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3 years ago

Can I please Get Help with this question ‍♂️ ?

Chemistry

1 answer:

Nitella [24]3 years ago

7 0

Answer:

Mg(NO4)2 is 180.3 g/mol

Explanation:

First find the substance formula.

Magnesium Nitrate.

Magnesium is a +2 charge.

Nitrate is a -1 charge.

So to balance the chemical formula,

We need 1 magnesium atom for every nitrate atom.

2(1) + 1(-2) = 0

So the substance formula is Mg(NO4)2.

Now find the molar mass of Mg(NO4)2.

Mg = 24.3 amu

N = 14.0 amu

O = 16.0 amu

They are three nitrogen and twelve oxygen atoms.

So you do this: 24.3 + 14.0(2) + 16.0(8) = 180.3 g/mol

So the molar is mass is 180.3 g/mol.

The final answer is Mg(NO4)2 is 180.3 g/mol

Hope it helped!

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In 1993 the Minnesota Department of Health set a health risk limit for chloroform in groundwater of 60.0 g/L Suppose an analytic

tangare [24]

Answer:

4.74 × 10³ mg

Explanation:

Given data

Health risk limit for chloroform in groundwater: 60.0 g/L

Volume of the sample of groundwater: 79.0 mL = 79.0 × 10⁻³ L

The maximum mass of chloroform that there could be in the sample of groundwater to meet the standards are:

79.0 × 10⁻³ L × 60.0 g/L = 4.74 g

1 gram is equal to 10³ milligrams. Then,

4.74 g × (10³ mg/1 g) = 4.74 × 10³ mg

6 0

3 years ago

The only pair that would result in a recessive phenotype is:

Inessa05 [86]

I believe cc would result in a recessive pair.

4 0

3 years ago

Read 2 more answers

Entalpy of vaporization of water is 41.1k/mol. if the vapor pressure of water at 373k is 101.3 kpa, what is the vapor pressure o

allsm [11]

Answer: The vapor pressure of water at 298 K is 3.565kPa.

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at 298 K = ?

$P_2$ = final pressure at 373 K = 101.3 kPa

$\Delta H_{vap}$ = enthalpy of vaporisation = 41.1 kJ/mol = 41100 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 298 K

$T_2$ = final temperature = 373 K

Now put all the given values in this formula, we get

$\log (\frac{101.3}{P_1})=\frac{41100}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{373K}]$

$\frac{101.3}{P_1}=antilog(1.448)$

$P_1=3.565kPa$

Therefore, the vapor pressure of water at 298 K is 3.565kPa.

3 0

4 years ago

Calculate the density of sulfuric acid if 35.4 ml of the acid has a mass of 65.14 grams. (Please show how you got this answer)

vesna_86 [32]

Answer:

d = 1.85 g/cm³

Explanation:

Given data:

Volume of sulfuric acid = 35.4 mL

Mass of sulfuric acid = 65.14 g

Density of sulfuric acid = ?

Solution:

1 ml = 1cm³

Formula:

d = m/v

d = 65.14 g / 35.2 cm³

d = 1.85 g/cm³

7 0

3 years ago

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