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alexandr402 [8]
3 years ago
10

Please help! Will give brainiest for best answer! Please Explain how you got there as well. Thanks.

Chemistry
1 answer:
jeka943 years ago
5 0

(ノ☉。☉)ノ (ノ☉。☉)ノ (ノ☉。☉)ノ (ノ☉。☉)ノ

it's in order and the answer is also there, just read it thanks ☜(^o^☜()

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Most enzymes in living things are made up of __________.
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The answer is c, protiens
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2 HClO(aq) + Ca(OH)2(aq) = 2 H2O(l) + Ca(ClO)2(aq) is the balanced equation

4 0
3 years ago
A scientist is studying a chemical reaction. She predicts that after the reaction the product will have less chemical potential
Nady [450]
Hello!

A reaction where the products have less chemical potential energy than the reactants had originally is called an exothermic reaction. Although the scientist didn't observe light or sound, an observation that would support her prediction is the rise in temperature after the reaction has completed. Exothermic reactions release energy in the form of heat, and the temperature would rise in an exothermic reaction.

Have a nice day!
3 0
4 years ago
The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins
saul85 [17]
The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>t_{1/2} = 2.5 min

We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{10 min}{2.5 min}
⇒ n=4<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>(1/2)^{n} = x
</span>⇒ x=(1/2)^4
<span>⇒x= \frac{1}{16}</span>
3 0
3 years ago
An ideal gas has a temperature of 25 oC, a pressure of 1.7 atm, and volume of 345 mL, how many moles of gas are present in the c
Sholpan [36]

Answer: There are 0.024 moles of gas are present in the container.

Explanation:

Given: Temperature = 25^{o}C = (25 + 273) K = 298 K

Pressure = 1.7 atm

Volume = 345 mL (1 mL = 0.001 L) = 0.345 L

Formula used is as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T= temperature

Substitute the values into above formula as follows.

PV = nRT\\1.7 atm \times 0.345 L = n \times 0.0821 L atm/mol K \times 298 K\\n = \frac{1.7 atm \times 0.345 L}{0.0821 L atm/mol K \times 298 K}\\= \frac{0.5865}{24.4658} mol\\= 0.024 mol

Thus, we can conclude that there are 0.024 moles of gas are present in the container.

6 0
3 years ago
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