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2 years ago

Match the landform to its description.

Chemistry

2 answers:

yuradex [85]2 years ago

5 0

Answer:

cinder cone: made of pieces of lava

shield volcano: wide summit, gentle slope

volcanic soil: rich in nutrients

caldera: bowl shaped depression

lava plateau: material fills in valleys

Explanation:

it’s right

nikitadnepr [17]2 years ago

5 0

Answer:

Caldera, Bowl-Shaped depression

Shield volcano, Wide summit, Gentle slope

Lava plateau, Material fills in valleys

Volcanic soil, Rich nutrients

Cinder Cone, Made of pieces of lava

Explanation:

Got it right on edge

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Radioactive gold-198 is used in the diagnosis of liver problems. the half-life of this isotope is 2.7 days. if you begin with a

Neporo4naja [7]

Answer:

See explanation below

Explanation:

To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:

m = m₀e^-kt (1)

In this case, k will be the constant rate of this element. This is calculated using the following expression:

k = ln2/t₁/₂ (2)

Let's calculate the value of k first:

k = ln2/2.7 = 0.2567 d⁻¹

Now, we can use the expression (1) to calculate the remaining mass:

m = 8.1 * e^(-0.2567 * 2.6)

m = 8.1 * e^(-0.6674)

m = 8.1 * 0.51303

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2 years ago

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

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2 years ago

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3 years ago

At 20 degrees celsius, how much sodium chloride could be dissolved in 2L of water

aalyn [17]

The source below says 35.89g of NaCl dissolve in 100 g of water at 20 °C.

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2 years ago

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Ganezh [65]

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2 years ago