B. HNO3 because it is copper and can not move the place of hydrogen's atom from the acid
Answer:
Either c or d.
Explanation:
Sorry probs not alot of help o.o
Answer:
H₂SO₄
Explanation:
We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.
Step 1: Calculate the total mass of the compound
Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g
Total mass = 23.139 g
Step 2: Determine the percent composition.
H: (0.475g/23.139g) × 100% = 2.05%
S: (7.557g/23.139g) × 100% = 32.66%
O: (15.107g/23.139g) × 100% = 65.29%
Step 3: Divide each percentage by the atomic mass of the element
H: 2.05/1.01 = 2.03
S: 32.66/32.07 = 1.018
O: 65.29/16.00 = 4.081
Step 4: Divide all the numbers by the smallest one
H: 2.03/1.018 ≈ 2
S: 1.018/1.018 = 1
O: 4.081/1.018 ≈ 4
The empirical formula of the compound is H₂SO₄.
Answer:
<h2>0.93 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 37.2 g
volume = 40 mL
We have
We have the final answer as
<h3>0.93 g/mL</h3>
Hope this helps you
Answer:
The hot tea should transfer <em>25.63 kJ</em> the surroundings to cool the tea.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat has to be transferred from the tea to the surroundings to cool the tea (Q = ??? J).
m is the mass of the hot tea (m = dV = (1.0 g/mL)(250 mL) = 250 g), suppose the density of water is the density of tea.
c is the specific heat of the hot tea (c = 4.10 J/°C.g).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 350 K - 375 K = -25°C).
<em>∴ Q = m.c.ΔT</em> = (250 g)(4.10 J/°C.g)(-25°C)) = <em>- 25630 J = - 25.63 kJ.</em>
<em>So, the hot tea should transfer 25.63 kJ the surroundings to cool the tea.</em>
