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3 years ago

What is true about past volcanic eruptions?

Chemistry

1 answer:

Nimfa-mama [501]3 years ago

4 0

Answer:

It is either

Past volcanic eruptions leave marks on rocks near the volcano.

Past volcanic eruptions leave marks in the atmosphere.

My guess is C.

Explanation:

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A scientist conducts an experiment to determine how the CO2 levels in ocean water affect the number of plants that can grow in t

Savatey [412]

Answer:

number of plants that grow in the ocean

Explanation:

The dependent variable would be the number of plants that grow in the ocean.

The dependent variable is the actual variable being measured in an experiment. The value of dependent variables is directly affected by another variable in an experiment - the independent variable. The latter is the manipulatable variable that is directly inputted during experiments.

In the illustration, the main aim of the experiment is to determine how CO2 levels in ocean water affect the number of plants that can grow in the ocean. The manipulatable input variable would be the CO2 while the outcome variable - the dependent variable - would be the number of plants that grow in the ocean.

3 0

3 years ago

What is the value of the equilibrium constant for this redox reaction?

aivan3 [116]

Correct answer: Option D, K = 5.04 × 10^52

Reason:
We know that,
Ecell = $\frac{0.0592}{n}log(K)$ ,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.

Also, Ecell = +1.56 v

Therefore, 1.56 = $\frac{0.0592}{2}log(K)$
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52

8 0

3 years ago

Read 2 more answers

Sea water has about 56.0 grams of NaCl for every 2.0 liters of water. what is the molarity?

Effectus [21]

5.60 cuz I I’m dumb vendbsnksnddn

3 0

3 years ago

Aluminum reacts with chlorine gas to produce aluminum chloride according to the following equation. Al + Cl2 → AlCl3 Which of th

anzhelika [568]

<h3>Answer;</h3>

B. 3/2

<h3> Explanation;</h3>

Balance the chemical equation

2Al + 3Cl2 → 2AlCl3

We want to convert moles of AlCl3 to moles of Cl2

The conversion factor is 2 mol AlCl3/3 mol Cl2.

We choose the one that makes the units cancel:

x mol AlCl3 x (3 mol Cl3)/(2mol AlCl3) = x mol Al

The fraction for the molar ratio is 3/2.

4 0

3 years ago

Read 2 more answers

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago