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Masteriza [31]
3 years ago
10

Need help looking for the rest of the words

Chemistry
2 answers:
Elan Coil [88]3 years ago
8 0
What words are you looking for so i dont put them down 
photoshop1234 [79]3 years ago
4 0
Mechanicalwave is 8 from top going right to left...
Transversewave is 6 from the bottom going from right to left...
Energy is right above transversewave going left to right....
Crest is 8 from top, 6 < going diagonally up left...
Longitudinalwave is 4 from bottom going diagonally up left...
Vibratons is 8 from bottom 3 > going up
Trough is 3 rows v going right to left

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Which of the following statements is false?
hodyreva [135]
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5 0
3 years ago
What are the products of the combustion of a hydrocarbon?
Oxana [17]

Answer:

I think carbon and hydrogen

7 0
3 years ago
Read 2 more answers
Cobalt-60 is a strong gamma emitter that has a half- life of 5.26 yr. The co balt-60 in a radiotherapy unit must be replaced whe
Alenkasestr [34]

<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5.26 years

Putting values in above equation, we get:

k=\frac{0.693}{5.26yrs}=0.132yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.132yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 75) = 25 grams

Putting values in above equation, we get:

0.132=\frac{2.303}{t}\log\frac{100}{25}\\\\t=10.5yrs

The original sample was purchased in June 2016

As, June is the 6th month of the year, which means the time period will be 2016+\frac{6}{12}=2016.5

Adding the time in the original time period, we get:

2016.5+10.5=2027

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027

3 0
3 years ago
What is the freezing point, in °C, of a solution made with 1.31 mol of CHCl₃ in 530.0 g of CCl₄?
lutik1710 [3]

Answer: 73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.A

Explanation: brainliest?

7 0
3 years ago
When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of
taurus [48]

Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where  ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C  and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c  or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

3 0
3 years ago
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