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3 years ago

What will the change in temperature be when 90 J are applied to 15 g of gold. (Cgold = 0.126 J/gºC)

Chemistry

2 answers:

luda_lava [24]3 years ago

8 0

Answer:

48 °C

Explanation:

q = 90J

m = 15 g

C = 0.126 J/gºC

q = mct

90 = (15)(0.126)t

t = 47.62 °C = 48 °C

Vikki [24]3 years ago

7 0

Answer:

Explanation:

Formula

E = m * c * Δt

Δt is what you seek.

Givens

E = 90 joules

m = 15 grams

c = 0.126

Solution

90 = 15 * 0.126 * Δt Divide by 15

90/15 = 15/15 * 0.126 * Δt

6 = 0.126 * Δt Divide by 0.126

6/0.126 = Δt

47.619 = Δt

Δt = 48 oC

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How many kilograms of NH3 are needed to produce 1.90x10^5 of (NH4)2SO4? Answer in scientific notation

leva [86]

Molar mass :

NH₃ = 17.0 g/mol

(NH₄)₂SO₄ = 132 g/mol

2 NH₃ + H₂SO₄ = (NH₄)₂SO₄

2 x 17 g NH₃ ------------ 132 g (NH₄)₂SO₄
( mass NH₃) ------------ 1.90 x 10⁵ (NH₄)₂SO₄

( mass NH₃) = (1.90 x 10⁵) x 2 x 17 / 132

( mass NH₃) = 6460000 / 132

( mass NH₃) = 48939.39 g

in kg :

48939.39 / 1000 = 48.93 kg = 4.893 x 10¹ kg

hope this helps!

6 0

3 years ago

Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +

nikklg [1K]

Answer:

For 1: The value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2: The value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

Explanation:

For the given chemical equation:

$H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)$

For 1:

To calculate the $\Delta G$ for given value of equilibrium constant, we use the relation:

$\Delta G=-RT\ln K_p$ .....(1)

where,

$\Delta G$ = ? kJ/mol

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

For 2:

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}$

We are given:

$p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm$

Putting values in above equation, we get:

$K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52$

Now, calculating the value of $\Delta G$ by using equation 1:

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

3 0

3 years ago

Read 2 more answers

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NemiM [27]

Answer:

Explanation:

Closer to equator means temp high more humidity and more hot

Further form equator cold temp less sun more cold

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3 years ago

Bryan is doing a science experiment on the stomata of plants. Bryan has a tomato plant. He uses wax to cover all the stomata on

skad [1K]

Answer: The answer is B

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2 years ago

Which product is used on the natural nail prior to product application to assist in adhesion and serves to chemically bond the e

ira [324]

The product that is used on the natural nail prior to application to assist in adhesion and serves to chemically bond the enhancement product to the natural nail is known as nail primer.

<h3>What is a nail primer?</h3>

A nail primer is a chemical agent used in esthetic centers before applying a colored polish to the nails and serves as an adhesive product.

The nail primers are also very useful for improving the cleaning efficiency of the product before its application.

Nail care products include different types of chemical formulations such as, for example, creams that reinvigorate the cuticle.

In conclusion, the chemical formulation employed on the natural nail that is capable of enhancing and also assisting adhesion is called the nail primer.

Learn more about nail esthetic products here:

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2 years ago