answer:The balanced chemical equation for the dissociation of ammonium bromide (NH₄Br) in water is: {eq}NH_4Br(s) + H_2O(l) rightarrow NH_4^+(aq) + Br^-(aq){/eq} Subsequently, the ammonium ion (NH₄⁺) reacts with water to produce ammonia (NH₃) and a hydronium ion (H₃O⁺), making the solution acidic: {eq}NH_4^+(aq) + H_2O(l) rightleftharpoons NH_3(aq) + H_3O^+(aq){/eq} Ammonium bromide is a salt derived from the weak base ammonium hydroxide (NH₄OH) and the strong acid hydrogen bromide (HBr). The dissociation of the ammonium ion in water releases hydrogen ions (H⁺) or hydronium ions (H₃O⁺), resulting in an acidic solution.

answer:An equilateral triangle is a triangle in which all of the side lengths are equal to each other. If we have two equilateral triangles that have one pair of corresponding bases that are congruent, then we can state that the bases are equal in length to each other. Since both triangles are equilateral, we can state that all side lengths of both triangles are equal to each other. Hence, we have two triangles that have the same number of edges and all of them are equivalent to each other correspondingly, so they are congruent. Yes, two equilateral triangles with congruent bases are always congruent.

answer:The integral will evaluate to 0 without any further calculations. This is because the integrand contains a factor of cos θ sin θ, which integrates to 0 over the interval [0, 2π]. Therefore, the volume of the region is 0.

answer:The definite integral of {eq}displaystyle int_{a}^{b} (2x+1) dx {/eq} can be computed directly from the definition as follows: {eq}begin{align*} int_{a}^{b} (2x+1) dx &= lim_{n to infty} sum_{i=1}^{n} f(x_i) Delta x[2ex] &= lim_{n to infty} sum_{i=1}^{n} (2x_i+1) frac{b-a}{n}[2ex] &= lim_{n to infty} frac{b-a}{n} sum_{i=1}^{n} (2x_i+1)[2ex] &= lim_{n to infty} frac{b-a}{n} left[ sum_{i=1}^{n} 2x_i + sum_{i=1}^{n} 1 right][2ex] &= lim_{n to infty} frac{b-a}{n} left[ 2 sum_{i=1}^{n} x_i + n right][2ex] &= lim_{n to infty} frac{b-a}{n} left[ 2 cdot frac{n(a+b)}{2} + n right][2ex] &= lim_{n to infty} frac{b-a}{n} left[ (b+a) + n right][2ex] &= lim_{n to infty} left[ (b-a) + frac{b-a}{n} n right][2ex] &= b-a + lim_{n to infty} (b-a)[2ex] &= b-a + (b-a)[2ex] &= 2(b-a)[2ex] end{align*} {/eq} Therefore, the value of the definite integral is {eq}bf{2(b-a)} {/eq}.