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Angelina_Jolie [31]
4 years ago
5

Cryosurgery uses ________ to remove skin lesions. cold heat chemicals all of these

Chemistry
1 answer:
True [87]4 years ago
5 0
Extreme cold which is usually produced by liquid nitrogen. Hope that helps!
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Please help; chemistry
Diano4ka-milaya [45]

Answer:

204/82 Pb ( /1.37%)

Explanation:

lead

4 0
3 years ago
Hi:)) how to do 5? I’m not sure if it’s hydrogen carbonate or carbonate
zysi [14]

Answer:

copper(ll) carbonate

Explanation:

Since the product is a salt which is copper(II) carbonate, water and carbon dioxide, this reaction is an acid + metal carbonate reaction.

Looking at the salt, Cu²⁺ has to be part of the reactants.

Hence, the missing compound there has to be copper(ll) carbonate, CuCO₃.

The balanced chemical equation would be:

CuCO₃ + 2HNO₃➙ Cu(NO₃)₂ +H₂O +CO₂

P.s. You left out CO₂ as a product in Q2 ;)

Just a recap of the main reactions you would've learnt:

1) Acid + base/ alkali ➙ salt + water

2) Acid + metal ➙ salt + hydrogen gas

3) Acid + metal carbonate ➙ salt + H₂O + CO₂

8 0
4 years ago
2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at 760.0 mmHg
Leviafan [203]

1.7960L

Explanation:

the mass of the gas is constant in both instances

pv/T=constant(according to pv=nRT)

745mmHg*2L/298K=760mmHg*v/273K

v=1.7960L

5 0
3 years ago
HELPP nowww plssss!!!!
yKpoI14uk [10]
The answer is #2: gas
4 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
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