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In-s [12.5K]
3 years ago
7

Which three of the following are possible consequences of malnutrition?

Chemistry
1 answer:
Tema [17]3 years ago
4 0
Three possible consequences if malnutrition are stunted growth, loss of vision, and scurvy
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How much mass do 2.0 moles of uranium contain
jasenka [17]
The answer is going to be 476.06.
8 0
3 years ago
How many grams are there in 2.00 moles of copper (ii) oxide
tatuchka [14]

Answer: 159 grams

Explanation:

Copper (ii) oxide has the chemical formula CuO.

Now given that:

Mass of CuO in grams = ? (let unknown value be Z)

Number of moles = 2.00 moles

Molar mass of CuO = ?

For the molar mass of CuO: Atomic mass of Copper = 63.5g ; Oxygen = 16g

= 63.5g + 16g

= 79.5 g/mol

Apply the formula:

Number of molecules = (mass in grams/molar mass)

2.00 moles = (Z / 79.5 g/mol)

Z = 79.5 g/mol x 2.00 moles

Z = 159g

Thus, there are 159 grams in 2.00 moles of copper (ii) oxide

7 0
3 years ago
If the reaction is at dynamic equilibrium at 500 K, which statement applies to the given chemical system?
Alenkinab [10]

The correct option is this: THE CONCENTRATION OF THE PRODUCTS AND THE REACTANTS DO NOT CHANGE.

A reversible chemical reaction is said to be in equilibrium if the rate of forward reaction is equal to the rate of backward reaction. At this stage, the concentrations of the products and the reactants remain constant, that is, there is no net change in the concentration even though the reacting species are moving between the forward and the backward reaction.

8 0
3 years ago
Read 2 more answers
The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J
Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

<em>A = 1,13x10¹⁰</em>

<em></em>

I hope it helps!

6 0
3 years ago
If the half-life of a radioisotope is 10,000 years, the amount remaining after 20,000 years is
Cloud [144]

Answer:

Considering the half-life of 10,000 years, after 20,000 years we will have a fourth of the remaining amount.

Explanation:

The half-time is the time a radioisotope takes to decay and lose half of its mass. Therefore, we can make the following scheme to know the amount remaining after a period of time:

Time_________________ Amount

t=0_____________________x

t=10,000 years____________x/2

t=20,000 years___________x/4

During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.

6 0
3 years ago
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