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tankabanditka [31]
3 years ago
9

Petals are surrounded by

Chemistry
1 answer:
Talja [164]3 years ago
8 0
Petals surround the main part of the flower           the bud
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How do I find x mm? This is for my chemistry class.
Evgesh-ka [11]
Im guessing it might be 98.4x0.58, when you rearrange the pressure formula.
7 0
3 years ago
The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb
Nimfa-mama [501]

Answer:

\boxed{\text{254 g}}

Explanation:

\begin{array}{rcl}\text{\% m/V} & = & \dfrac{\text{Mass of sucrose}}{\text{Volume of solution}}\\\\\text{Let m}& = &\text{mass of sucrose}\\\dfrac{\text{35.0 g}}{\text{100 mL}}& = & \dfrac{m}{\text{725 mL}}\\\\m & = &\dfrac{\text{35.0 g}\times 725}{100}\\\\ & = &\textbf{254 g}\\\end{array}\\\text{You need $\boxed{\textbf{254 g}}$ of sucrose}

3 0
3 years ago
I am a metalloid. I have 7 valence electrons. I have 6 energy levels.
Alexxx [7]
1 electron Bc of the electron you have
3 0
3 years ago
He reaction produces 1.6 g of gas in 30 seconds.Calculate the mean rate of the reaction in the first 30 seconds.Use the equation
devlian [24]

Answer:

Mean rate of reaction produced = 0.533 g/sec (approx.)

Explanation:

Given:

Reaction produced = 1.6 gram

Time taken = 30 sec

Find:

Mean rate of reaction produced

Computation:

Mean rate of reaction produced = Reaction produced / Time taken

Mean rate of reaction produced = 1.6 / 30

Mean rate of reaction produced = 0.533 g/sec (approx.)

5 0
2 years ago
If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h
frutty [35]

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

3 0
3 years ago
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