What is a siesmograph

Write a word or phrase beginning with each letter of the word gold that described the properties of these gold coins ???? Plzz h

just olya [345]

Glittering in the light
Opaque, not transparent
Light to the hold
D-block element

8 0

4 years ago

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The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

3 years ago

If a gun is found in water, what should investigators do to transport it to the lab?

GREYUIT [131]

Most items of evidence will be collected in paper containers such as packets, envelopes, and bags. Liquid items can be transported in non-breakable, leakproof containers. Arson evidence is usually collected in air-tight, clean metal cans. Only large quantities of dry powder should be collected and stored in plastic bags. Moist or wet evidence (blood, plants, etc.) from a crime scene can be collected in plastic containers at the scene and transported back to an evidence receiving area if the storage time in plastic is two hours or less and this is done to prevent contamination of other evidence. Once in a secure location, wet evidence, whether packaged in plastic or paper, must be removed and allowed to completely air dry. That evidence can then be repackaged in a new, dry paper container. UNDER NO CIRCUMSTANCES SHOULD EVIDENCE CONTAINING MOISTURE BE PACKAGED IN PLASTIC OR PAPER CONTAINERS FOR MORE THAN TWO HOURS. Moisture allows the growth of microorganisms which can destroy or alter evidence.

8 0

4 years ago

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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

Typically, water runs through the baseboard copper tubing and, therefore, fresh hot water is constantly running through the pipi

Ugo [173]

Answer:

-3.617 °C

Explanation:

Step 1: Given data

Mass of water (m): 210.0 g

Energy released in the form of heat (Q): -3178 J (the minus sign corresponds to energy being released)

Specific heat of water (c): 4.184 J/g.°C

Temperature change (ΔT): ?

Step 2: Calculate the temperature change

We will use the following expression.

Q = c × m × ΔT

-3178 J = 4.184 J/g.°C × 210.0 g × ΔT

ΔT = -3.617 °C

6 0

3 years ago