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2 years ago

Which measurement contains three significant figures?

2 answers:

8 0

The answer is C because you don’t have to worry about the number in front of the decimal unless it is something greater than zero. A- Doesn’t have there sig figs. B- The zero after the 5 doesn’t change it. D- The three at the end doesn’t matter because it does not round it up.

nadezda [96]2 years ago

3 0

A i think a is right 0.05

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What is the buffer component ratio, (bro-)/(hbro) of a bromate buffer that has a ph of 8.08. ka of hbro is 2.3 x 10-9?

Alla [95]

pKa= -log(ka)

= -log(2.3*10^-9)

= 8.64

Now pH can be calculated using equation:

pH=pka+log(BrO-)/(HBrO)

8.08 =8.64+log(BrO-)/(HBrO)

log(BrO-)/(HBrO)=8.08-8.64

= -0.56

(BrO-)/(HBrO)= 10^-0.56

=0.275

3 0

3 years ago

The following reaction was carried out in a 3.00 L reaction vessel at 1100 K:

Nady [450]

Answer:

$Q=0.840$

Explanation:

Hello,

In this case, since the given undergoing chemical reaction is correctly balanced, the reaction quotient is computed as well as the equilibrium constant but in terms of the given concentrations that are:

$C_{C}=\frac{5.25mol}{3.00L} =1.75M\\C_{H_2O}=\frac{12.2mol}{3.00L} =4.07M\\C_{CO}=\frac{3.90mol}{3.00L}=1.30M\\C_{H_2}=\frac{7.90mol}{3.00L} =2.63M$

In such a way, the reaction quotient turns out:

$Q=\frac{C_{CO}C_{H_2}}{C_{H_2O}}=\frac{1.30M*2.63M}{4.07M}\\ Q=0.840$

Taking into account that carbon is not included since it is solid.

Best regards.

7 0

2 years ago

A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve

mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

PV = nRT

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

= $\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}$

= 0.0052 mol

Also, No. of moles = $\frac{mass}{\text{molar mass}}$

0.0052 mol = $\frac{mass}{2 g/mol}$

mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at $25^{o}C$

$P_{\text{water vapor}}$ = 24 mm Hg

= $\frac{24}{760}$ atm

= 0.03158 atm

Now, P = $\frac{756}{760} - 0.03158$

= 0.963 atm

Hence, n = $\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}$

= 0.0056 mol

So, mass of $H_{2}$ = 0.0056 mol × 2

= 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

= $\frac{0.01013 g}{0.0104 g} \times 100$

= 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

3 0

3 years ago

Why is zinc not extracted from ZnO through reduction using CO?

Jet001 [13]

The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO2 from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO

6 0

1 year ago

Which planet is the most massive? A.Jupiter B.Mercury C.Venus D.Neptune

Ainat [17]

Jupiter is the biggest planet in are solar system

7 0

3 years ago

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