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3 years ago

Plzzzzzzzzzzz help what is a font!

Chemistry

1 answer:

spayn [35]3 years ago

6 0

Answer:

1. a receptacle in a church for the water used in baptism, typically a freestanding stone structure.

2. A type of writing or text style

Explanation:

There are mutiple definitions of font

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Give the number of significant figures 0.025

grandymaker [24]

There are 2 significant figures.
"Rule 3: Trailing zeros are significant if a decimal point is shown in the number, but may or may not be significant if no decimal point is shown."

7 0

3 years ago

What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?

Alika [10]

Answer:- B: $1.10g H_2$ is the right answer.

Solution:- The balanced equation is:

$O_2(g)+2H_2(g)\rightarrow 2H_2O(g)$

We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.

From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.

So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.

The complete set up looks as:

$8.75g O_2(\frac{1mole}{32g})(\frac{2mole H_2}{1mole O_2})(\frac{2.02g}{1mole})$

= $1.10g H_2$

Hence, the right option is B: $1.10g H_2$ .

7 0

3 years ago

Read 2 more answers

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: