Ask question

Ask question

All categories

3 years ago

10

HELP ASAPPP!!!!’........

1 answer:

a_sh-v [17]3 years ago

3 0

Answer:

naoh and h2

Explanation:

hope this helps! brainliest please? if not its ok! stay safe and have a good day love <3

You might be interested in

How many elements are there in CH3 CH2COONa

poizon [28]

Answer:

There are

Explanation:

There are 10 elements in CH3CH2COONa.

4 0

3 years ago

What can you infer about the volume of a gas as absolute zero is approached?

schepotkina [342]

Answer:

The volume of a gas approaches zero as the temperature approaches absolute zero.

Step-by-step explanation:

You may have done a <em>Charles' Law experiment</em> in the lab, in which you measured the volumes of a gas at various temperatures.

You plotted them on a graph, and perhaps you were asked to extrapolate the graph to lower temperatures.

Your graph probably looked something like the one below.

There is clearly an x-intercept at some low temperature.

Inference: The volume of a gas approaches zero as the temperature approaches absolute zero.

4 0

3 years ago

HELP Vertical columns in the periodotic table are known as

Alex Ar [27]

Both b and c for the vertical clomns

4 0

4 years ago

Read 2 more answers

Lanthanides and actinides are kept separately in the periodic table. why?

liberstina [14]

The lanthanides and actinides are separated from the rest of the periodic table, usually appearing as separate rows at the bottom. The reason for this placement has to do with the electron configurations of these elements.

6 0

3 years ago

During a routine check of the fluoride content of Gotham City\'s water supply, the following results were obtained from replicat

Aleksandr-060686 [28]

Answer:

The mean is $x=0.503\frac{mg}{L}$

The 90% confidence interval is:

$i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}]$

Explanation:

1. First organize the data:

$x_{1}=0.487$

$x_{2}=0.487$

$x_{3}=0.511$

$x_{4}=0.511$

$x_{5}=0.519$

As there are 5 data, the sample size (n) is n=5

2. Calculate the mean x:

The mean is calculated adding up all the data and divide them between the sample size.

$x=\frac{0.511+0.487+0.511+0.487+0.519}{5}$

$x=0.503\frac{mg}{L}$

3. Find 90% confidence interval.

The formula to find the confidence interval is:

$i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})]$ (Eq.1)

where x is the mean, d is the standard deviation and n is the sample size.

And

$1-\alpha=0.90$

$\alpha=0.10$

$\frac{\alpha}{2}=0.05$

$z_{0.05}=1.645$

4. Find the standard deviation

$d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}$

$d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}$

$d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}$

$d=\sqrt{2.24*10^{-4}}$

$d=0.015$

5. Replace values in (Eq.1):

$i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})]$

For the addition:

$i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.514$

For the subtraction:

$i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.492$

The 90% confidence interval is:

$i_{0.90}=[0.492,0.514]$

4 0

3 years ago