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3 years ago

This is the substance(s) formed in a chemical reaction.

2 answers:

7 0

Answer:A chemical reaction is the process in which atoms present in the starting substances rearrange to give new chemical combinations present in the substances formed by the reaction. These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products.

Explanation:

katrin2010 [14]3 years ago

5 0

Answer:

The answer is products

Explanation:

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Help me please) attached the screen below. Thanks

Vitek1552 [10]

Answer:

1) d = 2.4 g/cm³

2) m = 25 g

3) v = 126.7 cm³

Explanation:

Given data:

Mass of material = 24 g

Volume of material = 10 cm³

Density of material = ?

Solution:

Formula:

d = m/v

by putting value,

d = 24 g / 10 cm³

d = 2.4 g/cm³

2) Given data:

Density of material = 5 g/cm³

Volume of material = 5 cm³

Mass of material = ?

Solution:

Formula:

d = m/v

5 g/cm³ = m / 5 cm³

m = 5 g/cm³×5 cm³

m = 25 g

3)Given data:

Density of material = 3 g/cm³

Mass of material = 380 g

Volume of material = ?

Solution:

Formula:

d = m/v

3 g/cm³ = 380 g / v

v = 380 g /3 g/cm³

v = 126.7 cm³

7 0

3 years ago

How much energy is required to raise the temperature of 10.6 grams of gaseous neon from

Alona [7]

Answer:

Approximately $1.95 \times 10^{2}\; \rm J$ .

Explanation:

Look up the specific heat of gaseous neon:

$c = 1.03 \; \rm J \cdot g^{-1} \cdot K^{-1}$ .

Calculate the required temperature change:

$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Let $m$ denote the mass of a sample of specific heat $C$ . Energy required to raise the temperature of this sample by $\Delta T$ :

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

For the neon gas in this question:

$c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}$ .
$m = 10.6\; \rm g$ .
$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

Calculate the energy associated with this temperature change:

$\begin{aligned}Q &= c \cdot m \cdot \Delta T \\ &= 1.03\; \rm J \cdot g^{-1}\cdot K^{-1} \times 10.6\; \rm g \times 17.9\; \rm K \\ &\approx 1.95 \times 10^{2}\; \rm J\end{aligned}$ .