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3 years ago

A defibrillator delivers about 300 joules per shock calculate the number of calories

Chemistry

1 answer:

Tanya [424]3 years ago

4 0

Answer:- 71.7 calories

Solution:- it is a unit conversion problem where we need to convert joules into calories.

The unit conversion for these energy units is:

1 calorie = 4.184 joule

The given energy is 300 joules and has to be converted to calories. We will show the set up using dimensional analysis.

$300joules(\frac{1calorie}{4.184joules})$

= 71.7 calories

So, the energy in calories is 71.7.

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Helium is the lightest noble gas and the second most abundant element (after hydrogen) in the universe. The mass of a helium−4 a

dmitriy555 [2]

Answer: The fraction of atom's mass contributed by nucleus is 0.99

Explanation:

Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.

The isotopic symbol of Helium-4 atom is $_2^4\textrm{He}$

Number of electrons = 2

Number of protons = 2

Number of neutrons = 4 - 2 = 2

We are given:

Mass of He-4 atom = $6.64648\times 10^{-24}g$

Mass of 1 electron = $9.10939\times 10^{-28}g$

Calculating the mass contributed by the nucleus = $m_{He}-2(m_e)$

Mass of the nucleus of He-4 atom = $6.64648\times 10^{-24}-(2\times (9.10939\times 10^{-28}))=(6.64648-0.0018219)\times 10^{-24}=6.64466\times 10^{-24}$

To calculate the fraction of atom's mass contributed by the nucleus, we use the equation:

$\text{Fraction of atom's mass contributed by nucleus}=\frac{\text{Mass of nucleus}}{\text{Mass of atom}}$

Putting values in above equation, we get:

$\text{Fraction of atom's mass contributed by nucleus}=\frac{6.64466\times 10^{-24}g}{6.64648\times 10^{-24}g}=0.99$

Hence, the fraction of atom's mass contributed by nucleus is 0.99

8 0

3 years ago

Which of the following is NOT an example of a molecule A: H2O2 B:NCI3 C:F D:O3

timama [110]

C. F because it's an atom not a molecule

6 0

3 years ago

Read 2 more answers

Answer my my science homework please

ludmilkaskok [199]

1. D

2. 83.5 million years ago - 66 million years ago

3. It had armor and could rotate to keep its caudal bludgeon facing the enemy.

hope this helped :)
alisa202

5 0

3 years ago

Find the volume of a box with length 25 cm, height 25 cm and width 1.0 m. Volume

mariarad [96]

The volume of a box with length 25 cm, height 25 cm and width 1.0 m is 0.0625m³.

Volume of the box which is a cuboid can be calculated by multiplying the length, breadth and height of the given box.

Volume of the box is given by the product of the length of the box, Height of the box and Breadth or width of the box.

Since, the box is a cuboid, hence the formula is given by the products of length, breadth and height.

Given,

length of the box= 25cm = 0.25m

Height of the box =25cm = 0.25m

width of the box= 1m

Volume = length × width × height of box

Volume = 0.25 × 0.25 × 1

Volume = 0.0625m³

The volume of the box is 0.0625m³.

Learn more about Volume here, brainly.com/question/23118276

#SPJ9

4 0

2 years ago

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$