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3 years ago

Do you think it matters if you

Chemistry

1 answer:

vovikov84 [41]3 years ago

5 0

Answer:

yes I do think they mean the same thing

Explanation:

150 grams and 150 grams is the same thing and adding 0 to the end of a decimal does not change its value, you could even put 150.0000000 grams and it would still be equivalent to the other numbers

You might be interested in

What is the atomic number for an element whose mass number is 78,

Ann [662]

Answer:

protons = atomic numbermass number = protons + neutronsatomic number = mass number - neutrons = 78-38 = 40

Explanation:

5 0

2 years ago

Read 2 more answers

Mass has to do with how many particles

Ganezh [65]

The law of definite proportions agrees with Dalton atomic theory.

What is Dalton atomic theory?

It state that all matters is made of very tiny particles called atom. atoms are individual particles which can not be created or be destroyed in a chemical reactions. Atoms of given elements are identical in mass and chemical properties. Atoms of

different elements have different masses and chemical properties.

The law of definite proportions also known as proust's law ,state that a chemical compound contain the same proportion of elements by mass.this law is one of the stoichiometry .

Thus ,

This is the reason why it is agrees with dalton atomic theory.

To know more about Dalton atomic theory click-

brainly.com/question/13157325

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6 0

1 year ago

Show

SIZIF [17.4K]

Answer: 0.9375 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $HCl$ solution = 0.75 M

Volume of $HCl$ solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles$

$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)$

According to stoichiometry :

2 moles of $HCl$ require = 1 mole of $CaCO_3$

Thus 0.01875 moles of $HCl$ will require= $\frac{1}{2}\times 0.01875=0.009375moles$ of $CaCO_3$

Mass of $CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g$

Thus 0.9375 g of $CaCO_3$ is required to react with 25.0 ml of 0.75 M HCl

6 0

3 years ago

I need an answer now please asap and you will me marked brainiest please it's missing I need it now.

BaLLatris [955]

I really hope this helps I’ve been there too and I’m just doing this so I get the 20 words and I have the picture

8 0

3 years ago

You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt

Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the pKa closer to the desired pH value. When we find the pKa values we will obtain:

$pKa_1=-Log[6.9x10^-^3]=2.16$

$pKa_2=-Log[6.2x10^-^8]=7.20$

$pKa_3=-Log[4.8x10^-^13]=12.31$

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which $H_2PO_4^-^1$ is the acid and $HPO_4^-^2$ is the base. Therefore the answer for the first question is B and the answer for the second question is C.

8 0

3 years ago