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2 years ago

5

To the nearest gram, what is the mass of of one spoonfull of sugar? g

1 answer:

Vaselesa [24]2 years ago

4 0

Answer:

the mass of one spoonful of sugar to nearest gram is 10g

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What is the smallest unit that makes up matter? A. molecules B. chemicals C. atoms D. alloys

Kamila [148]

Hello there!

An Atom would be considered to be the smallest unit that would make up matter. If were to ever see a picture of an atom, they would consists of a proton, neutron, electron, and a nucleus. And all this would be called a Atom.

Your answer: Atom

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3 years ago

Read 2 more answers

Cci(4) will be soluble in

Romashka-Z-Leto [24]

LIKE DISSOLVES LIKE. Since Ccl4 is non-polar, it'll be soluble in any non-polar solvent. Hope this helps you!

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3 years ago

If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN

viktelen [127]

Answer : The concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is $SCN^-$ and $Fe^{3+}$ is excess reagent.

First we have to calculate the moles of KSCN.

$\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}$

$\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol$

Moles of KSCN = Moles of $K^+$ = Moles of $SCN^-$ = $1.08\times 10^{-5}mol$

Now we have to calculate the concentration of $[Fe(SCN)]^{2+}$

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}$

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

$\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M$

Thus, the concentration of $[Fe(SCN)]^{2+}$ is, $4.32\times 10^{-4}M$

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2 years ago

Match the terms with their correct definition:

geniusboy [140]

<h2><em>1. A</em></h2><h2><em>3. B</em></h2><h2><em>4. C</em></h2><h2><em>7. E</em></h2><h2><em>5. F</em></h2>

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3 years ago

You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p

Elenna [48]

Answer:

<u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

$PV=nRT$

Where:

P = pressure = 1.25 atm
V = volume = 25.2 liter
R = Universal constant of gases = 0.08206 atm-liter/K-mol
T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
n = number of moles

Solving for n:

$n=\frac{PV}{RT}$

Substituting:

$n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol$

8 0

3 years ago