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Strike441 [17]
3 years ago
12

Pt5 science...... Select correct answer ♡ !

Chemistry
2 answers:
arlik [135]3 years ago
4 0

Answer: what answer? wahhhhhhhat

Explanation:

Oksi-84 [34.3K]3 years ago
4 0

Answer: the answer is A

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2.524 g (5.1 × 106 g) ÷ (6.85 × 103 g) = ? How many significant figures should the result have?
irina1246 [14]
The correct answer is Three. Solving the given expression we get,

2.524 g (5.1 × 106 g) ÷ (6.85 × 103 g) = 1364.47÷ (6.85 × 103 g) = 1364.47÷ 705.55 = 1.93 
Here, the number of significant figures is three. Thus, the result should have three <span>significant figures.</span>
6 0
3 years ago
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Calculate the mass of 1.23×1024 helium atoms.
just olya [345]

6.023*1023 He atoms=1 mole He

1.23×1024 helium atoms=1.23×1024/6.023*1023 =2.042 mole He

one mole He=4.0 gm

2.042 mole He=2.042*4.0=8.168 gm

answer in grams of Helium is 8.168 gm

7 0
3 years ago
What describes the change in oxidation states of the following reaction? Cu2+ reduces to Cu, and Al oxidizes to AI3+. Cu2+ oxidi
Scilla [17]

The reaction is

Cu⁺²(aq) ---> Cu(s)  [reduction]

The species which undergoes reduction is able to oxidize other one thus will called as "oxidizing agent"

Thus Cu⁺² is an oxidizing agent

or Cu is a reducing agent

Al(s)  ---> Al⁺³(aq)

Aluminium is undergoing oxidation

Thus is able to reduced others and hence is a reducing agent

Al is a reducing agent

or Al⁺³ is an oxidizing agent

Answer: Cu is the reducing agent, and AI3+ is the oxidizing agent.

7 0
3 years ago
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the length of a string is 16cm.Its length becomes 20cm when supporting a weight of 5N.Calculate the length of the string when su
guapka [62]

Answer:

<h2>20.8cm</h2>

Explanation:

According to hooks law "provide the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force F"

F= ke

where F= applied force

           k= spring constant

           e= extension

Given data

length of string l = 16cm

extension  e = 20-16= 4cm

applied force = 5N

we need t o first calculate the spring constant k

apply the formula

F= ke

5=k*4\\k= \frac{5}{4} \\k= 1.25N/cm

we can now calculate the extension of the string when supporting a 6N weight

F= ke\\6=1.25*e\\e= \frac{6}{1.25} \\e= 4.8cm

The length of the string when supporting a 6N weight is

= length+ extension\\= 16+4.8\\= 20.8cm

<em>COMMENT </em><em>:According to the analysis an extra  weight of 1N will cause add 0.8cm to the length of the string</em>

4 0
3 years ago
How many molecules are in 13.5g of sulfur dioxide, SO2?
iVinArrow [24]
There are 24 molecules idk it maybe right

3 0
3 years ago
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