Answer:
21 g/mL
Explanation:
To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.
Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:

Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:



<em>k ≈ 9,56x10³ s⁻¹</em>
I hope it helps!
Purified Ara h 6 will be helpful for research into the immunological mechanisms underlying peanut allergy, including molecular and cellular studies, diagnostic IgE antibody testing, and clinical trials.
1.4% of children and 0.6% of adults in the United States suffer from the food allergy peanut. Approximately 1.8% of youngsters in the UK are allergic to peanuts. Food allergies to milk, eggs, and wheat are often outgrown by children, while allergies to peanuts are more persistent and frequently last into adulthood. As little as 0.4 g of peanut is required to cause milder allergic reactions like rashes, angioedema, and gastrointestinal problems. However, peanut is also one of the leading causes of severe, sometimes fatal anaphylactic reactions.
Learn more about immunological here-
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Answer:
T2 = 29.79°C
Explanation:
Equliibrium signifies that heat loss = heat gained
Heat gained by Ice;
H = ML
Mass, M = Number of moles * Molar mass = 1 * 18 = 18g
l = 6.01 k J m o l = 334 J/g
C = 4.186 J/g
H = 18(334)
H = 6012
Heat lost by water
H = MCΔT
H = 18 * 4.186 * (50 - T2)
H = 3767.4 - 75.348T2
Since H = H, we have;
6012 = 3767.4 - 75.348T2
- 75.348T2 = 3767 - 6012
T2 = 2245 / 75.348
T2 = 29.79°C
<span>rutherfordium element # 104</span>