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3 years ago

What is the sum of 2.7 g and 2.47g expressed in the correct number of significant digits

Chemistry

2 answers:

Anastasy [175]3 years ago

6 0

Answer : The answer will be, 5.2 g

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule apply for the addition and subtraction is :

The least precise number present after the decimal point determines the number of significant figures in the answer.

As we are given that the sum of 2.7 g and 2.47 g.

$2.7g+2.47g$

$\Rightarrow 5.17g$

In the given expression, 2.7 has 2 significant figures and 2.47 has 2 significant figures. From this we conclude that least precise number present after the decimal point is 1.

Thus, the answer will be $\Rightarrow 5.2g$

NeTakaya3 years ago

5 0

In order to get the sum of 2.7 and 2.47, we will add both numbers. So, 2.7 plus 2.47 would be 5.17. And the correct number of significant digits in this number is still the same, 5.17, so we have three significant numbers. Why? There are three rules in identifying the significant figures: 1. Non zeros are always significant (which this applies in our sum above). 2. Any zeros in between significant numbers are always significant. 3. The final zero or zeros in the decimal portion only are significant.

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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago

What are the 4 properties of hydrogen bonds?

tiny-mole [99]

Answer:

Solubility, Volatility, Viscosity and Surface Tension.

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2 years ago

Choose the FALSE statement.

Mariana [72]

Answer:

C. The half-life of C-14 is about 40,000 years.

Explanation:

The only false statement from the options is that the half-life of C-14 is 40,000yrs.

The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.

All living organisms contain both stable C-12 and the unstable isotope of C-14
The lower the C-14 compared to the C-12 ratio in an organism, the older it is.

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3 years ago

¿El agua salada es una mezcla homogénea o heterogénea?

Sedbober [7]

The question is in another language, so the English translation of the question is as follows:

Is salt-water a hom.ogeneous or heterogeneous mixture?

Answer:

Hom.ogenous mixture

Explanation:

There are two types of mixtures hom.ogenous and hetergenous. a heterogeneous mixture has two or more visible phases while a hom.ogeneous mixture is composed of a single visible phase.

The salt-water is hom.ogeneous because the solve dissolve evenly in throughout the entire salt-water sample and gives visible phase.

Hence, the correct option is "hom.ogenous mixture".

8 0

3 years ago

What fraction of a Sr-90 sample remains unchanged after 87.3 years

jolli1 [7]

The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
 $t_{1/2}$ - half-life
t - total time elapsed
n - a number of half-lives

The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
 $t_{1/2}$ = 28.8 years

We need:
n = ?
x = ?

We could first use the second equation, to calculate n:
If:
$t_{1/2} = \frac{t}{n}$ ,
Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{87.3 years}{28.8 years}$
⇒ $n=3.03$
⇒ n ≈ 3

Now we can use the first equation to calculate the remained amount of the sample.
 $(1/2)^{n} = x$
⇒ $x=(1/2)^3$
⇒ $x= \frac{1}{8}$

8 0

3 years ago