Please help me I’ll mark you the brainiest! Please do NOT answer if you don’t know it.

What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? (5 points)

konstantin123 [22]

C3 H4 N1 ~~that’s what I think

8 0

3 years ago

An intravenous saline drip has 9.8 g of sodium chloride per liter of water. by definition, 1 ml = 1 cm3.

ELEN [110]

Missing question: Express the salt concentration in kg/m³.
Answer is: the salt concentration is 9.8 kg/m³.
m(NaCl) = 9.8 g ÷ 1000 g/kg.
m(NaCl) = 0.0098 kg.
V(solution) = 1 L = 1 dm³.
V(solution) = 1 dm³ ÷ 1000 dm³/m³.
V(solution) = 0.001 m³.
d(solution) = m(NaCl) ÷ V(solution).
d(solution) = 0.0098 kg ÷ 0.001 m³.
d(solution) = 9.8 kg/m³.

4 0

3 years ago

Seawater has a ph of 8.1. what is the concentration of oh–?

lara31 [8.8K]

PH scale is used to determine how acidic or basic a solution is.
pH can be calculated as follows;
by knowing the ph we can calculate pOH
pH + pOH = 14
pOH = 14 - 8.1
pOH = 5.9
pOH is used to calculate the hydroxide ion concentration
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 1.26 x 10⁻⁶ M
therefore hydroxide ion concentration is 1.26 x 10⁻⁶ M

4 0

3 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.