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lisabon 2012 [21]
2 years ago
11

Is this equation right? ​

Chemistry
2 answers:
m_a_m_a [10]2 years ago
8 0

Answer:

no

Explanation:

because the characters are invalid

loris [4]2 years ago
4 0

Answer:

no

Explanation:

because they are not valid

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When measuring the volume of an object you are finding the?
marta [7]

Answer:

C. Amount of 3 dimensional space the object occupies

Explanation:

Single dimension is a laminar

Amount of 2 dimension is area

Amount of 4 dimension is negligible

.

7 0
3 years ago
What is the mass in milligrams of 4.30 moles of sodium?use âeâ for âÃ10" and use significant figures?
Elis [28]

First, calculate the mass of sodium in g with the help of molar mass and number of moles.

Number of moles = \frac{given mass in g}{molar mass}   (1)

Molar mass of sodium =23 .0 g/mol

Substitute the given value of number of moles and molar mass of sodium in formula (1)

4.30 moles = \frac{given mass in g}{23.0 g/mol}   (1)

4.30 moles\times 23.0 g/mol = mass in g   (1)

mass of sodium in g = 98.9 g

Now, according to conversion factor, 1 g = 1000 mg

So, 98.9 g of sodium = 98.9 \times 1000  mg

= 98900 mg of sodium

Thus, mass of sodium in mg  =9.89 \times 10^{4} mg






8 0
3 years ago
Most seeds germinate and start to grow in the dark. Why don’t they need light right away?
White raven [17]

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HD video is the best way too much for a TV now

8 0
3 years ago
1. TRUE OR FALSE the best way to deal with groundwater pollution is to avoid letting it happen to start with
ivann1987 [24]
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3 0
2 years ago
Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

5 0
2 years ago
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