Answer:
The answer to your question is: letter B
Explanation:
Data
3.58 x 10⁴ cal to joules
1 calories --------------- 4.184 joules
3.58 x 10⁴ cal----------- x
x = ( 3.58 x 10⁴ x 4.184) / 1
x = 149787.2 joules
x = 1.5 x 10⁵ J
112.2 milliliters volume of water will drip into the bucket in 44 minutes.
<h3>What is volume?</h3>
How much space an object or substance takes up. • Measured in cubic meters (m3), liters (L) & milliliters (mL).
Total drop = Drops per minute X time
= 51 drops per minute X 44 minutes
=2244 drop
Volume of water will drip into the bucket in 44 minutes
=Total drop X Volume of each drop
=0.050 milliliters X 2244 drop
=112.2 milliliters
Hence, 112.2 milliliters volume of water will drip into the bucket in 44 minutes.
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Answer:
B6Br4
Explanation:
Chemical formula gives a representation of the number of atoms that each element in a compound or molecule. In this question which is asking to give the formula of hexaboron tetrabromide, the hexa means 6 while the tetra means four.
Since the number of atoms of each element is written as a prefix before the main element, this means that the chemical formula of hexaboron tetrabromide will be B6Br4.
Answer:
The biggest risk with recharging alkaline batteries is leakage. As you probably know, alkaline batteries leak even under normal circumstances. Internal off gassing, made worse by heat, creates pressure that can breach battery seals. Therefore, the risk of leakage is an even bigger risk when recharging.
Answer:
HI.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
Rate of effusion ∝ 1/√molar mass.
- <em>(Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).</em>
- An unknown gas effuses at one half the speed of that of oxygen.
∵ Rate of effusion of unknown gas = 1/2 (Rate of effusion of O₂)
∴ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = 2.
Molar mass of O₂ = 32.0 g/mol.
∵ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).
∴ 2.0 = (√molar mass of unknown gas) / √32.0.
(
√molar mass of unknown gas) = 2.0 x √32.0
By squaring the both sides:
∴ molar mass of unknown gas = (2.0 x √32.0)² = 128 g/mol.
∴ The molar mass of sulfur dioxide = 80.91 g/mol and the molar mass of HI = 127.911 g/mol.
<em>So, the unknown gas is HI.</em>
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