All categories

2 years ago

If a body does not have enough potassium, how might that affect neuronal firing?

1 answer:

8 0

Potassium is fundamental for neuron activity. If a body does not have enough potassium, then the neuron will struggle to fire because there will not be enough positively charged ions to trigger the firing of the neuron.

The potassium (K+) is maintained at a high level within the neuron at rest, whereas sodium (Na+) is maintained at a high level outside of the cell.

Neurons contain K+ and Na+ leakage channels that allow the two ions to diffuse in favor of a concentration gradient.

Depolarization opens both K+ and Na+ channels in the membrane, thereby these ions can flow out and into the axon, respectively.

Learn more in:

brainly.com/question/20293974?referrer=searchResults

You might be interested in

G the hydrogen generated in this lab was a product of the reaction between magnesium and hydrochloric acid. which of these react

Colt1911 [192]

The limiting reactant is determined by the supply and stoichiometric equation.

<h3>Limiting reactants</h3>

They are reactants that are limited in availability, and thus, determine how far reactions can go in terms of producing products.

In a reaction involving magnesium and hydrochloric acid to produce hydrogen gas as follows:

$Mg + 2HCl --- > MgCl_2 + H_2$

The number of moles of HCl is twice that of Mg. Thus, if both reactants are supplied in the required proportion, no reactant will be limiting.

However, if the number of moles of HCl supplied is not up to twice that of Mg, HCl will become limiting. Also, if the amount of Mg supplied is not equal to half of that of HCl supplied, Mg will be limiting.

More on limiting reactants can be found here: brainly.com/question/14225536

#SPJ1

3 0

2 years ago

The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g

Yuki888 [10]

<u>Answer:</u> The volume of CO formed is 254.43 L.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Ni(CO)_4$ = 444 g

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol$

For the given chemical reaction:

$Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)$

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = $\frac{4}{1}\times 2.60=10.4mol$ of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of the gas = $22^oC=(273+22)K=295K$

Putting values in above equation, we get:

$752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L$

Hence, the volume of CO formed is 254.43 L.

5 0

3 years ago

You are examining decomposition of corn leaves following the growing season on a farm. To do this, you place 33 g of corn leaves

m_a_m_a [10]

Answer:

c. 9.94 g

Explanation:

From the question,

Using

mt = m₀e⁻kt.................... Equation 1

Where mt = mass of the leaf remaining in the bag, m₀ = original mass of leave that was placed in the bag, k = decay constant, t = time.

Given: m₀ = 33 g, k = 0.04, t = 30 days.

Substitute into equation 1

mt = 33(e⁻(0.04ˣ30))

mt = 33e⁻¹²/¹⁰

mt = 33/e¹²/¹⁰

mt = 33/3.320

mt = 9.94 g.

Hence the right answer is c. 9.94 g

8 0

4 years ago

Which statement best explains why atoms form chemical bonds with other atoms?

dedylja [7]

Sodium loses an electron and chlorine gains an electron. Simply put. It's a little more complicated, but that's what i think they're looking for.

6 0

4 years ago

How many grams of carbon are present in 45.0 g of CCl4?

lisabon 2012 [21]

To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:

45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present

Hope this answers the question. Have a nice day.

5 0

3 years ago