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kodGreya [7K]
3 years ago
12

The specific heat of water is 4.184 J/g·K. How much energy must be added to 100.0 g of water to raise the temperature of water f

rom 22.0 ºC to 90.0 ºC?
Chemistry
1 answer:
OLEGan [10]3 years ago
4 0

Answer:

10.9C digrey celsios and im sorry

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PLEASE HELPPPPPPPPPPPPPP​
melamori03 [73]
It changes from a liquid from to a solid.

I hope this helps!
3 0
2 years ago
The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
Karolina [17]

<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 3.45

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

8 0
3 years ago
I need help with this problem !
Sloan [31]

Answer:

3

Explanation:

7 0
3 years ago
Given the following balanced chemical reaction:
Gnom [1K]

Answer:

Explanation:

according to balance chemical equation

  3 A2 moles produced 2 moles of A3B

so 12 moles A2 will produced moles of A3B= 12*2/3=24/3= 8

therefore 12 moles of A2 produced 8 moles of A3B

6 0
3 years ago
24 grams of iron is added to a graduated
Vanyuwa [196]

Answer:

d = 8 g/mL

Explanation:

Given data:

Mass of metal = 24 g

Volume of eater = 45 mL

Volume of water + metal = 48 mL

Density of iron metal = ?

Solution:

Volume of metal:

Volume of metal = volume of water+ metal - volume of water

Volume of metal = 48 mL - 45 mL

Volume of metal = 3 mL

Density of metal:

d = m/v

d = 24 g/ 3 mL

d = 8 g/mL

4 0
3 years ago
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