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a_sh-v [17]
3 years ago
13

A 1.3 g sample of a substance is heated from 0°C to 45°C and is found to have absorbed 45 j of heat. What is the specific heat o

f this substance?
Chemistry
1 answer:
wlad13 [49]3 years ago
5 0

Q=mcat

45=(1.3)(c)(45)

45=58.5c

.769=c

~.77

The answer is .77

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What is the balenced equation of Cu+H2SO4---> CuSO4+H2O+SO2
algol [13]
Balanced chemical equation :

1 Cu + 2 H2SO4 = 1 CuSO4 + 2 H2O + 1 SO<span>2</span>

hope this helps!.

8 0
3 years ago
An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t
topjm [15]

Answer:

\boxed{3}

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )

where

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\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}

Data:

n_{f} = 2

λ = 657 nm

Calculation:  

\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}

7 0
3 years ago
BE FIRST TO ANSWER FOR SUM GOOD!!!!!:D
kipiarov [429]
Answer is B can you like btw??
3 0
3 years ago
Answers fast please Gets oxygen into your body and carbon dioxide out
yKpoI14uk [10]
B respiratory is correct
4 0
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Does sauce either rhyme with avoid, crowd,false,hall,or hawk​
Alenkinab [10]

Answer:

false

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3 years ago
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