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3 years ago

Explain how natural resources are identified and why natural recourses are in evenly distributed.

Chemistry

1 answer:

mixer [17]3 years ago

3 0

Answer:

Natural resources are not evenly distributed all over the world. Some places are more endowed that others — for instance, some regions have lots of water (and access to ocean and seas). Others have lots of minerals and forestlands. Others have metallic rocks, wildlife, fossil fuels and so on.

Explanation:

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Priscilla was building a circuit that used copper wires to connect a battery to a light bulb. As she connected the final wire fr

algol [13]

Answer:

The voltage or potential difference

Explanation:

What makes current flow in a circuit is the voltage or the potential difference.

This force is supplied by the battery or the mains electrical circuit.

Every circuit requires the voltage to drive current through
When a circuit is complete, the battery is able to overcome any resistance by the generating enough voltage which is the force to drive the current through.

7 0

3 years ago

An enzyme is discovered that catalyzes the chemical reaction:SAD -------->HAPPY

jekas [21]

Answer: Km = 10μM

Explanation: Michaelis-Menten constant (Km) measures the affinity a enzyme has to its substrate, so it can be known how well an enzyme is suited to the substrate being used. To determine Km another value associated to an eznyme is important: Turnover number (Kcat), which is the number of time an enzyme site converts substrate into product per unit time.

Enzyme veolcity is calculated as:

$V_{0} = \frac{E_{t}.K_{cat}.[substrate]}{K_{m}+[substrate]}$

where Et is concentration of enzyme catalitic sites and has to have the same unit as velocity of enzyme, so Et = 20nM = 0.02μM;

To calculate Km:

$V_{0}*K_{m} + V_{0}*[substrate] = E_{t}.K_{cat}.[substrate]$

$K_{m} = \frac{E_{t}.K_{cat}.[substrate]-V_{0}*[substrate]}{V_{0}}$

$K_{m} = \frac{0.02*600*40-9.6*40}{9.6}$

Km = 10μM

The Michaelis-Menten for the substrate SAD is 10μM.

3 0

3 years ago

Calculate the mass of water produced when 1.36 g of butane reacts with excess oxygen.

ANTONII [103]

Answer:

Mass of water produced= 1.8 g

Explanation:

Given data:

Mass of water produced = ?

Mass of butane = 1.36 g

Mass of oxygen = excess

Solution:

Chemical equation:

2C₄H₁₀ +13 O₂ → 8CO₂ + 10H₂O

Number of moles of butane:

Number of moles = mass/molar mass

Number of moles = 1.36 g/ 58.12 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of butane with water.

C₄H₁₀ : H₂O

2 : 10

0.02 : 10/2×0.02 = 0.1 mol

Mass of water produced:

Mass = molar mass × molar mass

Mass = 0.1 mol × 18 g/mol

Mass = 1.8 g

6 0

3 years ago

what electronic transition in a hydrogen atom starting from n=7 energy level will produce infrared radiation with energy of 55.1

Lostsunrise [7]

Answer:

(n = 7) ⟶ (n = 4)

Explanation:

1. Convert the energy to joules per mole of electrons.

E = 55.1 × 1000 = 55 100 J/mol

2. Convert the energy to joules per electron

E = 55 100/(6.022 × 10²³)

E = 9.150 × 10⁻²⁰ J/electron

3. Use the Rydberg equation to calculate the transition

Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy. The formula then becomes

$\Delta E = R_{\text{H}} (\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ n_{i}^{2}})$

where

$R_{\text{H}}$ = the Rydberg constant = 2.178 × 10⁻¹⁸ J

$n_{i}$ and $n_{f}$ are the initial and final energy levels.

$9.150 \times 10^{-20} = 2.178 \times 10^{-18}(\frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 7^{2}})$

$\frac{9.150 \times 10^{-20} }{2.178 \times 10^{-18}} = \frac{ 1}{ n_{f}^{2}} - \frac{ 1}{ 49}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01} + \frac{1 }{49 }$

$\frac{ 1}{ n_{f}^{2}} = \text{0.042 01 + 0.020 41}$

$\frac{ 1}{ n_{f}^{2}} = \text{0.062 42}$

$n_{f}^{2} = \frac{1 }{ \text{0.062 42}}$

$n_{f}^{2} = 16.02$

$n_{f} = \sqrt{16.02}$

$n_{f} = 4.003 \approx 4$

4 0

3 years ago

PLS PLS PLS HELP I WILL MARK YOU BRAINLYEST PLS

Aleksandr [31]

Answer:

A: Endoplasmic reticulum

B: Chroloplast

C: Mitochondrion

D: Cytoplasm

E: Cell wall

Explanation:

8 0

3 years ago