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Fiesta28 [93]
3 years ago
15

Chemical formula for copper I acetate

Chemistry
1 answer:
lions [1.4K]3 years ago
6 0
Cu(CH3COO)2 is your formula
Glad to help
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Angry pomeranian x bones? broken\broccoli boy anyone?
mezya [45]
Makes zero sense. What’s the question?
7 0
3 years ago
A balloon filled with 1.22 L of gas at 286 K is heated until the
TiliK225 [7]

Answer: 670K

Explanation:

Given that,

Original volume of gas V1 = 1.22 L

Original temperature T1 = 286 K

New volume V2 = 2.86 L

New temperature T2 = ?

Since volume and temperature are involved while pressure is constant, apply the formula for Charles law

V1/T1 = V2/T2

1.22 L/286 K = 2.86 L/ T2

Cross multiply

1.22 L x T2 = 286 K x 2.86 L

1.22T2 = 817.96

Divide both sides by 1.22

1.22T2/1.22 = 817.96/1.22

T2 = 670.459 K (Round to the nearest whole number as 670 K)

Thus, the temperature of the gas is 670 Kelvin

4 0
3 years ago
What is the speed of light of an orange light with a wavelength of 600nm and a frequency of 5.00x1014?
vovangra [49]

Answer:

3 × 10^8 m/s

Explanation:

The wavelength, can be calculated by using the following formula;

λ = v/f

Where;

λ = wavelength (m)

v = velocity/speed of light (m/s)

f = frequency (Hz)

According to the provided information in this question, λ = 600nm i.e. 600 × 10^-9m, f = 5.00 x 10^14 Hz

Hence, using λ = v/f

v = λ × f

v = 600 × 10^-9 × 5.00 x 10^14

v = 6 × 10^-7 × 5.00 x 10^14

v = 30 × 10^(-7 + 14)

v = 30 × 10^ (7)

v = 3 × 10^8 m/s

3 0
3 years ago
You'll be given 100 points if you answer this question!!!!!!!!!!!!
storchak [24]

Answer:conducts ,steel, polystyrene, thermal insulation

Explanation:

5 0
3 years ago
Calculate the mass in grams in nine molecules of CH3COOH? Please show how you got your answer.
WINSTONCH [101]
M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>

m 9CH₃COOH: 60u×9 = <u>540u</u>

<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
<u>X = 896,4</u></span><span><u>·10⁻²⁴g


</u>
</span>
7 0
3 years ago
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