All categories

3 years ago

1.) When blood flow is gone from the extremities for too long, the cells start to die. Depending on the severity of the

Chemistry

1 answer:

valkas [14]3 years ago

5 0

Answer:

its A

Explanation:

because frostbite kills the nerves

You might be interested in

I need the answer fast plzzz!! NO LINKS

Arturiano [62]

Answer:

your answer is 12 hope it's correct answer

4 0

3 years ago

When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200

GaryK [48]

Answer:

$2.023 m^3$ is the total displacement (volume) of the submarine.

Explanation:

Mass of water carried by submarine at 1000 ft depth = m = 2100 kg

The density of seawater at 1000 ft depth = d = $1033 kg/m^3$

Volume of the water displaced = V= ?

Total displacement of the submarine = Volume of the water displaced = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3$

$2.023 m^3$ is the total displacement (volume) of the submarine.

6 0

3 years ago

Which represents the balanced nuclear equation for the beta plus decay of C-11?

slavikrds [6]

Nuclear reaction: ¹¹C → ¹¹B + e⁺(positron) + ve(electron neutrino).
Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.
There are two types of beta decay: beta minus and beta plus. In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino and in beta plus decay, a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.

8 0

3 years ago

Read 2 more answers

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

3 years ago

What is the purpose of attaching the clip/ clamp with graphite of the the pencil to the battery?

Sonja [21]

The pencil attached to the negative terminal of the battery collects hydrogen gas while the one connected to the positive terminal collects oxygen.

4 0

3 years ago