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Vesna [10]
3 years ago
6

In healthcare settings where sterilization is an absolute necessity, what method of sterilization will typically be used?

Chemistry
1 answer:
tester [92]3 years ago
6 0

Answer:

Disinfecting (or boiling)

Explanation:

They'd want to disinfect everything so that it is germless. Boiling is a possible answer as well because you can boil certain things to rid them of germs.

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The unequal sharing of electrons between the oxygen and hydrogen atoms within a water molecule makes water a ____________ molecu
Ivanshal [37]

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Polar

Explanation:

8 0
2 years ago
What type of ions do nonmetals naturally form?
tensa zangetsu [6.8K]
Non-metal atoms gain an electron, or electrons, from another atom to become >negatively charged ions.
8 0
3 years ago
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Calculate the speed of a marble that rolls 9 cm in 4 seconds
cupoosta [38]

Answer:

2.25

Explanation:

9/4

6 0
3 years ago
What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express
Galina-37 [17]
First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

Fe+2 + 2e- --> Fe                  potential= -0.44
Mg+2 + 2e- --> Mg                potential= -2.37


Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.
Download pdf
8 0
3 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
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