Help pleaseeeeee dont waste my points

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

A chemist prepares a solution by adding 258 mg of K2Cr2O7 (MW = 294.19 g/mol ) to a volumetric flask, and then adding water unti

Maksim231197 [3]

Answer: Molarity of the prepared solution is $1.75\times 10^{-3}mole/L$

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution in ml = 500 ml

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=\frac{0.258g}{294.19g/mol}=8.77\times 10^{-4}mol$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{8.77\times 10^{-4}moles\times 1000}{500ml}=1.75\times 10^{-3}mole/L$

Thus molarity of the prepared solution is $1.75\times 10^{-3}mole/L$

3 0

3 years ago

A 996.9 g sample of ethanol undergoes a temperature change of -70.98 °C while releasing 62.9

Volgvan

Answer:

$c=3.71\ J/g^{\circ} C$

Explanation:

Given that,

Mass of sample, m = 996.9 g

The change in temperature of the sample, $\Delta T=-70.98^{\circ}C$

Heat produced, Q = 62.9 calories = 263173.6 J

The heat released by a sample due to change in temperature is given by :

$Q=mc\Delta T$

Where

c is the specific heat capacity

So,

$c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{263173.6}{996.9\times 70.98}\\\\c=3.71\ J/g^{\circ} C$

So, the specific heat of ethanol is equal to $3.71\ J/g^{\circ} C$ .

3 0

4 years ago

Calcular la normalidad de 1 Kg de sulfuro de aluminio en 5000 ml de solucion.

Troyanec [42]

Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.

Normality comes out to be 8.11

<h3> Given </h3>

Mass of solute: 1000g
Volume of solution (V): 5000 ml = 5 liters
Equivalent mass of solute (E) = molar mass / n-factor

n-factor for $Al_{2}S_{3}$ is 6 and molar mass is 148g

So, on calculating equivalent mass is equal to 24.66g

FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)

N= $\frac{1000}{24.66*5}$

Therefore, normality of 1 kg aluminum sulfide is 8.11

Learn more about normality here brainly.com/question/25507216

#SPJ10

7 0

2 years ago

What element is NaCl?

ale4655 [162]

Sorry but NaCl is a ionic compound not an element though it's made up of two elements sodium and chlorine

4 0

3 years ago

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