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3 years ago

Where are chemicals found in the home?

Chemistry

2 answers:

Thepotemich [5.8K]3 years ago

8 0

Chemicals are find in C. In every room

vodomira [7]3 years ago

6 0

Answer:

Explanation:

chemicals can be found in every part of our lives

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8. Rewrite each of the following as an "ordinary” decimal

saveliy_v [14]

Answer:

????????

Explanation:

3 0

3 years ago

Identify one trait in a plant that could be changed with selective breeding and explain how the change would be useful? Science.

saul85 [17]

The answers is 67 don’t ask me I don’t know but trying to get free points

3 0

3 years ago

2) Convert 2.65*10^ ^ 25 atoms of Chlorine to moles of CI

Irina18 [472]

Answer:

22 mol

Explanation:

Given data:

Number of atoms of Cl = 2.65×10²⁵ atom

Number of moles of Cl = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

2.65×10²⁵ atom × 1 mol / 6.022 × 10²³ atoms

0.44×10² mol

22 mol

7 0

2 years ago

Which of these expressions are correct variations of the Combined Gas Law?

mafiozo [28]

Answer:

Both

Explanation:

The combined gas law is also known as the general gas law.

From the ideal gas law we assume that n = 1;

So;

PV = nRT

and then;

$\frac{P_{1}V_{1} }{T_{1} }$ = $\frac{P_{2}V_{2} }{T_{2} }$

If we cross multiply;

P₁V₁T₂ = P₂V₂T₁

So;

T₁ = $T_{2} \frac{P_{1}V_{1} }{P_{2} V_{2} }$

Also;

V₂ = $V_{1} \frac{P_{1} T_{2} }{P_{2} T_{1} }$

So from the choices both are correct

3 0

3 years ago

4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60

Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

t = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Half-life of Co-60 = 5.3 years

Input the value :

$\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}$

8 0

2 years ago