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2 years ago

Which of the following contains most of the water on earth?

Chemistry

2 answers:

lions [1.4K]2 years ago

8 0

The answer is A because most of the earth is made up of oceans.

mrs_skeptik [129]2 years ago

4 0

Answer:

Explanation:

most of the world is made of water because the oceans.

You might be interested in

Solute and solvent of C12H22O11

nasty-shy [4]

Answer:

sugar is the solute and water is the solvent

8 0

3 years ago

Which of these following is true about the mass of the reactants in a chemical reaction?

yuradex [85]

They will be stay the same all the way through. your product must be the same mass

7 0

3 years ago

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.220 × 10-3 m3. Into this cavity is placed 1.100 × 10-

juin [17]

Answer:

Δ should be 0.1009

Explanation:

The change in the units volume when temperature change can be expressed as:

∆v = v0Δ

with v0 = the initial volume

with = the volumetric temperature expansion coefficient

with Δ = the change of temperature.

To calculate the final volume vf we'll get:

v = v0 + ∆ = v0(1 + Δ)

The liquid just begins to spill out if v(benzene) = :

v()(1 + Δ) = = v() (1 + Δ)

(v(cavity)-v(benzene))/(((benzene) -(copper)) = Δ

((1.22*10^-3)-(1.1*10^-3))/((1240*10^-6)-(51*10^-6)) = Δ

Δ = 0.1009

Δ should be 0.1009

5 0

3 years ago

Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: SO2(g) + O2(g) → SO3(g)

kipiarov [429]

Answer:

$m_{SO_3}=18.93gSO_3$

$V_{SO_3}=5.3LSO_3$

Explanation:

Hello,

STP conditions are P=1 atm and T=273.15 K, thus, the reacting moles are:

$n_{SO_2}=\frac{5.3L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2366molSO_2\\n_{O_2}=\frac{4.7L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2098molO_2$

Now, the balanced chemical reaction turns out into:

$2SO_2(g) + O_2(g) --> 2SO_3(g)$

Thus, the exact moles of oxygen that completely react with 0.2366 moles of sulfur dioxide are (limiting reagent identification):

$0.2366molSO_2*\frac{1molO_2}{2molSO_2}=0.1182molO_2$

Since 0.2098 moles of oxygen are available, we stipulate the oxygen is in excess and the sulfur dioxide is the limiting reagent. In such a way, the yielded grams of sulfur trioxide turn out into:

$m_{SO_3}=0.2366molSO_2*\frac{2molSO_3}{2molSO_2}*\frac{80gSO_3}{1molSO_3} \\m_{SO_3}=18.93gSO_3$

By using the ideal gas equation, one computes the volume as:

$V_{SO_3}=\frac{mRT}{MP}=\frac{18.93g*0.082\frac{atm*L}{mol*K} *273.15K}{80g/mol*1atm}\\V_{SO_3}=5.3LSO_3$

It has sense for volume since the mole ratio is 2/2 between sulfur dioxide and sulfur trioxide.

Best regards.

6 0

3 years ago

3350 J of heat is required to raise the temperature of a sample of AlF3 from 250C to 800C. What is the mass of the sample?

Aleonysh [2.5K]

Answer:

Look up the specific heat capacity of AlF₃
Calculate ΔT
Calculate the mass of AlF₃

Explanation:

The formula for for the heat (q) absorbed by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample. and

ΔT = the change in temperature

1. What you must do

Look up the specific heat capacity of AlF₃
Calculate ΔT
Calculate the mass of AlF₃

2. Sample calculation

For this example, I assume that the specific heat capacity of AlF₃ is 1.16 J·K⁻¹mol⁻¹ .

(a) Calculate ΔT

$\Delta T = T_{\text{f}} - T_{\text{i}} = 800 \, ^{\circ}\text{C} -250 \, ^{\circ}\text{C} = 550 \, ^{\circ}\text{C}$

(b) Calculate m

$\begin{array}{rcl}\text{3350 J} & = & m \times 1.16 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}\times \text{550 K}\\3350 & = & m \times \text{638 g}^{-1}\\m & = &\dfrac{3350}{\text{638 g}^{-1}}\\\\ & = & \text{ 5.2 g}\\\end{array}\\$

6 0

3 years ago