT + HVap*mass

80cal/g*50.0g + 1.00cal/g°C*50.0g*(100°C-0°C) + 540cal/g*50g

3.60x10⁵cal

Explanation:

Using the HFusion we can find the heat needed to convert the ice to liquid water.

With specific heat capacity we can find the heat needed to increase the temperature of water from 0 to 100°C.

With HVap we can find the heat to convert the liquid water into steam.

The equations are:

<h3>HFusion*mass + Spec.Heat*mass*ΔT + HVap*mass</h3><h3 />

Computing the values:

<h3>80cal/g*50.0g + 1.00cal/g°C*50.0g*(100°C-0°C) + 540cal/g*50g</h3>

36000cal =

<h3>3.60x10⁵cal</h3>
8 0
3 years ago
The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H2O, H2, and O2
jenyasd209 [6]

Answer:

1.742\times 10^{-5} is the value of the equilibrium constant at this temperature.

Explanation:

2H_2O\rightleftharpoons 2H_2+O_2

We are given:

Partial pressure of H_2O=p^o_{H_2O}=0.0750 atm

Partial pressure of H_2=p^o_{H_2}=0.00700 atm

Partial pressure of O_2=p^o_{O_2}=0.00200 atm

The expression of K_p for the given chemical equation is:

K_p=\frac{p^o_{H_2}^2\times p^o_{O_2}}{p^o_{H_2O}^2}

Putting values in above equation, we get:

K_p=\frac{(0.00700 atm)^2\times 0.00200 atm}{(0.0750 atm)^2}\\\\K_p=1.742\times 10^{-5}

1.742\times 10^{-5} is the value of the equilibrium constant at this temperature.

3 0
4 years ago
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