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3 years ago

Most metals are in the liquid state at room temperature.

1 answer:

7 0

That statement is False.

Most metals are in the solid state at room temperature. The only metal that is in liquid state in the room temperature is Mercury, (Hg)

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Describe the motion of objects that are viewed from your reference frame both inside and outside while you travel inside a movin

Nutka1998 [239]

Answer:

2 c

Explanation:

8 0

3 years ago

Predict the products of the reaction below. That is, complete the right-hand side of the chemical equation. Be sure your equatio

Furkat [3]

Answer:

HNO₃ (aq) —> H⁺ (aq) + NO₃¯ (aq)

Explanation:

From the question given above

HNO₃ + H₂O —> ?

Nitric acid, HNO₃ reacts with water, H₂O to form aqueous solution of nitric acid as illustrated below:

HNO₃ + H₂O —> HNO₃ (aq)

Nitric acid is a strong acid and, so will ionised completely when dissolved in water. This is illustrated below:

HNO₃ (aq) —> H⁺ (aq) + NO₃¯ (aq)

6 0

3 years ago

If 200. mL of 0.60 M MgCl2(aq) is added to 400 mL of distilled water, what is the concentration of Mg and Cl in the resulting so

sladkih [1.3K]

Answer:

C. 0.20 M Mg ion & 0.40 M Cl ion

Explanation:

MgCl₂ is a ionic salt which is dissociated as this

MgCl₂ → Mg²⁺ + 2Cl⁻

First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M

Molarity . volume = moles.

0.6 mol/l . 0.2l = 0.12 mol

MgCl₂ → Mg²⁺ + 2Cl⁻

0.12mol 0.12 0.24

This moles are also in 400mL of water, so the new concentration is

[Mg²⁺] = 0.12 m/0.6L = 0.2M

[Cl⁻] = 0.24 m/0.6L = 0.4M

Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)

8 0

2 years ago

Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,

ludmilkaskok [199]

Answer: The empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

Explanation:

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = $\frac{5.80}{1.44}=4.03\approx 4$

For Oxygen = $\frac{1.44}{1.44}=1$

For Nitrogen = $\frac{1.44}{1.44}=1$

For Chlorine = $\frac{1.44}{1.44}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

3 0

3 years ago

A 0.288 g sample of an unknown monoprotic acid is dissolved in water and titrated with a 0.115 M NaOH solution. After the additi

Sholpan [36]

Answer:

74.0 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

At the equivalence point, 33.83 mL of 0.115 M NaOH react.

0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol

Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.

Step 4: Calculate the molar mass of the acid

3.89 × 10⁻³ moles of HA have a mass of 0.288 g.

M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol

4 0

3 years ago