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kumpel [21]
3 years ago
10

Which type of molecule is shown below?

Chemistry
1 answer:
Colt1911 [192]3 years ago
7 0
C. Alkene..............
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This is my first question on Brainly, Can someone explain Molar mass to me? I am not very good with it..
suter [353]
Hey,
A mole is a unit of measure that helps us compare particles of any given substance and its mass. ... The molar mass, also known as molecular weight, is the sum of the total mass in grams of all the atoms that make up a mole of a particular molecule. The unit used to measure is grams per mole.
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3 years ago
Sarah is making a strawberry banana smoothie. She combines one banana, five strawberries, and a cup and a half of milk in a
NeTakaya

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homogeneous substance

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A solution was prepared by mixing 50.0 g
frez [133]

Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em></em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

4 0
4 years ago
Write down your observations of what the pieces of cabbage in the images look like
N76 [4]
They all look like there decaying and was left out or sat inside too long
5 0
2 years ago
Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.
TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

HCl+KOH\rightarrow KCl+H_2O

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

n_{KOH}=0.00768mol-0.00192mol=0.00576mol

And the resulting concentration of KOH and OH ions as this is a strong base:

[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M

And the resulting pH is:

pH=14+log(0.131)\\\\pH=13.1

Regards!

3 0
3 years ago
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