Please help me!! (brainliest)

You’re beautiful *heart*

Georgia [21]

I’m not your just blind. I’m sorry

8 0

3 years ago

The names of monatomic anions are formed by replacing the ending of the name of the element with

ziro4ka [17]

The answer is - ide.

That is how you name the anions in a ionic compound, you replace the ending of the name of the element with ide.

Some examples:

Ionic compound            anion            name of the anion

NaCl                                Cl-                 chloride

KF                                    F-                  fluoride

NaI                                   I-                    iodide

CaBr2                              Br-                 bromide

5 0

4 years ago

How much energy must be added to a 1 kg piece of gold with a specific heat of 130 j/kg to increase its temperature from 20 c to

Lorico [155]

$Q = 10.4 \; \text{kJ}$

The specific heat of a material gives the energy it takes to increase the temperature of one unit mass of the material by one unit temperature. The SI unit for specific heat $c$ is therefore $\text{J} /\text{kg} \cdot \textdegree{\text{C}}$ .

The formula relating the energy required to raise the temperature of $m$ grams of a substance with specific energy $c$ by $\Delta T$ degrees is

$Q= c \cdot m \cdot \Delta T$

The question provides the following information:

$c = 130\; \text{J} /\text{kg} \cdot \textdegree{\text{C}}$ for gold
$m= 1 \; \text{kg}$
$\Delta T = 100 \textdegree{\text{C}} - 20 \textdegree{\text{C}} = 80 \textdegree{\text{C}}$

Apply the formula:

$\begin{array}{ccc} Q & = & c \cdot m \cdot \Delta T\\ & = & 130 \; \text{J} \cdot \text{kg}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1 \; \text{kg} \times 80 \; \textdegree{\text{C}}\\ & = & 1.04 \times 10^{4} \;\text{J}\\ & = & 10.4 \; \text {kJ} \end{array}$

7 0

4 years ago

If u pass a curve of 3A in a time of 90mins. how much mg will be deposited.

Maslowich

Answer:

$\boxed{\text{2.0 g}}$

Explanation:

1. Write the equation for the half-reaction.

Mg²⁺ + 2e⁻ ⟶ Mg

The number of electrons transferred (z) is 2 mol per mole of Mg.

2. Calculate the number of coulombs

Q = It

$\text{t} = \text{90 min} \times \dfrac{\text{60 s}}{\text{1 min}} = \text{5400 s}\\\\Q = \text{3 C/s} \times \text{ 5400 s} = 16 000 C$

3. Mass of Mg

We can summarize Faraday's laws as

$m = \dfrac{QM}{zF}\\\begin{array}{rcl}m & = &\dfrac{16 000 \times 24.30}{2 \times 96 485}\\& = & \textbf{2.0 g}\\\end{array}\\\text{The mass of Mg deposited is }\boxed{\textbf{2.0 g}}$

8 0

3 years ago

Why do scientist sometimes discard theories?

STALIN [3.7K]

Contradictory observations are found

6 0

4 years ago