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Schach [20]
3 years ago
9

Valence electrons (electrons in the outermost shell – available for bonding)

Chemistry
1 answer:
Anastaziya [24]3 years ago
8 0
Electrons and newtrons is that what ur asking for
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The Environmental Protection Agency has determined that safe drinking
dmitriy555 [2]

Answer:

b) \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

c) Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

Explanation:

We assume that part a is test the claim. And we can conduct the following hypothesis test:

Null hypothesis: \mu =7

Alternative hypothesis \mu \neq 7

The statistic is to check this hypothesi is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

We know the following info from the problem:

\bar X = 6.4 , s=0.5, n =30

Replacing we got:

t = \frac{6.4-7}{\frac{0.5}{\sqrt{30}}}= -6.573

And the p value would be:

p_v= 2*P(Z

Since the p value is very low compared to the significance assumed of 0.05 we have enough evidence to reject the null hypothesis that the true mean is equal to 7 moles/liter

Part b

The confidence interval is given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

Part c

Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

6 0
3 years ago
A pool is 60.0 m long and 35.0 m wide. How many mL of water are needed to fill the pool to an average depth of 6.35 ft? Enter yo
ra1l [238]

Answer:

4.07 × 10⁹ mL

Explanation:

Step 1: Given data

Length of the pool (L): 60.0 m

Width of the pool (W): 35.0 m

Depth of water in the pool (D): 6.35 ft

Step 2: Convert "D" to m

We will use the relationship 1 m = 3.28 ft

6.35ft \times \frac{1m}{3.28ft} = 1.94m

Step 3: Calculate the volume of water (V)

We will use the following expression.

V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}

Step 4: Convert "V" to mL

We will use the relationship 1 m³ = 10⁶ mL.

4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL

7 0
3 years ago
I need some really interesting facts about Pumice and Sandstone 
stepladder [879]
Facts about Pumice:
The lightest rock on earth.
Formed in Molten Lava.
Most of the earths crust is made from rocks such as Pumice (Igneous Rocks).
Contains many minerals that help plants to grow.
Facts about Sandstone:
It can form on land or in the sea.
It has been used as far back as prehistoric times to make house ware items and shelter contruction.
They can take thousands of years to form.
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3 0
4 years ago
What energy transfer is occurring when a plant uses energy from the sun to make sugar in fruit? Radiant energy is converted to n
IrinaK [193]

The right answer is Radiant energy is converted to stored chemical energy.

Photosynthesis aims to create energy (in the form of carbohydrate) from the light energy from the sun.

Solar energy is used to oxidize water and reduce carbon dioxide to synthesize organic substances (carbohydrates). This phenomenon occurs in chloroplasts, a plant-specific organelle, at the level of the thylakoid membranes where photosystems I and II and cytochromes are located.

7 0
3 years ago
Read 2 more answers
The half-life of gold-198 is 2.7 days. After
Pie

Answer: 8.1 days

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= \frac{x}{4} 

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{2.7days}=0.257days^{-1}

b) for completion of one fourth of reaction

t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}

t=\frac{2.303}{0.257}\log{4}

t=8.1days

Thus after 8.1 days , one fourth of original amount will remain.

8 0
4 years ago
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